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Chemistry 5 Marks Questions

Equilibrium · CBSE STD 11 Science (CBSE - English Medium). 107 questions.

Questions · Page 2 of 3

5 Marks Questions

Question 515 Marks
The ionization constant of propanoic acid is $1.32 \times 10^{–5}$​​​​​​​. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?
Answer
Let the degree of ionization of propanoic acid be α.
Then, representing propionic acid as HA, we have:
$\text{HA}$ $+$ $\text{H}_2\text{O}$ $\leftrightarrow$ $\text{H}_3\text{O}^+$ $+$ $\text{A}^-$
$(.05-0.0\alpha)\approx.05$ $.05\alpha$   $.05\alpha$
$\text{K}_\text{a}=\frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA]}}$
$=\frac{(.05\alpha)(.05\alpha)}{0.05}=.05\alpha^2$
$\alpha=\sqrt{\frac{\text{K}_\text{a}}{.05}}=1.63\times10^{-2}$
Then, $[\text{H}_3\text{O}^+]=.05\alpha=0.5\times1.63\times10^{-2}=\text{K}_\text{b}.15\times10^{-4}\text{M}$
$\therefore\ \text{pH}=3.09$
In the presence of 0.1M of HCl, let α´ be the degree of ionization.
$\text{Then, }[\text{H}_3\text{O}^+]=0.01$
$[\text{A}^-]=005\alpha'$
$[\text{HA]}=.05$
$\text{K}_\text{a}=\frac{0.01\times0.5\alpha'}{.05}$
$1.32\times10^{-5}=.01\times\alpha'$
$\alpha'=1.32\times10^{-3}$
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Question 525 Marks
What do you understand by following?
  1. Dissociation constant of an acid.
  2. Buffer solution.
  3. Solubility product.
Answer
  1. Dissociation Constant of an Acid $K_a$: It measures the ability of an acid to lose [$H^+$] experimentally.
$\text{CH}_3\text{COOH}(\text{l})+\text{H}_2\text{O}(\text{l})\\\rightleftharpoons\text{CH}_3\text{COO}^-(\text{aq})+\text{H}_3\text{O}^+(\text{aq})$
$\text{K}_{\text{a}}=\frac{[\text{CH}_3\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{CH}_3\text{COOH}]}$
  1. Buffer Solution: The solution whose pH does not change by adding small amount of H+ or $OH^+$ is called buffer solution. e.g. mixture of $CH_3COOH$ and $CH_3COONa$ is a buffer solution.
  2. Solubility Product: It is a product of molar concentration of ions formed in a saturated solution at a given temperature raised to the power equal to the number of each ions formed by 1 mole of sparingly soluble compound.
$\text{e.g. AgCl}(\text{s})\rightleftharpoons\text{Ag}^+(\text{aq})+\text{Cl}^-\text{(aq)};$
$\text{K}_{\text{sq}}=[\text{Ag}^+][\text{Cl}^-]$
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Question 535 Marks
An equilibrium mixture at 300K contains $N_2O_4$ and $NO_2$​​​​​​​ at 0.28 and 1.1atm pressure respectively. If the volume of the container is doubled, calculate the new equilibrium pressure of two gases.
Answer
$\begin{matrix}&\text{N}_2\text{O}_4(\text{g})&\rightleftharpoons&2\text{NO}_2(\text{g})\\\text{perssure at equilibrium }&0.28&&1.1\end{matrix}$
$\text{K}_{\text{p}}=\frac{\text{p}(\text{NO}_2)^2}{\text{p}(\text{N}_2\text{O}_4)}$
$=\frac{(1.1)^2}{(0.28)}=4.32\text{atm}$
If volume of the container is doubled, the pressure will be reduced to half.
$\begin{matrix}&\text{N}_2\text{O}_4(\text{g})&\rightleftharpoons&2\text{NO}_2(\text{g})\\\text{New perssure }&\Big(\frac{0.28}{2}-\text{p}\Big)&&\Big(\frac{1.1}{2}+2\text{p}\Big)\end{matrix}$
$\text{K}_{\text{p}}=\frac{\Big(\frac{1.1}{2}+2\text{p}\Big)^2}{\Big(\frac{0.28}{2}-\text{p}\Big)}=4.32$
$\text { On solving } p=0.045$
$\therefore p\left(N_2 \mathrm{O}_4\right)=0.14-0.045=0.095 \mathrm{~atm}$
$p\left(\mathrm{~N}_2\right)=0.55+0.045=0.64 \mathrm{~atm}$.
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Question 545 Marks
i. The equilibrium constant of a reaction is $2 \times 10^{-3}$ at $25^{\circ} \mathrm{C}$ and $2 \times 10^{-2}$ at $50^{\circ} \mathrm{C}$ Is the reaction endothermic or exothermic?
ii. The solubility of $\mathrm{CaF}_2$ in water at 298 K is $1.7 \times 10^{-3}$ gram per 100 ml of the solution.
Calculate solubility product of $\mathrm{CaF}_2$.
Answer
 
  1. The reaction is endothermic because equilibrium constant is increasing with increase in temperature.
  2. $\text{s}=1.7\times10^{-3}\text{g/ 100ml}$
$=\frac{1.7\times10^{-3}}{100}\times1000$
$=1.7\times10^{-2}\text{g L}^{-1}$
$=\frac{1.7\times10^{-2}\text{g L}^{-1}}{(40+38)\text{g mol}^{-1}}$
$=\frac{1.7\times10^{-2}\text{g L}^{-1}}{78\text{g mol}^{-1}}$
$=2.18\times10^{-4}\text{mol L}^{-1}$
$\text{CaF}_2\rightleftharpoons\text{Ca}^{2+}+\text{2F}^-$
$\text{K}_{\text{sp}}=[\text{Ca}^{2+}][\text{F}^-]^2$
$=(2.18\times10^{-4})(2\times2.18\times10^{-4})^2$
$=41.44\times10^{-12}$
$=4.144\times10^{-11}$
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Question 555 Marks
Match the following equilibria with the corresponding condition.
Column I
Column II
i.
Liquid ⇌ Vapour
a.
Saturated solution
ii.
Solid ⇌ Liquid
b.
Boiling point
iii.
Solid ⇌ Vapour
c.
Sublimation point
iv.
Solute(s) ⇌ Solute (solution)
d.
Melting point
 
