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Chemistry Answer the following in brief.

Ionic Equilibria · Maharashtra Board STD 12 Science (MSBSHSE - English Medium). 33 questions.

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Answer the following in brief.

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Question 13 Marks
It is enough to add a few mL of a buffer solution to maintain its pH. Which property of buffer is used here ?
Answer
The important property of reserve acidity and reserve basicity of a buffer solution is used to maintain constant pH. Weak acid or weak base along with ions (cations or anions) from salt react with excess of added acid (H+) or base [OH–] and maintain pH constant.
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Question 23 Marks
Home made jams and jellies without any added chemical preservative additives spoil in a few days whereas commercial jams and jellies have a long shelf life. Explain. What role does added sodium benzoate play ?
Answer
Sodium benzoate added to jams and jellies in commercial products maintains the pH constant and acts as a preservative. Hence jams and jellies are not spoiled for a very long time unlike homemade products.
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Question 33 Marks
Suppose that pH of monobasic and dibasic acid is the same. Does this mean that the molar concentrations of both acids are identical ?
Answer
Even if monobasic acid and dibasic acid give same pH , their molar concentrations are different. One mole of monobasic acid like HCl gives 1 mol of $H ^{+}$while one mole of dibasic acid gives 2 mol of $H ^{+}$in solution. Hence the concentration of dibasic acid will be half of the concentration of monobasic acid. For example, for same pH .
[Monobasic acid] $=[$ Dibasic acid $] / 2$
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Question 43 Marks
In NaOH solution $\left[ OH ^{-}\right]$is $2.87 \times 10^{-4}$. Calculate the pH of solution.
Answer
$\text { Given : }\left[ OH ^{-}\right]=2.87 \times 10^{-4} M , pH =?$
$pOH =-\log _{10}\left[ OH ^{-}\right]$
$=-\log _{10} 2.87 \times 10^{-4}$
$=-(\overline{4} .4579)$
$=(4-0.4579)$
$=3.5421$
$\because pH + pOH =14$
$\therefore pH =14- pOH =14-3.5421=10.4579$
$pH =10.4579 .$
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Question 53 Marks
pH of a weak monobasic acid is 3.2 in its 0.02 M solution. Calculate its dissociation constant.
Answer
$ \text { Given: } pH =3.2 ; C =0.02 M ; K _{ a }=?$
$pH =-\log _{10}\left[ H ^{+}\right]$
$\therefore\left[ H ^{+}\right]=\text {Antilog }- pH$
$=\text { Antilog }-3.2$
$=\text { Antilog } \overline{4} .8$
$=6.31 \times 10^{-4} M$
$HA _{( aq )} \rightleftharpoons H ^{+}+ A ^{-}$
$c(1-\alpha) \ \ \ c \alpha \ \ \ \ \ \ c \alpha$
$\because\left[ H ^{+}\right]=c \alpha$
$\alpha=\frac{\left[ H ^{+}\right]}{c}=\frac{6.31 \times 10^{-4}}{0.02}=0.0315$
$K _{ a }= c \alpha^2$
$=0.02 \times(0.0315)^2$
$=1.984 \times 10^{-5} $ Dissociation constant $= K _{ a }=1.984 \times 10^{-5}$
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Question 63 Marks
Aqueous solution of sodium carbonate is alkaline whereas aqueous solution of ammonium chloride is acidic. Explain.
Answer
(A) (i) Sodium carbonate is a salt of weak acid and strong base.
(ii) In aqueous solution it undergoes hydrolysis.Image
(iii) Strong base dissociates completely while weak acid dissociates partially since $\left[ OH ^{-}\right]>\left[ H _3 O ^{+}\right]$, the solution is basic.
(B) (i) Ammonium chloride is a salt of strong acid and weak base.
(ii) In aqueous solution it undergoes hydrolysis
Image
(iii) Since $\left[ H ^{+}\right]$or $\left[ H _3 O ^{+}\right]>\left[ OH ^{-}\right]$the solution is acidic.
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Question 73 Marks
Derive the relation pH + pOH = 14.
Answer
The ionic product of water, Kw is given by, $KW =\left[ H _3 O ^{+}\right] \times\left[ OH ^{-}\right]$
At $298 K _{,} K _{ w }=1 \times 10^{-14}$
$\therefore pK _{ w }=-\log _{10} K _{ w }=\log _{10} 1 \times 10^{-14}=14$
$\because\left[ H _3 O ^{+}\right] \times\left[ OH ^{-}\right]=1 \times 10^{-14}$
Taking logarithm to base 10 of both sides, $\log _{10}\left[ H _3 O ^{+}\right]+\log _{10}\left[ OH ^{-}\right]=\log _{10} 1 \times 10^{-14}$
Multiplying both the sides by -1 ,
$-\log _{10}\left[ H _3 O ^{+}\right]-\log _{10}\left[ OH ^{-}\right]=-\log _{10} 1 \times 10^{-14}$
$\because pH =-\log _{10}\left[ H _3 O ^{+}\right] ; pOH =-\log _{10}\left[ OH ^{-}\right] ;$
pKw $=-\log _{10} K _{ w }$
$\therefore pH + pOH = pK _{ w }$
OR $ pH + pOH =14$
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Question 83 Marks
Acetic acid is 5% ionised in its decimolar solution. Calculate the dissociation constant of acid.
Answer
Given : $C =0.1 M ;$ Dissociation $=5 \%, K _{ a }=2$ Percent dissociation
$
\begin{aligned}
\alpha & =\frac{\text { Percent dissociation }}{100} \\
& =\frac{5}{100}=0.05 \\
K_{ a } & =\frac{C \alpha^2}{1-\alpha}=\frac{0.1 \times(0.05)^2}{1-0.05} \\
& =\frac{0.1 \times(0.05)^2}{0.95} \text { } \\
& =2.63 \times 10^{-4}
\end{aligned}
$
Dissociation constant of acid $= K _{ a }=2.63 \times 10^{-4}$Given : $C =0.1 M ;$ Dissociation $=5 \%, K _{ a }=2$ Percent dissociation
$
\begin{aligned}
\alpha & =\frac{\text { Percent dissociation }}{100} \\
& =\frac{5}{100}=0.05 \\
K_{ a } & =\frac{C \alpha^2}{1-\alpha}=\frac{0.1 \times(0.05)^2}{1-0.05} \\
& =\frac{0.1 \times(0.05)^2}{0.95} \text { } \\
& =2.63 \times 10^{-4}
\end{aligned}
$
Dissociation constant of acid $= K _{ a }=2.63 \times 10^{-4}$
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Question 93 Marks
Ammonia serves as a Lewis base whereas $AlCl _3$ is Lewis acid. Explain.
Answer
  • Since ammonia molecule, NH3 has a lone pair of electrons to donate it acts as a Lewis base.
  • $AlCl _3$ is a molecule with incomplete octet hence it is electron deficient and acts as a Lewis acid.
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Question 103 Marks
Write a reaction in which water acts as a base.
Answer

