Question 12 Marks
Calculate pH of 0.01 M HCOONa solution, if value of $K$ is $2 \times 10^{-4}$
AnswerHCOONa, is a salt formed by weak acid and strong base. Hence pH of its solution :
$
\begin{aligned}
pH & =\frac{1}{2} pK_{w}+\frac{1}{2} pK_{a}+\frac{1}{2} \log C \\
K_{w} & =1 \times 10^{-14} \text { Hence } pK_{w}=-\log 10^{-14}=14 \\
pK_{a} & =-\log K_{a}=-\log 10^{-4}-\log 2 \\
pK_{a} & =4-0.3010 \\
pK_{a} & =3.6990
\end{aligned}
$
Hence,
$
\begin{array}{l}
pH=\frac{1}{2}(14)+\frac{1}{2}(3.6990)+\frac{1}{2} \log 10^{-2} \\
pH=7+1.8495+\frac{1}{2}(-2) \\
pH=8.8495-1 \\
pH=7.8495
\end{array}
$
View full question & answer→Question 22 Marks
If in 500 ml of buffer solution 0.004 mole HCl are added then its pH is decreased by 0.04 , then what will be the buffer capacity of buffer solution?
AnswerMoles of HCl formed in 500 ml solution
$
=0.004
$
Hence moles of HCl in 1 lit volume
$
\begin{array}{l}
=2 \times 0.004 \\
=0.008
\end{array}
$
Hence, Buffer capacity $=\frac{\text { Added moles }}{\text { Charge in } pH }$
$
=\frac{0.008}{0.04}=0.2
$
View full question & answer→Question 32 Marks
The dissociation constant of HCOOH and $CH _3 COOH$ is $2 \times 10^{-4}$ and $1.8 \times 10^{-5}$ respectively then calculate the isohydric concentration of HCOOH with $0.03 N CH _3 COOH$.
AnswerFor isohydric solutions :
$
\begin{aligned}
\frac{C_1}{C_2} & =\frac{K_2}{K_1} \\
\frac{C_1}{0.03} & =\frac{1.8 \times 10^{-5}}{2 \times 10^{-4}}
\end{aligned}
$
Concentration of $HCOOH \left( C _1\right)=\frac{0.03 \times 1.8 \times 10^{-5}}{2 \times 10^{-4}}$
$
\begin{array}{l}
\quad=\frac{5.4 \times 10^{-7}}{2 \times 10^{-4}} \\
C_1=2.7 \times 10^{-3}=0.0027 N
\end{array}
$
View full question & answer→Question 42 Marks
Calculate pH of 0.1 M solution of $HCOONH _4$, if $pK _{ a }$ of $HCOOH =3.6$ and $pK _{ b }$ of $NH _4 OH =4.8$.
Answer$HCOONH _4$ is a salt formed by weak acid and weak base. Hence pH of solution :
$
\begin{array}{l}
pH=7+\frac{1}{2} pK_{n}-\frac{1}{2} pK_{b} \\
pH=7+\left(\frac{1}{2} \times 306\right)-\frac{1}{2}(4.8) \\
pH=7+1.8-2.4 \\
pH=6.4
\end{array}
$
View full question & answer→Question 52 Marks
The $K_{\text {sp }}$ of PbS is $4 \times 10^{-28}$. If $\left[ Pb ^{+2}\right]=4 \times$ $10^{-27}$ then what will be the minimum concentration of $S ^{-2}$ to precipitation PbS ?
Answer Where minimum word is used, there ionic product is taken equal to $K _{ sp }$.
$
\begin{aligned}
PbS & \rightleftharpoons Pb^{+2}+S^{-2} \\
K_{sp} & =\left[Pb^{+2}\right]\left[S^{-2}\right] \\
{\left[S^{-2}\right] } & =\frac{K_{sp}}{\left[Pb^{+2}\right]} \\
{\left[S^{-2}\right] } & =\frac{4 \times 10^{-28}}{4 \times 10^{-27}}=10^{-1}
\end{aligned}
$
View full question & answer→Question 62 Marks
If the $K _{\text {sp }}$ of AgCl is $4 \times 10^{-8}$ then in 15 lit of its saturated solution, how much AgCl is added?