 
e.
Unsaturated solution
Answer
Column I
Column II
i.
Liquid ⇌ Vapour
b.
Boiling point
ii.
Solid ⇌ Liquid
d.
Melting point
iii.
Solid ⇌ Vapour
c.
Sublimation point
iv.
Solute(s) ⇌ Solute (solution)
a.
Saturated solution
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Question 565 Marks
The solubility product of $Al ( OH )_3$ is $2.7 \times 10^{-11}$. Calculate its solubility in $gL ^{-1}$ and also find out pH of this solution. (Atomic mass of $Al =27 u$ ).
Answer
Let S be the solubility of $Al(OH)_3$​​​​​​​
  $\text{Al(OH})_3$ $\rightleftharpoons$ $\text{Al}^{3+}\text{(aq)}$ $+$ $3\text{OH}^-\text{(aq)}$
Concentration of species at t = 0 $1$   $0$   $0$
Concentration of various species at equilibrium $1-\text{S}$   $\text{S}$   $3\text{S}$
$\text{K}_\text{sp}=[\text{Al}^{3+}][\text{OH}^-]^3=\text{(S)(3S)}^3=27\text{S}^4$
$\text{S}^4=\frac{\text{K}_\text{sp}}{27}=\frac{27\times10^{-11}}{27\times10}=1\times10^{-12}$
$\text{S}=1\times10^{-3}\text{mol L}^{-1}$
Solubility of $Al(OH)_3​​​​​​​$​​​​​​​
Molar mass of $Al(OH)_3 is 78g$. Therefore,
Solubility of $Al(OH)_3 in g L^{-1} = 1 \times 10^{-3} \times 78g L^{-1}$
$= 78 \times 10^{-3}g L^{-1}$
$= 7.8 \times 10^{-2}g L^{-1}​​​​​​​$​​​​​​​
pH of the solution
$S = 1 \times 10^{-3}mol L^{-1}$
$[OH^-] = 3S = 3 \times 1 \times 10^{-3}= 3 \times 10^{-3}$
$pOH = 3 -\log 3$
$pH = 14 -pOH = 11 + \log 3 = 11.4771$
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Question 575 Marks
What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78M?
$2\text{ICI (g)}\rightleftharpoons\text{I}_2\text{ (g) + Cl}_2\text{ (g)};\text{ K}_{\text{c}}=0.14$
Answer
  $2\text{ICl}_{\text{(g)}}$ $\rightleftharpoons$ $\text{I}_{2\text{(g)}}$ $+$ $\text{Cl}_{2\text{(g)}}$ $;$ $\text{K}_{\text{c}}$ $=$ $0.14$
Initial molar conc. $0.78$   $0$   $0$        
Eqm. molar conc. $0.78-2\text{x}$   $\text{x}$   $\text{x}$        
Applying law of chemical equilibrium,
$\text{K}_{\text{c}}=\frac{[\text{I}_2][\text{Cl}_2]}{[\text{ICl}]^2}\Rightarrow0.14=\frac{\text{x.x}}{(0.78-2\text{x})^2}$
$\text{x}^2=0.14(0.78-2\text{x})^2$
$>$
$\text{or }\frac{\text{x}}{0.78-2\text{x}}=\sqrt{0.14}=0.374$
$\text{or }\text{x}=0.292-0.748\text{x}$
$\text{or }1.748\text{x}=0.292$
$\text{or }\text{x}=0.167$
Hence at equilibrium, $[\text{I}_2]=[\text{Cl}_2]=0.167\text{ M}$
$[\text{ICl}]=0.78-2\times0.167=0.446\text{ M}$
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Question 585 Marks
What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298K? (For calcium sulphate, $K_{sp}$ is $9.1 \times 10^{–6}$​​​​​​​).
Answer
$\text{CaSO}_\text{4(s)}\leftrightarrow\text{Ca}^{2+}_\text{(aq)}+\text{SO}^{2-}_\text{4(aq)}$
$\text{K}_\text{sp}=[\text{Ca}^{2+}][\text{SO}^{2-}_4]$
Let the solubility of $CaSO_4$ be s.
Then,
$\text{K}_\text{sp}=\text{s}^2$
$9.1\times10^{-6}=\text{s}^2$
$\text{s}=3.02\times10^{-3}\text{mol/L}$
Molecular mass of $CaSO_4$​​​​​​​ = 136 g/mol
Solubility of $CaSO_4$​​​​​​​​​​​​​​ in gram/L = $3.02 \times 10^{–3} \times 136= 0.41g/L$
This means that we need 1L of water to dissolve 0.41g of $CaSO_4$​​​​​​​​​​​​​​
Therefore, to dissolve 1g of $CaSO_4$​​​​​​​​​​​​​​ we require $=\frac{1}{0.41}\text{L}=2.44\text{L}$ of water.
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Question 595 Marks
The ionization constant of $HF , HCOOH$ and HCN at 298 K are $6.8 \times 10^{-4}, 1.8 \times 10-4$ and $4.8 \times 10^{-9}$ respectively. Calculate the ionization constants of the corresponding conjugate base.
Answer
It is known that,
$\text{K}_\text{b}=\frac{\text{K}_\text{w}}{\text{K}_\text{a}}$
Given,
$K_a$ of $HF = 6.8 \times 10^{–4}$
Hence, $K_b$_ of its conjugate base $F^–$​​​​​​​
$=\frac{\text{K}_\text{w}}{\text{K}_\text{a}}$
$=\frac{10^{-14}}{6.8\times10^{-4}}$
$=1.5\times10^{-11}$
Given,
$K_a $ of $ HCOOH = 1.8 \times 10^{–4}$​​​​​​​
Hence,$ K_b $of its conjugate base $HCOO^–$​​​​​​​
$=\frac{\text{K}_\text{w}}{\text{K}_\text{a}}$
$=\frac{10^{-14}}{1.8\times10^{-4}}$
$=5.6\times10^{-11}$
Given,
$K_a$ of $ HCN = 4.8 \times 10^{–9}$​​​​​​​
Hence, $K_b$​​​​​​​ of its conjugate base $CN^–$​​​​​​​
$=\frac{\text{K}_\text{w}}{\text{K}_\text{a}}$
$=\frac{10^{-14}}{4.8\times10^{-9}}$
$=2.08\times10^{-6}$
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Question 605 Marks
What is the pH of 0.001M aniline solution? The ionization constant of aniline can be taken from Table 7.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
Answer
$K_b = 4.27 \times 10^{–10} c$ = 0.001M pH =? α =? $\text{K}_\text{b}=\text{c}\alpha^2$ $4.27\times10^{-10}=0.001\times\alpha^2$ $4270\times10^{-10}=\alpha^2$ $65.34\times10^{-5}=\alpha=6.53\times10^{-5}$ $\text{Then [anion]}=\text{c}\alpha=.001\times65.34\times10^{-5}$ $=.065\times10^{-5}$ $\text{pOH}=-\log(.065\times10^{-5})$ $=6.187$ $\text{pH}=7.813$Now,
$\text{K}_\text{a}\times\text{K}_\text{b}=\text{K}_\text{w}$ $\therefore\ 4.27\times10^{-10}\times\text{K}_\text{a}=\text{K}_\text{w}$ $\text{K}_\text{a}=\frac{10^{-14}}{4.27\times10^{-10}}$ $=2.34\times10^{-5}$Thus, the ionization constant of the conjugate acid of aniline is $2.34 \times 10^{–5}$​​​​​​​.
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Question 615 Marks
Calculate the volume of water required to dissolve 0.1g lead (II) chloride to get a saturated solution.
($K_{sp}$ of $PbCl_2 = 3.2 \times 10^{–8}$ , atomic mass of $Pb = 207u$).
Answer
  $\text{PbCl}_2$ $\rightleftharpoons$ $\text{Pb}^{2+}$ $+$ $\text{2Cl}^-$
Conc. at t = 0 $1$   $0$   $0$
Conc. at equilibrium $1-\text{S}$   $\text{S}$   $2\text{S}$
$\text{K}_\text{sp}=\text{[Pb}^{2+}][\text{Cl}^-]^2=\text{(S)(2S)}^2=4\text{S}^3$
$\text{S}^3=\frac{\text{K}_\text{sp}}{4}=\frac{3.2\times10^{-8}}{4}8\times10^{-9}\text{mol L}^{-1}$
$\text{S}=\sqrt[3]{8\times10^{-9}}=2\times10^{-3}\text{mol L}^{-1}$
Solubility of $\text{PbCl}_4=2\times10^{-3}\times278$ (molar mass of $PbCl_2) = 556 \times 10^{-3}g L^{-1}$
$= 0.556 g L^{-1}$
To get saturated solution, $0.556g\ PbCl_2$ is dissolved in 1L water.
0.1g of $PbCl_2$, i disobed in $\frac{0.1}{0.556}=0.1798\text{L}$ water.
To make a saturated solution, 0.1g $PbCl_2$ is dissolved in $0.1798\approx0.2\text{L}$ water.
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Question 625 Marks
A reaction between ammonia and boron trifluoride is given below:
: $NH_3 + BF_3 → H_3N: BF_3$
Identify the acid and base in this reaction. Which theory explains it? What is the hybridisation of B and N in the reactants?
Answer
Although $\mathrm{BF}_3$ does not have a proton but acts as Lewis acid as it is an electron deficient compound. It reacts with $\mathrm{NH}_3$ by accepting the lone pair of electrons from $\mathrm{NH}_3$ and completes its octet. The reaction can be represented by.
$\mathrm{BF}_3+: \mathrm{NH}_3 \rightarrow \mathrm{BF}_3 \leftarrow: \mathrm{NH}_3$
Lewis electronic theory of acids and bases can explain it. Boron in $\mathrm{BF}_3$ is $\mathrm{sp}^2$ hybridised, whereas N in $\mathrm{NH}_3$ is $\mathrm{sp}^3$ hybridised.
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Question 635 Marks
For the reaction: $\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g})$at 400K, $K_p = 41$. Find the value of $K_p$ for each of the following reactions at the same temperature.
  1. $\text{2NH}_3(\text{g})\rightleftharpoons\text{N}_2(\text{g})+3\text{H}_2(\text{g})$
  2. $\frac{1}{2}\text{Na}(\text{g})+\frac{3}{2}\text{H}_2(\text{g})\rightleftharpoons\text{NH}_3(\text{g})$
  3. $2\text{N}_2(\text{g})+6\text{H}_2(\text{g})\rightleftharpoons4\text{NH}_3(\text{g})$
Answer
  1. $\text{K}_{\text{p}}=\frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}=41$
$2\text{NH}_3\text{(g)}\rightleftharpoons\text{N}_2(\text{g})+3\text{H}_2(\text{g})$

$\text{K}'_{\text{p}}=\frac{\text{[N}_2][\text{H}_2]^3}{[\text{NH}_3]^2}$

$=\frac{1}{\text{K}_{\text{p}}}=\frac{1}{41}=0.024$
  1. $\frac{1}{2}\text{N}_2(\text{g})+\frac{3}{2}\text{H}_2\rightleftharpoons\text{NH}_3(\text{g})$
$\text{K}''_{\text{p}}=\frac{[\text{NH}_3]}{[\text{NH}_2]^{\frac{1}{2}}[\text{H}_2]^{\frac{3}{4}}}$

$=\sqrt{\text{K}_{\text{p}}}=\sqrt{41}=\sqrt{6.4}$
  1. $2\text{N}_2\text{(g)}+6\text{H}_2(\text{g})\rightleftharpoons4\text{NH}_3(\text{g})$
$\text{K}'''_{\text{p}}=\frac{[\text{NH}_3]^4}{[\text{N}_2]^2[\text{H}_2]^6}$