Answer$\begin{array}{r}\text { Sol. : } K _{ sp }= s ^2, s=\sqrt{ K _{ sp }}=\sqrt{4 \times 10^{-8}}=2 \times 10^{-4} \\ s=\frac{\text { Mass of } 1 \text { lit. }}{\text { Molecular mass }}, 2 \times 10^{-4}=\frac{x gm }{143.5}\end{array}$
then $\quad x=143.5 \times 2 \times 10^{-4}\left( gm L ^{-1}\right)$
Hence, weight of AgCl in 15 lit. solution
$
\begin{array}{l}
=143.5 \times 2 \times 10^{-4} \times 15 \\
=4.305 \times 10^{-1}
\end{array}
$
View full question & answer→Question 72 Marks
What will be the pH of a mixture of $\frac{ N }{10} HCl$
$
(100 ml)+\frac{N}{2} 20 HNO_3(200 ml)+\frac{M}{60} H_2 SO_4(300 ml) ?
$
Answer$\frac{ M }{60} H _2 SO _4=\frac{1}{60} \times 2=\frac{1}{30} \times N$
all are strong bases. Hence,
$
\begin{array}{l}
NV=N_1 V_1+N_2 V_2+N_3 V_3 \\
N \times 600=\frac{1}{10} \times 100+\frac{1}{20} \times 200+\frac{1}{30} \times 300 \\
{[V=\text { Total volume }=600 ml] } \\
N \times 600=10+10+10 \\
N=\frac{30}{600}=\frac{1}{20}=0.05 \\
=5 \times 10^{-2}=\left[H^{+}\right] \\
pH=2-\log 5 \\
pH=2-0.6990=1.3010(\log 5=0.6990)
\end{array}
$
View full question & answer→Question 82 Marks
In 20 ml 0.2 N KOH , how much ml of 0.1 $N H _2 SO _4$ is added so that solution becomes neutral?
Answer$\begin{aligned} KOH & H _2 SO _4 \\ N_1 V_1 & = N _2 V_2 \\ 0.2 \times 20 & =0.1 \times V _2 \\ V_2 & =40 ml \end{aligned}$
View full question & answer→Question 92 Marks
Two solutions of $pH , 2$ and 3 are mixed in same volume then what will be the pH of resulting solution?
Answer $pH =2$, then $\left[ H ^{+}\right]=10^{-2}$
If $pH =3$, then $\left[ H ^{+}\right]=10^{-3}$
The volume on adding both solutions become doubled. Hence, concentration becomes half.
$
\begin{aligned}
{\left[H^{+}\right] } & =\frac{10^{-2}+10^{-3}}{2} \\
{\left[H^{+}\right] } & =\frac{10^{-2}\left(1+10^{-1}\right)}{2}=\frac{10^{-2} \times 1.1}{2} \\
{\left[H^{+}\right] } & =10^{-2} \times 0.55=10^{-3} \times 5.5 \\
pH & =-\log 10^{-3}-\log 5.5 \\
pH & =3-0.7404 \quad(\log 5.5=0.7404) \\
pH & =2.2596
\end{aligned}
$
View full question & answer→Question 102 Marks
Capsule of $1 ml ; 1 N HCl$ is breaked and added in a container and volume of solution becomes 1000 ml , then what will be the pH of solution?
Answer$\begin{aligned} \text { Sol. : } & N _1 V_1 \\ 1 \times & = N _2 V_2 \\ 1 \times 1 & = N _2 \times 1000 \\ & N_2\end{aligned}=10^{-3}=\left[ H ^{+}\right]$
View full question & answer→Question 112 Marks
Calculate pH of $10^{-8} N KOH$ solution.
AnswerKOH is a strong acid. For this
$
N=\left[OH^{-}\right]=10^{-8}
$
Hence
$
\begin{array}{l}
pOH=-\log \left[OH^{-}\right] \\
\text {pOH }=-\log 10^{-8} \\
\text { pOH }=8 \text { hence } pH=6
\end{array}
$
This is not possible because pH of base can't be less than 7.