$=(\text{K}_{\text{p}})^2=(41)^2=1681$

$=1.68\times10^3$
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Question 645 Marks
The solubility of silver chloride (AgCl) in water at $25^\circ C is 1.06 \times 10^{-5}mol L^{-1}.$ Calculate the solubility product of AgCl at this temperature.
Answer
The solubility equilibrium of AgCl is
$AgCl(s) \rightleftharpoons Ag^{+}(aq)+Cl^{-}(aq)$
One mole of AgCl in solution gives 1 mole of $Ag ^{+}$ions and 1 mole of $Cl ^{-}$ions.
Since, the solubility of AgCl is $1.06 \times 10^{-5} mol L ^{-1}$, it will gives $1.06 \times 15^{-5} mol L ^{-1}$ of $Ag ^{+}$ions and $1.06 \times 10^{-5} mol L ^{-}$, of $Cl ^{-}$ions. Therefore,
${\left[Ag^{+}\right]=1.06 \times 10^{-5} mol L^{-1},\left[Cl^{-}\right]=1.06 \times 10^{-5} mol L^{-1}}$
$\text { Now, } K_{sp}=\left[Ag^{+}\right]\left[Cl^{-}\right]=\left(1.06 \times 10^{-5}\right) \times\left(1.06 \times 10^{-5}\right)=1.12 \times 10^{-10}$
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Question 655 Marks
What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, $K_{sp} = 6.3 \times 10^{–18}$).
Answer
Let the maximum concentration of each solution be xmol/L. After mixing, the volume of the concentrations of each solution will be reduced to half i.e.,$\frac{\text{x}}{2}.$ $\therefore[\text{FeSO}_4]=[\text{Na}_2\text{S}]=\frac{\text{x}}{2}\text{M}$ $\text{Then, [Fe}^{2+}]=\text{[FeSO}_4]=\frac{\text{x}}{2}\text{M}$ $\text{Also, [S}^{2-}]=[\text{Na}_2\text{S}]=\frac{\text{x}}{2}\text{M}$ $\text{FeS}_\text{(x)}\leftrightarrow\text{Fe}^{2+}_\text{(aq)}+\text{S}^{2-}_\text{(aq)}$ $\text{K}_\text{sp}=[\text{Fe}^{2+}][{\text{S}^{2-}}]$ $6.3\times10^{-18}=\Big(\frac{\text{x}}{2}\Big)\Big(\frac{\text{x}}{2}\Big)$ $\frac{\text{x}^2}{4}=6.3\times10^{-18}$ $\Rightarrow\text{x}=5.02\times10^{-9}$If the concentrations of both solutions are equal to or less than $5.02 \times 10^{–9}M$, then there will be no precipitation of iron sulphide.
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Question 665 Marks
 The solubility product of AgCl is $1.5 \times 10^{-10}$. Predict whether there will be any precipitation by mixing $50 mL$ of $0.01 M NaCl$ and $50 mL$ of $0.01 M AgNO _3$ solution.
Answer
On mixing 50mL of $0.01M NaCl$ and 50mL pf $0.01M AgNO_3$, the total volume becomes 100mL.
Therefore,
Conc. of $NaCl$ in $100\text{mL}=\frac{0.01\times50}{100}=0.005\text{M}$
Conc. of$ AgNO_3 $in $100\text{mL}=\frac{0.01\times50}{100}=0.005\text{M}$
Now $NaCl(aq) = Na^+(aq) + Cl^-(aq)$and $\text{AgNO}_3 \text{aq}= \text{Ag}+ \text{(aq}) +\text{NO}_3^- \text{(aq)}$
$[Cl^-] = [NaCl] = 0.005N$
$[Ag^+] = [AgNO_3] = 0.005M$
$\therefore$ Ionic product of $[Ag^+][Cl^-] = 0.005 \times 0.005 = 2.5 \times 10^{-5}$​​​​​​​
Since ionic product is greater than its solubility product, precipitation will occur.
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Question 675 Marks
The solubility product of $Fe ( OH )_3$ is $1 \times 10^{-36}$. What is the minimum concentration of $OH ^{-}$ions required to precipitate $Fe ( OH )_3$ from a 0.001 M solution of $FeCl _3$ ?
Answer
$K _{\text {sp }}$ for $Fe ( OH )^3$ is $K _{ sp }=\left[ Fe ^{3+}\right]\left[ OH ^{-}\right]^3$
Precipitation will occur when ionic product,
$\left[ Fe ^{3+}\right]\left[ OH ^{-}\right]^3$ becomes greater than $K _{ sp }$
$\left[ Fe ^{3+}\right]=\left[ FeCl _3\right]=0.001 M$
The concentration of $OH ^{-}$ions required to start the precipitation is
$[\text{OH}^-]^3=\frac{\text{K}_{\text{sp}}}{[\text{Fe}^{3+}]}$
$=\frac{1\times10^{-6}}{0.001}=1\times10^{-33}$
$\therefore[\text{OH}^-]=(1\times10^{-33})^{\frac{1}{3}}$
$=1\times10^{-11}\text{mol L}^{-1}$
Thus, concentration of $OH^-$ required to start precipitation of $Fe(OH)_3 = 1 \times 10^{-11}mol L^{-1}$
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Question 685 Marks
Which of the following reactions involve homogeneous equilibria and which involve heterogeneous equilibria?
  1. $2\text{N}_2\text{O(g)}\rightleftharpoons2\text{N}_2(\text{g})+\text{O}_2\text{(g)}$
  2. $2\text{NH}_3\text{(g)}\rightleftharpoons\text{N}_2(\text{g})+3\text{H}_2\text{(g)}$
  3. $2\text{Cu}(\text{NO}_3)_2(\text{s})\rightleftharpoons2\text{CuO(s)}+4\text{NO}_2(\text{g})+\text{O}_2(\text{g})$
  4. $\text{CH}_3\text{COOC}_2\text{H}_5(\text{aq})+\text{H}_2\text{O}\text{l}\\\rightleftharpoons\text{CH}_3\text{COOH(aq)}+\text{C}_2\text{H}_5\text{OH(aq)}$
  5. $\text{Fe}^{3+}(\text{aq})+\text{3OH}^-\text{(aq)}\rightleftharpoons\text{Fe(OH)}_3(\text{s})$
Answer
  1. Homogeneous equilibriales.
  2. Homogeneous equilibria.
  3. Heterogeneous equilibria bond.
  4. Homogeneous equilibria.
  5. Heterogeneous equilibriato.
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Question 695 Marks
Calculate pH when $9.8g$ of $H_2SO_4$ is dissolved in $2L$ solution.
Answer
$\text{M}=\frac{\text{W}_{\text{B}}}{\text{M}_{\text{B}}}\times\frac{1}{\text{Volume of solution in Litres}}$
$W_B = 9.8g$ = mass of solute
MB = Molar mass of $H_2SO_4 = 98g mol^{-1}$
Volume of solution = 2L
$\text{M}=\frac{9.8}{98}\times\frac{1}{2}=0.05\text{mol L}^{-1}$
$\text{H}_2\text{SO}_4\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ }2\text{H}^++\text{SO}^{2-}_4$
$[H^+] = 2 \times$ Molarity of $H_2SO_4$
$= 2 \times 0.05 = 0.1mol L^{-1}$
$= 10^{-1}mol L^{-1}$​​​​​​​
$\text{pH}=-\log(\text{H}^+)=-\log10^{-1}$
$=+1\log10=1\times1=1$
$[\because\log10=1]$
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Question 705 Marks
$3.2mol$ of HI were taken in a sealed bulb at $440°C$ till the equilibrium state was reached. Its degree of dissociation was found to be $20\%$. Calculate the number of moles of HI. $H_2$ and $I_2$ present at equilibrium point and also determine the equilibrium constant.
Answer
$\begin{matrix}&\text{2HI}(\text{g})&\rightleftharpoons&\text{H}_2(\text{g})&+&\text{l}_2(\text{g})\\\text{Initial Conc. }&3-2\text{moles}&&0&&0\\\text{Final Conc.at equilibrium}&3.2-\frac{2\times20}{100}&&\frac{20}{100}&&\frac{20}{100}\\&2.8\text{mol}&&0.2\text{mol}&&0.2\text{mol}\end{matrix}$
$\text{K}_{\text{c}}=\frac{[\text{H}_2][\text{I}_2]}{[\text{HI}^2]}$
$[\text{HI}]=2.8\text{moles},$
$[\text{H}_2]=[\text{I}_2]=0.2\text{mol.}$
$\text{K}_{\text{c}}=\frac{0.2\times0.2}{(2.8)^2}=\frac{4}{100}\times\frac{100}{28\times28}$
$\text{K}_{\text{c}}=\frac{1000}{196}\times10^{-3}$
$=5.1\times10^{-3}$
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Question 715 Marks
Write a relation between $\Delta\text{G}$ and Q and define the meaning of each term and answer the following:
  1. Why a reaction proceeds forward when Q < K and no net reaction occurs when Q = K?
  2. Explain the effect of increase in pressure in terms of reaction quotient Q.
For the reaction,
Answer
The relation between $\Delta\text{G}$ and Q is
$\Delta\text{G}=\Delta\text{G}^{\ominus}+\text{RT InQ}$
$\Delta\text{G}=$ change in free energy as the reaction proceeds
$\Delta\text{G}^{\ominus}$ standard free energy
Q = reaction quotienten
R= gas constant
T = absolute temperature in Kual
  1. Since, $\Delta\text{G}^{\ominus}=-\text{RT In K}$
$\therefore\Delta\text{G}=-\text{RT In K}+\text{RT In Q;}$
$\Delta\text{G}=\text{RT In}\frac{\text{Q}}{\text{K}}$
will be negative and the reaction proceeds in the forward direction. $\text{Q}=\text{K},\Delta\text{G}=0$ If reaction is in equilibrium and there is no net reaction.
  1. $\text{CO}\text{(g)}+3\text{H}_2(\text{g})\rightleftharpoons\text{CH}_4\text{(g)}+\text{H}_2\text{O}(\text{g})$
$\text{K}_{\text{c}}=\frac{[\text{CH}_4][\text{H}_2\text{O}]}{[\text{CO}][\text{H}_2]^3}$
On increasing pressure, volume decreases. If we doubled the pressure, volume will be halved but the molar concentrations will be doubled. Then,
$\text{Q}_{\text{c}}=\frac{2[\text{CH}_4].2[\text{H}_2\text{O}]}{2[\text{CO}]\{2[\text{H}_2]\}^3}$
$\frac{1}{4}\frac{[\text{CH}_4][\text{H}_2\text{O}]}{[\text{CO}][\text{H}_2]^3}=\frac{1}{4}\text{K}_{\text{c}}$
Therefore, $Q_c$ is less than $K_C$ so $Q_c$ will tend to increase to re-establish equilibrium and the reaction will go in forward direction.
$\text{CO}\text{(g)}+3\text{H}_2(\text{g})\rightleftharpoons\text{CH}_4\text{(g)}+\text{H}_2\text{O}(\text{g})$
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Question 725 Marks
The pH of $0.005M$ codeine $(C_{18}H_{21}NO_3)$ solution is $9.95$. Calculate its ionization constant and $pK_b$.
Answer
c = 0.005
pH = 9.95
pOH = 4.05
pH = – log (4.105)
$4.05=-\log[\text{OH}^-]$
$[\text{OH}^-]=8.91\times10^{-5}$
$\text{c}\alpha=8.91\times10^{-5}$
$\alpha=\frac{8.91\times10^{-5}}{5\times10^{-3}}=1.782\times10^{-2}$
$\text{Thus, K}_\text{b}=\text{c}\alpha^2$
$=0.005\times(1.782)^2\times10^{-4}$
$=0.005\times3.1755\times10^{-4}$
$=0.0158\times10^{-4}$
$\text{K}_\text{b}=1.58\times10^{-6}$
$\text{Pk}_\text{b}=-\log\text{K}_\text{b}$
$=-\log(1.58\times10^{-6})$
$=5.80$
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Question 735 Marks
The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
Answer
c = 0.1M
pH = 2.34
$-\log[\text{H}^+]=\text{pH}$
$-\log[\text{H}^+]=2.34$
$[\text{H}^+]=4.5\times10^{-3}$
Also,
$[\text{H}^+]=\text{c}\alpha$
$4.5\times10^{-3}=0.1\times\alpha$
$\frac{4.5\times10^{-3}}{0.1}=\alpha$
$\alpha=45\times10^{-3}=.045$
Then,
$\text{K}_\text{a}=\text{c}\alpha^2$
$=0.1\times(45\times10^{-3})^2$
$=202.5\times10^{-6}$
$=2.02\times10^{-4}$
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Question 745 Marks
  1. Write the conjugate acid for Bronsted base of $HCOO^-$.
  2. Calculate the pH of a $1.0 \times 10^{-8}$ M solution of HCl.
  3. Calculate the solubility of $A_2X_3$ in pure water, assuming that neither kind of ion reacts with water. [The solubility product of $A_2X_3, K_{sp} = 1.1 \times 10^{-23}$]
Answer
  1. $HCOOH$ is conjugate acid of $HCOO^-$.
  2. $\text{H}_2\text{O}+\text{H}_2\text{O}\rightleftharpoons\text{H}_3\text{O}^++\text{OH}^-$
$\text{K}_\text{w}=1.0\times10^{-14}=[\text{H}_3\text{O}^+][\text{OH}^-]$ $(\because[\text{H}_3\text{O}^+]=[\text{OH}^-])$
$=1.0\times10^{-14}=[\text{H}_3\text{O}^+]^2$
$[\text{H}_3\text{O}^+]=\sqrt{1.0\times10^{-14}}=10^{-7}\text{mol L}^{-1}$
$\text{H}_2\text{O}+\text{HCl}\xrightarrow{\ \ \ \ }\text{H}_3\text{O}s^++\text{Cl}^-$
$10^{-8}$
Total $[\text{H}_3\text{O}^+]=10^{-8}+10^{-7}=10^{-7}(1+10^{-1})$
$=1.1\times10^{-7}\text{mol L}^{-1}$
$\text{pH}=-\log[\text{H}_3\text{O}^+]$
$=-\log1.1\times10^{-7}$
$=-0.04139+7.00=6.958$
  1. $\text{A}_2\text{X}_3\rightleftharpoons2\text{A}^{3+}+3\text{X}^{2-}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\text{s}\ \ \ \ \ \ \ \ \ \ \ 3\text{s}$
where solubility is 's' mol $L^{-1}$​​​​​​​
$\text{K}_\text{sp}=[\text{A}^{3+}]^2[\text{X}^{2-}]^3$
$1.1\times10^{-23}=(2\text{s})^2(3\text{s})^3=108\text{s}^5$
$\Rightarrow \sqrt{\frac{110}{108}\times10^{-25}}=1\times10^{-5}\text{mol L}^{-1}$
$\text{s}\simeq1\times10^{-5}\text{mol L}^{-1}$
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Question 755 Marks
  1. At 473K, equilibrium constant, K for decomposition of $PCl_5$ is $8.3 \times 10^{-3}$. If decomposition is depicted as
$\text{PCl}_5(\text{s})\rightleftharpoons\text{PCl}_3(\text{s})+\text{Cl}_2(\text{g});$
$\Delta_\text{r}\text{H}^\text{o}=124.0\text{kJ mol}^{-1}$
  1. Write an expression for $K_c$ for the reaction.
  2. What is the value of $K_c$ for the reverse reaction at same temperature.
  3. What would be the effect on $K_c$ if:
  1. The pressure is increased?
  2. The temperature is increased?
  1. Write equilibrium constant for the following reactions:
  1. $\text{BaCO}_3(\text{s})\rightleftharpoons\text{BaO(s)}+\text{CO}_2(\text{g})$
  2. $\text{CH}_4(\text{g})+2\text{O}_2(\text{g})\rightleftharpoons\text{CO}_2(\text{g})+2\text{H}_2\text{O(g)}$
Answer
  1. $\text{PCl}_5(\text{s})\rightleftharpoons\text{PCl}_3(\text{s})+\text{Cl}_2(\text{g});$
$\Delta _\text{r}\text{H}^\text{o}=124.0\text{kJ mol}^{-1}$
  1. $\text{K}_\text{c}=\frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}=8.3\times10^{-3}$
  2. $K_c$ for the reverse reaction $=\frac{1}{\text{K}_\text{c}\text{for the forward reaction}}$
$=\frac{1}{8.3\times10^{-3}}=120.48$
  1.  
  1. When pressure increases $K_c$ reamains unchanged.
  2. $K_c$ increses with increases in temprature because the reaction is endothermic.
  1.  
  1. $\text{K}=\frac{[\text{BaO(s)}][\text{CO}_2(\text{g})]}{[\text{BaCO}_3(\text{g})]}$
Since $[\text{BaCO}_3(\text{s})]=[\text{BaO(s)}]=1$
$\text{K}=[\text{CO}_2(\text{g})]$
  1. ​​​​​​​$\text{K}=\frac{[\text{CO}_2(\text{g})][\text{H}_2\text{O(g)}]^2}{[\text{CH}_2(\text{g})][\text{O}_2(\text{g})]^2}$
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Question 765 Marks
  1. Which of the following are Lewis acids?
$\text{H}_2\text{O},\text{BF}_3,\text{H}^+$ and $\text{NH}^+_4$