As it is very dilute solution, hence in it $\left[ OH ^{-}\right]$in water must be added which is $10^{-7}$.
$
\begin{array}{ll}
\text { Hence } & {\left[OH^{-}\right]=10^{-7}+10^{-8}} \\
& {\left[OH^{-}\right]=10^{-7}(1+0.1)} \\
& {\left[OH^{-}\right]=10^{-7}(1.1)} \\
& pOH=-\log 10^{-7}-\log 1.1 \\
& pOH=7-(0.0414)(\log 1.1=0.0414) \\
& pOH=7-0.0414 \\
& pOH=6.95 \\
& pOH e \\
& pH=14-pOH \\
& pH=14-6.95=7.05
\end{array}
$
View full question & answer→Question 122 Marks
The $\left[ OH ^{-}\right]$of two weak bases are $10^{-1}$ and $10^{-2} mol /$ lit respectively then what will be the ratio of dissociation constants of it?
Answer$
\begin{aligned}
\frac{\left[OH^{-}\right]_1}{\left[OH^{-}\right]_2}=\sqrt{\frac{K_{b_1}}{K_{b_2}}}=\frac{\left(10^{-1}\right)^2}{\left(10^{-2}\right)^2} & =\frac{K_{b_1}}{K_{b_2}} \\
& =\frac{100}{1}=100: 1
\end{aligned}
$
View full question & answer→Question 132 Marks
What will be the pOH of $\frac{ M }{100} \quad H _2 SO _4$ solution ?
Answer $\frac{ M }{100} H _2 SO _4=0.01 M H _2 SO _4$
By ionisation of $H _2 SO _4, 2 H ^{+}$ions are obtained.
Hence $0.01 M =0.01 \times 2=0.02 N$
Therefore
$
\begin{aligned}
{\left[H^{+}\right] } & =0.02=10^{-2} \times 2 \\
pH & =-\log 10^{-2}-\log 2 \\
pH & =2-0.3010 \quad(\log 2=0.3010) \\
pH & =1.6990
\end{aligned}
$
As $pH + pOH =14$
Hence
$
\begin{array}{l}
pOH=14-pH \\
pOH=14-1.6990 \\
pOH=12.3010
\end{array}
$
View full question & answer→Question 142 Marks
If the amount of ionization of 0.01 M aqueous solution of $NH _3$ is $2 \%$, then calculate its dissociation constant.
Answer The degree of dissociation of $NH _3\left( NH _4 OH \right)$ is a weak base. Hence its dissociation constant :
$
\begin{aligned}
K_{b} & =\frac{C \alpha^2}{1-\alpha} \\
C & =0.01 M \text { and } \alpha=\frac{2}{100}=0.02 \\
K_{b} & =\frac{0.01 \times(0.02)^2}{1-0.02} \\
& =\frac{\left(1 \times 10^{-2}\right) .\left(4 \times 10^{-4}\right)}{0.98} \\
K_{b} & =\frac{4 \times 10^{-6}}{0.98}=4.08 \times 10^{-6}
\end{aligned}
$
View full question & answer→Question 152 Marks
6.8 gm NaOH is dissolved in water to prepare
2 litre solution, then calculate pH of solution.
AnswerMolarity of $NaOH ( M )$
$
\begin{array}{l}
=\frac{\text { Mass of substance }(g)}{\text { Molar mass }} \times \text { Volume of solution }(L) \\
\text { Mass of substance }=8 g \\
\text { Volume of solution }=2 L \\
\text { Molar mass of } NaOH=23+16+1=40
\end{array}
$
Hence
$
\begin{array}{l}
M=\frac{8}{40 \times 2}=0.1 \\
M=0.1
\end{array}
$
Hence $N =0.1$ because acidity of $NaOH =1$
Hence
$
\begin{aligned}
{\left[OH^{-}\right] } & =0.1=10^{-1} \\
{\left[H^{+}\right] } & =\frac{K_{w}}{[\overline{OH}]}=\frac{10^{-14}}{10^{-1}}=10^{-13} \\
pH & =-\log \left[H^{+}\right]=-\log 10^{-13} \\
pH & =13
\end{aligned}
$
View full question & answer→Question 162 Marks
Calculate the following :
(i) ionization constant of water
(ii) number of $H ^{+}$in 1 litre water.
Answer (i) Ionic product of water $K _{ w }= K \times\left[ H _2 O \right]$
$
\begin{array}{l}
\text { Ionisation constant of water }(K)=\frac{K_w}{\left[H_2 O\right]} \\
K_{w}=1.0 \times 10^{-14} \text { and }\left[H_2 O\right]=55.5 M \\
\qquad K=\frac{1 \times 10^{-14}}{55.5}=1.8 \times 10^{-16}
\end{array}
$
$\begin{array}{l}\text { (ii) Concentration of } H ^{+} \text {ion in one litre water } \\ \quad \begin{aligned} & =1 \times 10^{-7} M \end{aligned} \\ \begin{array}{l} \text { Hence number of } H ^{+} \text {ions } \\ =\text { Concentration } \times \text { Avogadro's number } \\ =1 \times 10^{-7} \times 6.022 \times 10^{23} \\ =6.022 \times 10^{16}\end{array}\end{array}$
View full question & answer→Question 172 Marks
In a container of 5 L volume, in a vessel at 1250 K temperature following reaction takes place :
$
CO(g)+H_2 O(g) \rightleftharpoons CO_2(g)+H_2(g)
$
Initially one-one mole of CO and $H _2 O$ is taken and at equilibrium 0.4 mole $H _2 O$ remains then calculate equilibrium constant.
Answer $\quad CO ( g )+ H _2 O ( g ) \rightleftharpoons CO _2(g)+ H _2(g)$
Initial moles $1 \quad 1 \quad 0 \quad 0$
At equilibrium $0.4 mol H _2 O$ remains i.e. 0.6 mol $H _2 O$ react. Hence,
Moles at $\begin{array}{lllll}\text { equilibrium } & 1-0.6 & 1-0.6 & 0.6 & 0.6\end{array}$
Concentration at equilibrium
Mole/Volume $\frac{0.4}{5} \quad \frac{0.4}{5} \quad \frac{0.6}{5} \quad \frac{0.6}{5}$
Hence, Equilibrium concentration :
$
\begin{array}{l}
K_{c}=\frac{\left[CO_2\right]\left[H_2\right]}{[CO]\left[H_2 O\right]} \\
K_{c}=\frac{\left(\frac{0.6}{5}\right)\left(\frac{0.6}{5}\right)}{\left(\frac{0.4}{5}\right)\left(\frac{0.4}{5}\right)}=\frac{0.6 \times 0.6}{0.4 \times 0.4} \\
K_{c}=\frac{0.36}{0.16}=2.25
\end{array}
$
View full question & answer→Question 182 Marks
In reaction $P + Q \rightleftharpoons R + S$, the initial concentration $P$ is double of initial concentration of Q . At equilibrium the concentration $R$ is three times of $Q$, then calculate equilibrium constant for the reaction.
View full question & answer→Question 192 Marks
At 523 K temperature following reaction $PCl _5(g) \rightleftharpoons PCl _3(g)+ Cl _2(g)$ value of $K _{ p }$ is 2.35 atm then calculate value of $K_c \cdot\left(R=0.082 L\right.$ atm degree ${ }^{-1}$ $mol ^{-1}$.
Answer$
\begin{aligned}
K_{p} & =K_{c}(RT)^{\Delta n} \\
\Delta n & =(1+1)-1=1 \\
K_{c} & =\frac{K_{p}}{(RT)^{\Delta_{n}}}=\frac{2.35}{(0.082 \times 523)^1} \\
K_{c} & =\frac{2.35}{42.88}=5.48 \times 10^{-2}
\end{aligned}
$
View full question & answer→Question 202 Marks
For reaction $N _2(g)+3 H _2(g) \rightleftharpoons 2 NH _3(g)$ at 673 K temperature the value of $K _{ c }$ is 0.75 then calculate the value of $K _{ p ^*}\left(K=0.082 L\right.$ atm degree ${ }^{-1}$ $mol ^{-1}$.