  1. The pH of a sample of vinegar is $3.76$. Calculate the concentration of hydrogen ion in it.
  2. What is common ion effect? Explain its application in qualitative analysis of II group radicals.
Answer
  1. Lewis acids are electron deficient species. So, Lewis acids are $\text{BF}_3,\text{H}^+$ and $\text{NH}^+_4.$
  2. $\text{pH}=-\log[\text{H}^+]$
$\Rightarrow\log[\text{H}^+]=-\text{pH}=-3.76$
$\Rightarrow \log[\text{H}^+]=-3.76+1-1$
$[\text{H}^+]=\bar{4}.24\text{ Antilog}$
$[\text{H}^+]=1.738\times10^{-4}$
$=1.74\times10^{-4}\text{M}$
  1. Common ion effect is the suppression of the dissociation of weak electrolyte in the presence of a strong electrolyte having a common ion. Sulphides of II group radicals are precipitated by passing $H_2S$ gas in presence of HCl. $H_2S$ being weak electrolyte ionises slightly and HCl which is a strong electrolyte is completely ionised.
$\text{H}_2\text{S}\rightleftharpoons2\text{H}^++\text{S}^{2-}$
$\text{HCl}\xrightarrow{\ \ \ \ \ \ \ }\text{H}^++\text{Cl}^-$
Due to common ion effect the degree of dissociation of $H_2S$ decreases and concentration of $S^{2-}$ ions in solution becomes small enough to precipitate only II group cations and not group IV cations. The second group cations have lower solubility product than IV group cations.
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Question 775 Marks
$\text{K}_{\text{c}}\text{ for }\text{PCl}_5(\text{g})\rightleftharpoons\text{PCl}_3(\text{g})+\text{Cl}_2(\text{g})\text{ is }0.04\text{ at }25^\circ\text{C}.$
How much moles of $PCl_5$ must be added to 3L flask to obtain a chlorine concentration of $0.15M$.
Answer
$\begin{matrix}&\text{PCl}_5(\text{g})&\rightleftharpoons&\text{PCl}_3(\text{g})&+&\text{Cl}_2(\text{g})\\\text{Initial Conc. }&\text{x}&&0&&0\\\text{Final Conc.}&\text{x}-0.15&&0.15&&0.15\end{matrix}$
$\text{K}_{\text{c}}=\frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}$
$0.04=\frac{0.15\times0.15}{\text{x}-0.15}$
$\text{x}-0.15=\frac{0.15\times0.15}{0.04}$
$=\frac{225}{1000}\times\frac{100}{4}$
$=\frac{225}{400}=\frac{9}{16}$
$\text{x}=\frac{9}{16}+\frac{15}{100}=\frac{9}{16}+\frac{3}{20}$
$=\frac{180+48}{320}=\frac{228}{320}=0.7\text{mol L}^{-1}$
Number of moles of $PCl_5$ per litre = 0.7mol
Number of moles of $PCl_5$ in 3L flask = 0.7 × 3 = 2.1mol.
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Question 785 Marks
  1. Consider the following endothermic reaction:
$\text{CH}_4(\text{g})+\text{H}_2\text{O}(\text{g})\rightleftharpoons\text{CO(g)}+3\text{H}_2(\text{g})$
  1. Write expression for $K_p$ for the above reaction.
  2. How will the values of $K_p$ and composition of equilibrium mixture be affected by.
  1. increasing the pressure.
  2. increasing the temperature.
  3. using a catalyst?
  1. Calculate the pH of the resultant mixture of $10$ml of $0.1$M $H_2SO_4 + 10$ ml of $0.1$M KOH.
Answer
  1. $\text{CH}_4(\text{g})+\text{H}_2\text{O}(\text{g})\rightleftharpoons\text{CO(g)}+3\text{H}_2(\text{g})$
  1. $\text{K}_{\text{p}}=\frac{(\text{p}_{\text{CO}})(\text{p}_{\text{H}_2})^3}{(\text{p}_{\text{CH}_4})(\text{p}_{\text{H}_2\text{O}})}$
  2.  
  1. On increasing pressure, the reaction equilibria will shift in the backward direction.
  2. As the given reaction is endothermic, on increasing temperature the given equilibrium will shift in forward direction.
  3. There is no effect of catalyst in equilibrium composition, however the equilibrium will be attained faster.
  1. $10ml$ of $0.1M H_2SO_4$ is mixed with $10ml$ of $0.1$M KOH.
The reaction is
$2\text{KOH}+\text{H}_2\text{SO}_4\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\text{K}_2\text{SO}_4+2\text{H}_2\text{O}$
$10$ ml of $0.1 \mathrm{M} \mathrm{~H}_2 \mathrm{SO}_4=0.1 \times 10=1$ millimole
$10$ ml of $0.1 \mathrm{M} \mathrm{~KOH}=0.1 \times 10=1$ millimole
$1$ millimole of $KOH$ will react with $0.5$ millimole of $\mathrm{H}_2 \mathrm{SO}_4$
$\left[\because 0.5\right.$ millimole will produce 1 millimole of $\left.\mathrm{H}^{+}\right]$
$\therefore \mathrm{H}_2 \mathrm{SO}_4$ left $=1-0.5=0.5$ millimole
Volume of reaction mixtrue
$=10+10=20 \mathrm{ml}$
$\therefore$ Molarity of $x$ in the mixtrue.
$\therefore$ Molarity of × in the mixtrue.
$=\frac{0.5}{20}=2.5\times10^{-2}\text{M}$
$[\text{H}^+]=2\times2.5 \times10^{-2}$
$=5\times10^{-2}\text{M}$
$\text{pH}=-\log(5\times10^{-2})$
$=-\log5-\log10^{-2}$
$=-0.6990+2.0000=1.30$
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Question 795 Marks
The ionization constant of dimethylamine is $5.4 \times 10^{–4}$​​​​​​​. Calculate its degree of ionization in its $0.02M$ solution. What percentage of dimethylamine is ionized if the solution is also $0.1M$ in NaOH?
Answer
$\text{K}_\text{b}=5.4\times10^{-4}$
$\text{c}=0.02\text{M}$
$\text{Then, }\alpha=\sqrt{\frac{\text{K}_\text{b}}{\text{c}}}$
$=\sqrt{\frac{5.4\times10^{-4}}{0.02}}$
$=0.1643$
Now, if 0.1M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionization.
$\text{NaOH}_\text{(aq)}$ $\leftrightarrow$ $\text{Na}^+_\text{(aq)}$ $+$ $\text{OH}^-_\text{(aq)}$
    $0.1\text{M}$   $0.1\text{M}$
And,
$\text{(CH}_3)_2\text{NH}$ $+$ $\text{H}_2\text{O}$ $\leftrightarrow$ $(\text{CH}_3)_2\text{NH}_2^+$ $+$ $\text{OH}^-$
$(0.02-\text{x})$       $\text{x}$   $\text{x}$
$;0.02\text{M}$           $;0.1\text{M}$
Then, $[(\text{CH}_3)_2\text{NH}_2^+]=\text{x}$
$[\text{OH}^-]=\text{x}+0.1;0.1$
$\Rightarrow\text{K}_\text{b}=\frac{[(\text{CH}_3)_2\text{NH}_2^+][\text{OH}^-]}{[(\text{CH}_3)_2\text{NH]}}$
$5.4\times10^{-4}=\frac{\text{x}\times0.1}{0.02}$
$\text{x}=0.0054$
It means that in the presence of 0.1M NaOH, 0.54% of dimethylamine will get dissociated.
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Question 805 Marks
Consider the following equilibrium at 2773K
$\text{H}_2(\text{g})+\text{Cl}_2(\text{g})\rightleftharpoons2\text{HCl}(\text{g})$
Initially $0.25M H_2$ and $0.25MCl_2$ are introduced into a reaction vessel and the system in allowed to attain equilibrium. At equilibrium the concentrations of $H_2(g)$ and $Cl_2(g)$ became $0.0314M$. Calculate $K_c$ and $K_p$.
Answer
  1. $\begin{matrix}&\text{H}_2(\text{g})&+&\text{Cl}_2(\text{g})\rightleftharpoons&\text{2Hcl}(\text{g})\\\text{Initial Conc. }&0.25&&0.25\text{M}&0\\\text{Final Conc.at equilibrium}&0.0314\text{M}&&0.0314\text{M}&0\end{matrix}$
$2(0.25-0.0314)=0.2186\text{M}\times2$

$=0.219\text{M}\times2=0.438$

$\text{K}_{\text{c}}=\frac{[\text{HCl}]^2}{[\text{H}_2][\text{Cl}_2]}$

$=\frac{(0.438)^2}{0.0314\times0.314}=195$
  1. $\text{K}_{\text{p}}=\text{K}_{\text{c}}(\text{RT})^{\Delta\text{n}}$
$\Delta\text{n}=0$

$\text{K}_{\text{p}}=\text{K}_{\text{c}}=195$
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Question 815 Marks
$\mathrm{K}_p=0.04 \mathrm{~atm}$ at 899 K for the equilibrium shown below. What is the equilibrium concentration of $\mathrm{C}_2 \mathrm{H}_6$ when it is placed in a flask at 4.0atm pressure and allowed to come to equilibrium?
$\text{C}_2\text{H}_6(\text{g})\rightleftharpoons\text{C}_2\text{H}_4(\text{g})+\text{H}_2(\text{g})$
Answer
$\begin{matrix}&\text{C}_2\text{H}_6(\text{g})&\rightleftharpoons&\text{C}_2\text{H}_4(\text{g})&+&\text{H}_2(\text{g})\\\text{Intial pressure}&4.0\text{atm}&&0&&0\\\text{Equli pressure}&(4.0-\text{p})\text{atm}&&\text{p}&&\text{P}\end{matrix}$$\text{K}_{\text{p}}=\frac{\text{p}_{\text{C}_2\text{H}_4}.\text{p}_{\text{H}_2}}{\text{p}_{\text{C}_2\text{H}_6}}$
$=\frac{\text{p}\times\text{p}}{4.0-\text{p}}$
$0.04=\frac{\text{p}^2}{4.0-\text{p}}$
$\text{or }0.16-0.04\text{p}=\text{p}^2$
$\text{p}=-0.04\pm\frac{\sqrt{0.0016-4(-0.16)}}{2}$
$\text{p}=\frac{-0.04\pm0.80}{2}$
$\Rightarrow\text{p}=0.38$
(by taking positive value)
Hence, $\text{p}_{\text{C}_2\text{H}_6}=4.0-0.38=3.62\text{atm}.$
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Question 825 Marks
Calculate the pH of $1 \times 10^{-8}M$ solution of $HCl.$
Answer
If we use the relation, $\text{pH} = -\log [\text{H}_3\text{O}^+]$ we get pH equal to 8. But this is not correct because an acidic solution cannot have pH greater than 7. It may be noted that in very dilute acidic solution, when $H^+$ concentrations from acid and water are comparable, the concentration of$ H^+ $from water cannot be neglected. Therefore,
$[H^+]_{total}= [H^+]_{acid} + [H^+]_{water}$
water Since HCl is strong acid and is completely ionized
$[H^+]_{HCl}= 1.0 \times 10^{-8}$​​​​​​​
The concentration of $H^+$​​​​​​​^ from ionisation is equal to the$ [OH^-]$ from water,
$[\text{H}^+]_{\text{H}_2\text{O}}=[\text{OH}^-]_{\text{H}_3\text{O}}=\text{x say}$
$[H^+]_{total} = 1.0 \times 10^{-8} + x$
$But [H^+] [OH^-] = 1.0 \times 10^{-14}​​​​​​​$​​​​​​​
$\therefore$ $(1.0 \times 10^{-8}+ x) (x) = 1.0 \times 10^{-14}$
$\Rightarrow x^2 + 10^{-8} x - 10^{-14} = 0$
On solving for x, we get $x = 9.5 \times 10^{-8}​​​​​​​$​​​​​​​
$\therefore$ $[H^+] = 1.0 \times 10^{-8} + 9.5 \times 10^{-8}$
$= 10.5 \times 10^{-8}$
$= 1.05 \times 10^{-7}​​​​​​​$​​​​​​​
$\text{pH}=-\log[\text{H}^+]$
$=-\log(1.05\times10^{-7})=6.98$
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Question 835 Marks
 The solubility product of lead bromide is $8 \times 10^{-5}$ at 298 K . If the salt is $80 \%$ dissociated in saturated solution, calculate the solubility of the salt (in $g / L$ ).
Answer
The solubility equilibrium is,
$\text{PbBr}_2(\text{s})\rightleftharpoons\text{Pb}^{2+}(\text{aq})+2\text{Br}^-(\text{aq})$
Suppose the solubility of $PbBr_2$ is S moles per litre. Then, the concentrations of various species at equilibrium are
$[Pb^{2+}] = S, [Br^-] = 2S$
$Now, K_{sp} = [Pb^{2+}][Br^-]^2$
$8 \times 10^{-5} = (S) \times (2S)^2$
$8 \times 10^{-5} = (S) \times (2S)^2​​​​​​​$
$\text{or }\text{S}^3=\frac{8\times10^{-5}}{4}=2\times10^{-5}$
$\text{S} = (2.0 \times 10-5)^{\frac{1}{3}}$
$=2.714\times10^{-2}\text{mol L}^{-1}$
This is the solubility of PbBr, if it is 100% dissociated. Solubility of PbBr, if it is 80% dissociated
$=\frac{2.714\times10^{-2}\times80}{100}$
$=2.192\times10^{-2}\text{ mol L}^{-1}$
Molecular weight of $PbBr _2=207+2 \times 80=367$
Solubility of $PbBr _2=2.192 \times 10^{-2} \times 367=8.04 g L ^{-1}$.
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Question 845 Marks
  1. For the reaction:
$\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g}),$
The value of $K_p$ is $3.6 \times 10^{-2}$ at $500K$.
Colculate the value of $K_c$ for the reaction at the same temperature R = 0.083L bar $K^{-1}mol^{-1}$​​​​​​​.
  1. What is the effect of increasing pressure in the reactions? Give reason.
$\text{PCl}_5(\text{g})\rightleftharpoons\text{PCl}_3(\text{g})+\text{Cl}_2\text{(g)}$
$\text{N}_2(\text{g})+\text{O}_2(\text{g})\rightleftharpoons2\text{NO(g)}$
Answer
  1. The reaction is
$\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g})$
Given: $\text{K}_\text{p}=3.6\times10^{-2}$ at 500K
The reation between $K_p$ and $K_c$ is
$\text{K}_\text{p}=\text{K}_\text{c}(\text{RT})^{\Delta\text{n}}$
For the above reaction
$\Delta \text{n}=2-4=-2$
$\text{K}_\text{c}=\frac{\text{K}_\text{p}}{(\text{RT})^{\Delta\text{n}}}$ (R = 0.083bar L $K^{-1}mol^{-1}$​​​​​​​)
$=\frac{3.6\times10^{-2}}{(0.083\times500)^{-2}}$
$=3.6\times10^{-2}\times(0.083\times500)^2$
$=62$​​​​​​​
  1.  
  2. The equilibrium will shift in backward reaction because number of moles of products are more than reactants $\Delta \text{n}>0.$
  3. No effect because number of moles of reactants and products are equal, i.e., $\Delta \text{n}=0.$
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Question 855 Marks
  1. The reaction quotient of a reversible reaction is $Q_C$ and the equilibrium constant is $K_c$. What do you conclude for the reaction if $Q_c < K_c$?
  2. State Le Chatelier's principle.
  3. In qualitative analysis, $NH_4Cl$ is added before adding $NH_4OH$ solution for testing of III group radicals [$Fe^{3+}, Cr^{3+}$ and $Al^{3+}$]. Explain by using concept of common ion effect.
Answer
  1. If $Q_c < K_c$; the reaction tends towards forward direction to attain equilibrium.
  2. Le Chatelier's Principle: If a system in equilibrium is subjected to a change in concentration, temperature or pressure, the equilibrium shifts in a direction that tends to undo the effect of the change.
  3. Hydroxides of group III are precipitated by adding $NH_4OH$ in presence of $NH_4Cl$. The role of $NH_4Cl$ is to produce common ion effect.
$\text{NH}_4\text{OH}\rightleftharpoons\text{NH}^+_4+\text{OH}^-$
$\text{NH}_4\text{Cl}\xrightarrow{ \ \ \ \ \ \ \ \ \ \ \ }\text{NH}_4^++\text{Cl}^-$
Due to common ion effect, the degree of dissociation of $NH_4OH$ is suppressed and less $OH^‑$ are formed. This less concentration of $OH^-$ is sufficient to precipitate group III cations but not the cations of higher groups since the $K_{sp}$ of group III < subsequent groups.
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Question 865 Marks
The values of K of two sparingly soluble salts $Ni ( OH )_2$ and AgCN are $2.0 \times 10^{-15}$ and $6 \times 10^{-17}$ respectively. Which salt is more soluble? Explain.
Answer
Given: $K _{\text {sp }}$ of $Ni ( OH )_2=2 \times 10^{-15}$
$K_{sp} of AgCN = 6 \times 10^{-17}$
$K_{sp} = [Ni^2+] [OH-]^2 = 2 \times 10^{-15}​​​​​​​$​​​​​​​
$\text{AgCN}\rightleftharpoons\text{Ag}^++\text{CN}^-$
$K_{sp} = [Ag^+] [CN^-] = 6 \times 10^{-17}$
$Let [Ag^+] [CN^-] = s_1​​​​​​​$​​​​​​​
and $[Ni^{2+}] = s_2​​​​​​​$​​​​​​​
Hence $[OH^-] = 2s_2​​​​​​​$​​​​​​​
Since $\text{s}_1^2=6\times10^{-17}$
$\Rightarrow\text{s}_1=\sqrt{60\times10^{-18}}$
$\text{s}_1=7.8\times10^{-9}\text{M}$
Since $\text{s}_2\times(2\text{s}_2)^2=2\times10^{-15}$
$\Rightarrow4\text{s}^3_2=2\times10^{-15}$
$\Rightarrow\text{s}_2^3=0.5\times10^{-15}$
$\Rightarrow\text{s}_2^3=5\times10^{-16}$
$\Rightarrow\text{s}_2^3=500\times10^{-18}$
$\Rightarrow\text{s}_2^3=3\sqrt{500\times10^{-18}}$
$=7.9\times10^{-6}\text{M}$
$\text{so }\text{s}_2=7.9\times10^{-6}\text{M}$
Since $s_2 > s_1​​​​​​​$​​​​​​​ therefore $Ni(OH)_2​​​​​​​$​​​​​​​ is more soluble than AgCN.
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Question 875 Marks
  1. Hydrolysis of sucrose give
$\text{Sucrose}+\text{H}_2\text{O}\rightleftharpoons\text{Glucose}+\text{Fructose}$

Equilibrium constant $K_c$ for the reaction is $2 \times 10^{13}$ at $300K$. Calculate $\Delta\text{G}^\circ\text{ at }300\text{K.}(\log 2=0.3010)$
  1. The concentration of hydrogen in two sample of soft drinks A and B $4.0 \times 10^{-7}$ and $3.2 \times 10^{-6}$ respectively. Which of these two soft drinks has higher pH?
Answer
  1. $\text{Sucrose}+\text{H}_2\text{O}\rightleftharpoons\text{Glucose}+\text{Fructose}$
$\text{K}_{\text{c}}2\times10^{13},\text{T}=300\text{K},\Delta\text{G}^\circ=?$