AnswerWe know that
$
K_{p}=K_{c}(RT)^{An}
$
Hence $\quad \begin{aligned} \Delta n & =2-(1+3)=-2 \\ K_{ p } & =0.75 \times(0.082 \times 673)^{-2} \\ K_{ p } & =\frac{0.75}{(0.082 \times 673)^2} \\ K_{ p } & =\frac{0.75}{(55.186)^2}=\frac{0.75}{3045.49} \\ K_{ p } & =2.46 \times 10^{-4}\end{aligned}$
View full question & answer→Question 212 Marks
Why is dil. HCl added before passing $H _2 S$ gas into II group of alkaline radicals?
Answer On adding dil. HCl to II group of alkaline radicals, common ion effect due to which ionization of $H _2 S$ decreases, and obtained $S ^2$ concentration decreases and only II group radicals are precipitated otherwise with II group, IV group radicals get also precipitated, if present.
View full question & answer→Question 222 Marks
For the solution of salt made from weak acid and strong base, give the relationship between $K_b, K_w$ and $K_a$, the amount of hydrolysis and the formula to determine the pH of solution.
AnswerAn example of a salt made from a weak acid and strong base is $CH _3 COONa$. In its solution, acetate ion hydrolysed with water to form acetic acid and $OH ^{-}$. Hence it is called anionic hydrolysis.
$
CH_3 COO^{-}(aq)+H_2 O(l) \rightleftharpoons CH_3 COOH(aq)+OH^{-}(aq)
$
Due to $OH ^{-}$ions, the solution is alkaline and pH of solution is more than 7 .
For this hydrolysis :
$
K_{w}=\text { ionic product of water }
$
(i)
$
\begin{aligned}
K_{h}=\frac{K_{w}}{K_{a}} \quad K_{h} & =\text { hydrolysis constant } \\
K_{a} & =\text { acid dissociation }
\end{aligned}
$
(ii) Degree of hydrolysis :
$
h=\sqrt{\frac{K_{h}}{c}} \text { or } h=\sqrt{\frac{K_{w}}{K_{a \times c}}}
$
(iii) pH of solution :
$
pH=\frac{1}{2}\left(pK_{w}+pK_{a}+\log c\right)
$
View full question & answer→Question 232 Marks
What is buffer solution? It is of how many types?
Answer Buffer solution : The solution whose pH remains unchanged when some amount of acid or base is added on it is known as buffer solution.
Hence buffer resist the change in pH . This process is known as buffer mechanism.
Buffer solution are of two types :
(i) Mixed buffer and (ii) Simple buffer solution
(i) Mixed buffer solution : It is of two types :
(a) Acidic buffer (b) Basic buffer
(a) Acidic buffer : A mixture formed by weak acid and when this weak acid react with strong base, is known as acidic buffer.
Ex. : Mixture of $CH _3 COOH$ and $CH _3 COONa$.
(b) Basic buffer : A mixture formed by weak base and reaction of weak base and strong acid is basic buffer.
Ex. : Mixture of $NH _4 OH$ and $NH _4 Cl$.
(ii) Simple buffer solution : Buffer formed by weak acid and weak base is simple buffer.
Ex. : $CH _3 COONH _4$.
View full question & answer→Question 242 Marks
Write the equations which shows the amphoteric nature of water.
Question 252 Marks
Derive the necessary equation to calculate pH of weak acid.
AnswerThe ionisation of weak acid take place as :
HX $( aq )+ H _2 O (l) \rightleftharpoons H _3 O ^{+}( aq )+ X ^{-}( aq )$
Initial
concen-
$\begin{array}{lllll}\text { tration }( M ) & c & \text { Excess } & 0 & 0\end{array}$
If degree of ionisation $=\alpha$
Concentration at
$K_{a}=\frac{\left[H^{+}\right]\left[X^{-}\right]}{[HX]}$
$K_{a}=\frac{(c \alpha)(c \alpha)}{(c-c \alpha)}$
$K_{a}=\frac{c \alpha^2}{1-\alpha}$
$\alpha \ll 1$ hence $K_a=c \alpha^2$
$
\alpha=\sqrt{\frac{K_{a}}{c}}
$
$
pH=-\log \left[H^{+}\right]
$
To calculate $pH \quad\left[ H ^{+}\right]= c \alpha$.
$
\begin{array}{l}
{\left[H^{+}\right]=c \sqrt{\frac{K_{a}}{c}}} \\
{\left[H^{+}\right]=\sqrt{c \cdot K_{a}}}
\end{array}
$
or
$
pH=\frac{1}{2} p_{ka}-\frac{1}{2} \log c
$
View full question & answer→Question 262 Marks
Give the uses and example of acid base indicator.
AnswerPhenolphtalien, bromothimol blue are acidbase indicator. With help of this in acid-base Fitration, end point is calculated.
Ex. : Phenolphthalein is colourless in acidic medium and pink in basic medium.
$
\begin{array}{l}
Hph(aq)+H_2 O(l) \rightleftharpoons H_3 O^{+}(aq) \\
\text { Conjugate } \\
\text { Colourless }
\end{array}+\begin{array}{c}
\text { acid }
\end{array}
$
View full question & answer→Question 272 Marks
Explain the Lewis concept of acid and base? How it is different from Bronsted-Lowry concept? Explain.
Answer According to Lewis the substance which accept electron pair are known as acid and which donate lone pair electron are bases.
In this way definition of base is similar for BronstedLowry and Lewis concept as in both concepts base gives lone pair electron.
Hence according to Lewis acids are proton losing substances.
Generally between Lewis acid and base a coordination bond is formed.
$
\begin{array}{l}
BF_3+NH_3 \longrightarrow\left[BF_3 \longleftarrow NH_3\right] \\
\begin{array}{l}
\text { Lewis } \\
\text { acid }
\end{array} \\
\begin{array}{l}
\text { Lewis } \\
\text { base }
\end{array}
\end{array}
$
View full question & answer→Question 282 Marks
What will be the effect of catalyst on equilibrium? Explain.
Answer There will be no effect of catalyst on equilibrium because catalyst increase the rate of forward and backward reaction with same rate. It decreases the activation energy of forward and backward reaction uniformly.
View full question & answer→Question 292 Marks
In any reversible reaction what is the effect of pressure on it?
Answer The effect of pressure is applicable only on gaseous reaction.
(i) Gaseous reaction in which $\Delta n _{ g }=0$, they will not have any effect on change in pressure because in both sides the pressure remains same.
Example : $H _2(g)+ I _2(g) \rightleftharpoons 2 HI ( g )$
(ii) The gaseous reaction in which $\Delta n =+ ve$ or -ve i.e. there is a difference between moles of reactant and product in reaction then in that on increasing pressure reaction will move in that direction where gaseous moles are less.
$
\text { (a) } \begin{aligned}
& \Delta n=-ve \\
& CO(g)+3 H_2(g) \rightleftharpoons CH_4(g)+H_2 O(g) \\
& \Delta n=2-4=-2
\end{aligned}
$
In this reaction on increasing pressure reaction will move in forward direction.
(b)
$
\begin{array}{l}
\Delta n=+ve \\
C_{(s)}+CO_2(g) \rightleftharpoons 2 CO(g) \Delta n=(2-1)=+1
\end{array}
$
In this on increasing pressure reaction will move in backward direction as moles of reactant is less than that of product.
View full question & answer→Question 302 Marks
Explain Arrhenius theory of acid and base with example.
Answer According to Arrhenius acids are those substances which gives $H ^{+}$when they get dissociated in water while base gives $OH ^{-}$.
Example :
Acids : $HX ^{-}( aq ) \rightleftharpoons H ^{+}( aq )+ X ^{-}( aq )$
$
HCl(aq)+H_2 O(l) \longrightarrow H_3 O^{\oplus}(aq)+Cl^{( }(aq)
$
In aqueous solution $H ^{+}$do not remain free else it exist in form of hydronium ion $\left( H _3 O ^{+}\right)$.