$\Delta\text{G}^\circ=-2.30\text{ RT }\log\text{K}_{\text{c}}$

$=-2.303\times8.314\times300\log2\times10^{13}$

$=-19.147\times300(\log2+\log^{13})$

$=\frac{-19.147\times300\times13.3010\text{J}}{1000}$

$=-76.402\text{kJ mol}^{-1}$
  1. For A,
$\text{pH}=-\log[\text{H}_3\text{O}^+]$

$=-\log4.0\times10^{-7}$

$=-\log4.0-\log10^{-7}$

$=-0.6021+7.000=6.3979$
  1. For B,
$\text{pH}=-\log[\text{H}_3\text{O}^+]$

$=-\log3.2\times10^{-6}$

$=-\log3.2-\log10^{-6}$

$=-0.5050+6.000=5.4950$

pH of 'A' is higher than 'B'.
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Question 885 Marks
Calculate the pH of 0.4g of NaOH dissolved in water to give 200ml of solution.
Answer
$\text{M}=\frac{\text{W}_{\text{B}}}{\text{M}_{\text{B}}}\times\frac{1000}{\text{Volume of solution in ml}}$
where, Mass of solute, $\text{W}_{\text{B}}$ = 0.4
Volume of solution = 200mL
$=\frac{0.4}{40}\times\frac{1000}{200}$
$=\frac{2}{40}=\frac{1}{20}=0.05\text{M}$
$\text{pOH}=-\log[\text{OH}^-]$
$=-\log(5\times10^{-2})$
$=-\log5-\log(10^{-2})$
$=-0.6990+2.0000$
$=1.3010$
$\text{pH}=14-1.3010$
$=12.399$
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Question 895 Marks
The buffers of $X$ and $Y$ of pH $4.0$ and $6.0$ respectively are prepared from acid $HA$ and the salt $Na A$.
Both the buffers are $0.50M$ in $HA$. What would be the $pH$ of the solution obtained by mixing equal volumes of the two buffers?
$(K_{HA}= 1.0 \times 10^{-5})$
Answer
$K_{HA}= 1.0 \times 10^{-5}$
$\therefore \text{pK}_{\text{a}}=-\log(1.0\times10^{-5})=5.0$ Determination of conc. of saltin buffer X. $\text{pH}=\text{pK}_{\text{a}}+\log\frac{[\text{Salt}]}{[\text{Acid}]};$
$4.0=5.0+\log\frac{[\text{Salt}]}{0.5}$
$\log\frac{[\text{Salt}]}{0.5}=-1$
$\Rightarrow\frac{[\text{Salt}]}{0.5}=10^{-1}$ [salt] $= 0.5 \times 10^{-1} = 0.05M$ Determination of conc. of saltin buffer Y. $6.0=5.0+\log\frac{[\text{Salt}]}{0.5}$
$\log\frac{[\text{Salt}]}{0.5}=1.0$
$\Rightarrow\frac{[\text{Salt}]}{0.5}=10$[salt] $= 10 \times 0.5 = 5.0$
Conc. of salt in the mixture
$[\text{Salt}]_{\text{mix}}=\frac{0.5+5}{2}=2.75\text{M}$
$\text{pH}=\text{pK}_{\text{a}}+\log\frac{[\text{Salt}]}{[\text{Acid}]}$
$=5.0+\log\frac{2.75}{0.5}$
$=5.0+0.74=5.74$
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Question 905 Marks
An aqueous solution contains an unknown concentration of $Ba ^{2+}$. When 50 mL of 1 M solution of $Na _2 SO _4$ is added, $BaSO _4$ first begin to precipitate. The final volume is $500 mL$ . The solubility product of $BaSO _4$ is $1.0 \times 10^{-10}$. What is original concentration of $Ba ^{2+}$ ?
Answer
$\text{BaSO}_4\text{(s)}\rightleftharpoons\text{Ba}^{2+}+\text{SO}^{2-}_4\text{(aq)}$
For $Na_2SO_4$
$M_1V_1= M_2V_2$
$1M \times 50mL = M_2 \times 500mL$
$M_2 = 0.1M$
$[\text{SO}^{2-}_4]=0.1$
$\text{K}_{\text{sp}}=[\text{Ba}^{2+}][\text{SO}^{2-}_4]$
$\Rightarrow[\text{Ba}^{2+}]=\frac{\text{K}_{\text{sp}}}{[\text{SO}^{2-}_{4}]}$
$=\frac{1\times10^{-10}}{0.1}=1.0\times10^{-9}\text{M}$
$[Ba^{2+}] = 1.0 \times 10^{-9}M$ in $500mL$ of solution
$[Ba^{2+}] in 500mL = 1.0 \times 10^{-9}M$
$[\text{Ba}^{2+}]\text{in }1000\text{mL}=\frac{1.0\times10^{-9}}{500}\times1000$
$=2\times10^{-9}\text{mol}.$
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Question 915 Marks
  1. In qualitative analysis, on what basis cations are grouped?
  2. The value of $K_c$ in the reaction:
$2\text{A}\rightleftharpoons\text{B}+\text{C}$ is $2 \times 10^{-3}$. At a given time, the composition of reaction mixture is $[A] = [B] = [C] = 3 \times 10^{-4}M$. In which direction the reaction will proceed?
  1. The solubility of $Sr(OH)_2$ at 298K is 19.23g/L of solution. Calculate the concentration of strontium and hydroxyl ions and the pH of the solution.
Answer
  1. Cations are grouped on the basis of their solubility product $(K_{sp})$. i.e. value of $K_{sp}$ is closed to each other, e.g., $K_{sp}$ of sulphides of groups II cations are close to each other.
  2. For the reaction, the reaction quotient $Q_c$ is given by $\text{Q}_\text{c}=\frac{[\text{B}][\text{C}]}{[\text{A}]^2}$
as $[\text{A}]=[\text{B}]=[\text{C}]=3\times10^{-4}\text{M}$
$\text{Q}_\text{c}=\frac{(3\times10^{-4})(3\times10^{-4})}{(3\times10^{-4})^2}=1$
as $\text{Q}\text{c}>\text{K}_\text{c}$
So, the reaction will proceed in the reverse direction.
  1. Molar mass of $Sr(OH)_2$ = 87.6 + 34
$= 121.6g mol^{-1}$
Solubility of $Sr(OH)_2$ in mol $L^{-1}$
$=\frac{19.23\text{g L}^{-1}}{121.6\text{g mol}^{-1}}$
$\text{Sr(OH)}_2\xrightarrow{\ \ \ \ \ }\text{Sr}^{2+}+2\text{OH}^-$
$[\text{Sr}^{2+}]=0.1581\text{M},[\text{OH}^-]$
$=2\times0.1581=0.3162\text{M}$
$\text{p}[\text{OH}]=-\log0.3162=0.5$
$\Rightarrow \text{pH}=14-0.5=13.5$
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Question 925 Marks
How much volume of $0.1 \mathrm{M} \mathrm{~CH}_2\mathrm{COOH}$ should be added to 50 mL of $0.2 \mathrm{M} \mathrm{~CH}_3 \mathrm{COONa}$ solution to prepare a buffer solution of pH 4.91 . ( $\mathrm{pK}_{\mathrm{a}}$ of ACH is 4.76 ).
Answer
According to Henderson's equation
$\text{pH}=\text{pK}_{\text{a}}+\log\frac{[\text{Salt}]}{[\text{Acid}]};$
$\text{pH}=4.91,\text{pK}_{\text{a}}=4.76$
$4.91=4.76+\log\frac{[\text{Salt}]}{[\text{Acid}]}$
$\log\frac{[\text{NaAcl}]}{[\text{AcH}]}=4.91-4.76=0.15$
$\frac{[\text{NaAc}]}{[\text{AcH}]}=\text{antilog}(0.15)=1.41$
If V is the volume of 0.1M AcH required
$\frac{\text{NaAc}}{[\text{AcH}]}=\frac{\frac{0.2\times50}{1000}}{\frac{0.1\times\text{V}}{1000}}=1.41$
$=\frac{0.2\times50}{0.15\times\text{V}}=1.41$
$\text{or }\text{V}=\frac{0.2\times50}{0.1\times1.41}=70.92\text{mL}$
Volume of 0.1M acetic acid required = 70.92mL.
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Question 935 Marks
  1. Predict the acidic, basic or neutral nature of the following salts:
$NaCN, KBr, NaNO_2, NH_4NO_3$
  1. What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298K? (For calcium sulphate $K_{sp} is 9.1 \times 10^{-6}$).
  2. At $450K ; K_p = 2.0 \times 10^{10}bar^{-1}$ for the reaction at equilibrium:
​​​​​​​​​​​​​​$2\text{SO}_2(\text{g})+\text{O}_2(\text{g})\rightleftharpoons2\text{SO}_3(\text{g})$
What is $K_c$ at this temperature?
Answer
$NaCN, NaNO_2$ - solutions are basic as they are salts of strong base and weak acid. (HCN and $HNO_2$ are weak acids and NaOH is strong base).
KBr - this solution is neutral as it is salt of strong acid HBr and strong base KOH.
$NH_4NO_3$ - its solution is acidic as it is a salt of strong acid $(HNO_3)$ and weak base $(NH_4OH)$.
  1. $\text{CaSO}_4(\text{s})\rightleftharpoons\text{Ca}^{2+}(\text{aq})+\text{SO}^{2-}_4(\text{aq});$
(mol. mass of $CaSO_4 = 136g/ mol$)
Let the solubility of $CaSO_4$ in mol/L is x.
$\text{K}_{\text{sp}}=[\text{Ca}^{2+}][\text{SO}^{2-}_4]=\text{x}^2$
$\text{x}=\sqrt{\text{K}_\text{sp}}=\sqrt{9.1\times10^{-6}}$
$=3.02\times10^{-3}\text{mol/L}$
$=3.02\times10^{-3}\times136\text{g/L}$
$=0.411\text{g/L};$
For dissolving $0.411g$ of $CaSO_4$, water required is 1L
For dissolving 1g of $CaSO_4$, water required $=\frac{1}{0.411}\text{L}=2.43\text{L}$
  1. For the reaction:
$2\text{SO}_2(\text{g})+\text{O}_2\text{(g)}\rightleftharpoons2\text{SO}_3(\text{g})$
$\Delta\text{n}_\text{g}=2-3=-1$
$\text{K}_\text{p}=\text{K}_\text{c}(\text{RT})^{\Delta\text{n}}$
$\text{K}_\text{c}=\text{K}_\text{p}(\text{RT})^{-\Delta\text{n}}$
For this reaction
$\text{K}_\text{c}=\text{K}_\text{p}\text{RT}=(2\times10^{10}\text{bar}^{-1}) $
$(0.083\text{L bar K}^{-1}\text{mol}^{-1})\times450\text{K}$
$=7.48\times10^{11}\text{L mol}^{-1}$
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Question 945 Marks
  1. Define solubility product. Write solubility product expression for $Zr_3(PO_4)_4$.
  2. Calculate the pH of $0.01 M CH_3COOH$ solution. $[K_a(CH_2COOH) = 1.74 \times 10^{-51}$
  3. Explain why NaCl is precipitated when HCl(g) is passed through the saturated solution of NaCl.
Answer
  1. Solubility Product: It is defined as the product of molar concentrations of the ions (formed in the saturated solution at a given temperature) raised to the power equal to the number of times each ion occurs in the equation for solubility equilibrium,
  1. $=-\log\text{C}\alpha$
    $=-\log\sqrt{\text{K}_\text{a}\times\text{C}}$
    $=-\log\sqrt{1.74\times10^{-5}\times0.01}$
    $=-\log\sqrt{1.74\times10^{-7}}$
    $=-\log\sqrt{17.4\times10^{-8}}$
    $=-\log4.17\times10^{-4}$
    $=-\log4.17-\log10^{-4}$
    $=-0.6217+4.000$
    $=3.3783$
  2. It is due to common ion, $CI^-$ increase, therefore rate of backward reaction increases, solubility of NaCl decreases.