$
\text { Bases: } NaOH(aq) \xrightarrow{+H_2 O} Na^{+}(aq)+OH^{-}(aq)
$
View full question & answer→Question 312 Marks
(a) Differentiate between dissociation and ionization.
(b) What is the relationship between dielectric constant of solvent and electrostatic force between ions?
Answer(a) Dissociation : The solute which have ions in solid state also, and when dissolved in water their ions get separated is known as dissociation.
$
Na^{+} Cl^{-}(g) \xrightarrow{+H_2 O} Na^{+}(aq)+Cl^{-}(aq)
$
Ionization : The process in which neutral atom do not have ion initially but when dissolved in water it gives ions, is known as ionization.
$
\text { Example : } HCl(g) \stackrel{+H_2 O}{\rightleftharpoons} H_3 O^{+}(aq)+Cl^{-}(aq)
$
Nowadays there is no difference considered between ionisation and dissociation.
(b) The electrostatic force of attraction between two oppositely charged ions is inversely proportional to dielectric constant of the medium (solvent) i.e. the solvent which has higher dielectric constant is a good solvent for ionic substances. Water is a universal solvent whose dielectric constant is high (80).
View full question & answer→Question 322 Marks
Give important characters of equilibrium.
AnswerThe important characteristics of equilibrium constant are :
(i) The equilibrium constant expression will be useful when the concentration of reactant and product became constant at equilibrium.
(ii) The value of equilibrium constant do not depend on initial concentration of reactant and product.
(iii) The equilibrium constant of backward reaction is reciprocal of equilibrium constant of forward reaction.
(iv) The equilibrium constant ( K ) of a reaction is related to equilibrium constant of the corresponding reaction whose equation is obtained by multiplying or dividing the equation of original reaction by small integer.
(v) The value of equilibrium constant is fixed for any chemical reaction expressed by a balanced equation, i.e. it depends on the stoichiometry of the reaction and the nature of the reactants and products.
(vi) The value of equilibrium constant for any reaction is constant at a certain temperature, i.e. it depends on temperature.
View full question & answer→Question 332 Marks
Write the general characteristics of equilibrium in any physical process.
Answer Equilibrium in any physical process has the following characteristics :
i Equilibrium is established only in a cloud system at a certain temperature.
ii At equilibrium both the opposite reactions occur at the same rate. These have a dynamic but permanent state.
iii All measurable properties of the system are constant. When equilibrium is established in any physical process, then stability of melting point, vapour pressure etc., identify equilibrium.
iv The value of various quantities at any time shows to what extent the process has progressed before reaching equilibrium.
View full question & answer→Question 342 Marks
AnswerHenry's Law At constant temperature, in the given quantity of solvent, the amount of gas dissolved in a given volume is directly proportional to pressure of gas. On increasing temperature, the quantity decreases.
Hence:
$\begin{aligned} & m \propto p \\ & m = kp \end{aligned}$
where m quantity of dissolved gas
k = Henry's constant and p = Pressure of gas
View full question & answer→Question 352 Marks
What is saturated solution? In this solution which states have equilibrium between them?
Answer Saturated solution: At a certain temperature
if more solute can't be dissolved in solution, then this solution is known as saturated solution.
In saturated solution, there is a dynamic equilibrium between solid state and particles of solute of solution.
Example: Sugar (solution) $\rightleftharpoons$Sugar (solid) At equilibrium Rate of salvation of sugar = Rate of crystallisation of sugar.
View full question & answer→Question 362 Marks
Explain Solid-liquid equilibrium.
AnswerConversion of ice to water is an example of solid-liquid equilibrium.
$\underset{\text { ice }}{ H _2 O ( s )} \rightleftharpoons \underset{\text { water }}{ H _2 O (l)}$
In a container (thermos flask) at 273 K temperature and on one atmospheric pressure, ice and water are in equilibrium. 273 K , is the melting point of ice and freezing point of water. At equilibrium, with time there is no change in mass of ice and water. Because rate of formation of water from ice is equal to rate of formation of ice from water.
View full question & answer→