  3. $\text{Zr}_3(\text{PO}_4)_4\rightleftharpoons3\text{Zr}^{4+}+4\text{PO}^{3-}_4$
    $\text{K}_{\text{sp}}=[\text{Zr}^{4+}][\text{PO}^{3-}_4]^4$
  4. $\text{pH}=-\log[\text{H}_3\text{O}^+]$
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Question 955 Marks
  1. Write the conjugate acid of $NH_3$.
  2. Assign reason for the following:
  1. A solution of $NH_4Cl$ in water shows pH less than 7.
  2. In qualitative analysis $NH_4Cl$ is added before adding $NH_4OH$ for testing $Fe^{3+}$ or $AP^{3+}$ ions.
  1. Consider the reaction:
$\text{N}_2(\text{g})+3\text{H}_2\text{(g)}\rightleftharpoons2\text{NH}_3+\text{Heat}$
Indicate the direction in which the equilibrium will shift when:
  1. Temperature is increased.
  2. Pressure is increased.
Answer
  1. $\text{NH}^+_4$ is conjugate acid of $NH_3$.
  2.  
  1. $NH_4Cl$ is salt of weak base $NH_4OH$ and strong acid HCl, therefore $H^+$ ions are more than $OH^-$ ions thus, pH is less than 7.
  2. It is done to decrease $[OH^-]$ due to common ion effect, so that only group III radicals $Fe^{3+}$ or $Al^{3+}$ get precipitated and higher group radicals do not.
  1.  
  1. When temperature is increased equilibrium will shift to backward direction as reaction is exothermic.
  2. When pressure is increased rate of forward reaction will increase as there is decrease in number of moles from reactants to products.
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Question 965 Marks
  1. State Henry's Law.
  2. Assign reason for the following:
  1. A solution of $NH_4Cl$ in water shows pH less than 7.
  2. In qualitative analysis $NH_4Cl$ is added before adding $NH_4OH$ for testing $Fe^{3+}$ or $AP^{3+}$ ions.
  1. Consider the reaction:
$\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3+\text{Heat}$
Indicate the direction in which the equilibrium will shift when:
  1. Temperature is increased.
  2. Pressure is increased.
Answer
  1. The mass of gas dissolved in given mass of a solvent at any temperature in proportional to the pressure of the gas above the solvent.
  2.  
  1. $\mathrm{NH}_4 \mathrm{Cl}$ is salt of weak base $\mathrm{NH}_4 \mathrm{OH}$ and strong acid HCl , therefore $\mathrm{H}^{+}$ions are more than $\mathrm{OH}^{-}$ions thus, pH is less than 7.
  2. It is done to decrease $\left[\mathrm{OH}^{-}\right]$due to common ion effect, so that only group III radicals $\mathrm{Fe}^{3+}$ or $\mathrm{Al}^{3+}$ get precipitated and higher group radicals do not.
  1.  
  1. When temperature is increased equilibrium will shift to backward direction as reaction is exothermic.
  2. When pressure is increased rate of forward reaction will increase as there is decrease in number of moles from reactants to products.
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Question 975 Marks
  1. Define solubility product. Write solubility product expression in terms of molar solubility for $FeCl_3$
  2. What is the effect of temperature on solubility of gases in liquids?
  3. Equilibrium constant for the reaction is $4.0$. What will be the equilibrium constant for the reverse reaction.
  4. Calculate the pH of $10^{-8}M$ HCl solution.
Answer
  1. Solubility product is defined as the product of molar concentration of ions raised to the power the number of ions formed per formula of the compound.
$\text{FeCl}_3(\text{s})\rightleftharpoons\text{Fe}^{3+}+3\text{Cl}^-\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ '\text{s}' \ \ \ \ \ \ \ \ \ \ 3\text{s}$
$\text{K}_{\text{sp}}=[\text{Fe}^{3+}][\text{Cl}^-]^3$
$\text{K}_\text{sp}=(\text{s})(3\text{s})^3,$
where 's' $mol L^{-1}$ is solubility.
$\Rightarrow\text{K}_\text{sp}= 27\text{s}^4$
$\Rightarrow\text{s}=4\sqrt{\frac{\text{K}_\text{sp}}{27}}$
  1. Solubility of gases in liquids decreases with increase in temperature because force of attraction between gas and liquid decreases at high temperature.
  2. K for the reaction = 4
$\therefore \text{K}'$ for reverse reaction $=\frac{1}{4}=0.25$ [$\because \text{K}'=\frac{1}{ \text{K}}$ for reverse reaction]
  1. pH of $10^{-8}M$ HCl solution.
$\text{HCl}\xrightarrow{\ \ \ \ \ \ }\text{H}^++\text{Cl}^-\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{10^{-8}\text{M}}$
$\text{H}_2\text{O}\rightleftharpoons\text{H}^++\text{OH}^-$
$\text{K}_\text{w}=1\times10^{-14}$
$\Rightarrow [\text{H}^+][\text{OH}^-]=10^{-14}$
$\because[\text{H}^+]=[\text{OH}]$
$\therefore [\text{H}^+]^2=10^{-14}$
$\Rightarrow [\text{H}^+]=10^{-7}\text{mol L}^{-1}$
Total concentration of
$[\text{H}^+]=(10^{-8}+10^{-7})$
$=10^{-7}(1+0.1)$
$=1.1\times10^{-7}$
$\therefore \text{pH}=-\log[\text{H}^+]$
$=-\log1.1\times10^{-7}$
$=-\log1.1-\log10^{-7}$
$\Rightarrow \text{pH}=-0.0454+7.000$
$=6.9546$
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Question 985 Marks
Show that the degree of dissociation $(\alpha)$ for the dissociation of $PCl_5$ into PCl3 $\alpha=\Big(\frac{\text{K}_{\text{p}}}{\text{p}+\text{K}_{\text{p}}}\Big)^{\frac{1}{2}}$ and $Cl_2$ in pressure p is given by,
Answer
$\begin{matrix}&\text{PCl}_5&\rightleftharpoons&\text{PCl}_3&+&\text{Cl}_2\\\text{Initial moles }&1&&0&&0\\\text{Moles after diss.}&1-\alpha&&\alpha&&\alpha&(\text{Total }= 1+\alpha)\end{matrix}$
$\therefore\text{p}_{\text{PCl}}=\frac{1-\alpha}{1+\alpha}\times\text{p},\text{p}_{\text{PCl}_3}$ $=\frac{\alpha}{1+\alpha}\times\text{p},\text{p}_{\text{Cl}_2}=\frac{\alpha}{1+\alpha}\times\text{p},$ $\text{K}_{\text{p}}=\frac{\text{p}_{\text{Cl}_3}\times\text{p}_{\text{Cl}_2}}{\text{p}_{\text{PCl}_5}}$$=\frac{\Big(\frac{\alpha}{1+\alpha}\text{p}\Big)\Big(\frac{\alpha}{1+\alpha}\Big)}{\Big(\frac{1-\alpha}{1+\alpha}\Big)\text{p}}$
$=\frac{\alpha^2\text{p}}{1-\alpha^2}$ $\text{or }(1-\alpha^2)\text{K}_{\text{p}}=\alpha^2\text{p}$ $\text{or }(\text{p}+\text{K}_{\text{p}})\alpha^2=\text{K}_{\text{p}}$ $\text{or }\alpha=\Big(\frac{\text{K}_{\text{p}}}{\text{p}+\text{K}_{\text{p}}}\Big)^{\frac{1}{2}}$
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Question 995 Marks
Calculate the equilibrium constant for the following equilibrium system at 1120K. $\text{C}(\text{s})+\text{CO}_2(\text{g})+2\text{Cl}_2\stackrel{\text{K}_\text{p}}{\rightleftharpoons}2\text{COCl}_2(\text{g})$ Given the following equations and equilibrium constants: $\text{CO}(\text{s})+\text{Cl}_2(\text{g})\rightleftharpoons\text{COCl}_2(\text{g})$ $\text{Kp}_1 = 6.0 \times 10-3 \dots(1)$ $\text{CO}(\text{s})+\text{Cl}_2(\text{g})\rightleftharpoons\text{CO}_2(\text{g})$$\text{Kp}_2 = 1.3 \times 1014 \dots(2)$
Answer
$\text{Multiply Eq. (1) by 2}$
$2\text{CO}(\text{g})+2\text{Cl}_2(\text{g})\rightleftharpoons2\text{COCl}_2(\text{g})$
$\text{K}'_{\text{p}1}=\left(6.0 \times 10^{-3}\right)^2=36 \times 10^{-6}$
$\text{K}'_{\text{p}1}=\frac{[\text{COCl}_2]^2}{[\text{CO}]^2[\text{Cl}_2]^2}=36\times10^{-6}\dots(3)$
$\text{K}'_{\text{p}2}=\frac{[\text{CO}]^2}{[\text{CO}_2]}=36\times10^{14}\dots(4)$
$\text{Multiply K'p}_1 \text{ and } \text{K'p}_2$
$\text{K}'_{\text{p}}=\frac{[\text{COCl}_2]^2}{[\text{CO}_2][\text{Cl}_2]^2}$
$=46.8\times10^8=4.68\times10^9$
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Question 1005 Marks
The average concentration of $\mathrm{SO}_2$ in atmosphere over a city on a certain day is 10 ppm , when the average temperature is 298 K . Given that the solubility of $\mathrm{SO}_2$ in water at 298 K is $1.3653 \mathrm{~mol} / \mathrm{L}$ and the $\mathrm{pK}_{\mathrm{a}}$ of $\mathrm{H}_2 \mathrm{SO}_3$ is 1.92 , estimate the pH of acid rain on that day.
Answer
Amount of $SO_2$ in atmospheres $= 10\text{ ppm}=\frac{10}{10^6}10^{-5}$
Molar conc. of $SO_2$ in pressure of water = amount of $SO_2 \times$ solubility of $SO_2$ in water
$H_2SO_3$ dissociates as = $1.3653 \times 10^{-5}$
$\begin{matrix}&\text{H}_2\text{SO}_3&\rightleftharpoons&\text{H}^+&+&\text{HSO}^-_3\\\text{Intial conc.}&1.3653\times10^{-5}&&0&&0\\\text{Molar cons. of equiv.}&(1.3653\times10^{-5})&&\text{x}&&\text{x}\end{matrix}$
$\text{K}_{\text{a}}=\frac{\text{x}^2}{(1.3653\times10^{-5}-\text{x})}$
$\because\text{PK}_{\text{a}}=1.92$
$\therefore-\log\text{K}_{\text{a}}=1.92$
$\text{K}_{\text{a}}=12\times10^{-2}$
Substituting $1.2\times10^{-2}=\frac{\text{x}^2}{(1.3653\times10^{-5}-\text{x})}$
$\text{or }\text{x}^2=1.2\times10^{-2}(1.3653\times10^{-5}-\text{x})$
On solving we, get $x = 1.3664 \times 10^{-5}$
$\therefore\text{pH}=-\log(1.364\times10^{-5})$
$=4.865$
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5 Marks Questions - Page 2 - Chemistry STD 11 Science Questions - Vidyadip