Image

Since water accepts a proton, it acts as a base.

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Question 113 Marks
Label the conjugate acid-base pair in the following reactions a. $HCl + H _2 O \rightleftharpoons H _3 O ^{+}+ Cl ^{-}$ b. $CO _3^{2-}+ H _2 O \rightleftharpoons OH ^{-}+ HCO _3^{-}$
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Question 123 Marks
What is meant by conjugate acid-base pair?
Answer
Conjugate acid-base pair : A pair of an acid and a base differing by a proton is called a conjugate acid-base pair.

Image

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Question 133 Marks
What are acids and bases according to Arrhenius theory ?
Answer
According to Arrhenius theory: Acid: It is a substance which contains hydrogen and on dissolving in water produces hydrogen ions $\left( H ^{+}\right)$E.g. $HCl$ $ HCl _{( aq )} \rightleftharpoons H _{( aq )}^{+}+ Cl _{( aq )}^{-} $ Base : It is a substance which contains $OH$ group and on dissolving in water produces hydroxyl ions $\left( OH ^{-}\right)$. E.g. $NaOH$ $ NaOH _{( aq )} \rightleftharpoons Na _{( aq )}^{+}+ OH _{( aq )}^{-} $
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Question 143 Marks
Explain the effect of common ion on the dissociation of weak base.
Answer
(1) Consider the dissociation or ionisation of a weak base, $NH _4 OH$ in its dilute solution.
$NH _4 OH _{( aq )} \rightleftharpoons NH _{4( aq )}^{+}+ OH _{( aq )}^{-}$
The dissociation constant $K _{ b }$ for $NH _4 OH$ will be,
$K_{ b }=\frac{\left[ NH _4^{+}\right] \times\left[ OH ^{-}\right]}{\left[ NH _4 OH \right]}$
(2) If a strong electrolyte like salt $NH _4 Cl$ is added to the solution of $NH _4 OH$, then it gives common ion $NH _4^{+}$.
$NH _4 Cl \rightarrow NH _4^{+}+ Cl ^{-}$
(3) Due to common ion $NH _4^{+}$, overall concentration of $NH _4^{+}$is increased, which increases the ratio $\left[ NH _4^{+}\right] \times\left[ OH ^{-}\right] /\left[ NH _4 OH \right]$,
In order to keep this ratio constant, the equilibrium is shifted to the left hand side which satisfies Le Chatelier's principle.
(4) Thus the ionisation of a weak base is suppressed by a common ion.
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Question 153 Marks
Explain the common ion effect on dissociation of a weak acid.
Answer
(1) Consider the dissociation or ionisation of a weak acid, $CH _3 COOH$ in its solution.
$CH _3 COOH _{( aq )} \rightleftharpoons CH _3 COO _{( aq )}^{-}+ H _{( aq )}^{+}$
The dissociation constant $Ka$ for $CH _3 COOH$ will be,
$K _{ a }=\frac{\left[ CH _3 COO ^{-}\right] \times\left[ H ^{+}\right]}{\left[ CH _3 COOH \right]}$
$K _{ a }$ is constant for $CH _3 COOH$ at constant temperature.

(2) If a strong electrolyte like salt $CH _3 COONa$ is added to the solution of $CH _3 COOH$, then on dissociation it gives a common ion $CH _3 COO ^{-}$.
$CH _3 COONa \rightarrow CH _3 COO ^{-}+ Na ^{+}$

(3) Due to common ion $CH _3 COO ^{-}$, overall concentration of $CH _3 COO ^{-}$in the solution is increased which increases the ratio, $\frac{\left[ CH _3 COO ^{-}\right] \times\left[ H ^{+}\right]}{\left[ CH _3 COOH \right]}$. In order to keep this ratio constant, the concentration of $H ^{+}$is decreased, by shifting the equilibrium to the left hand side according to Le Chatelier's principle.

(4) Thus the ionisation of a weak acid is suppressed by a common ion.
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Question 163 Marks
Explain common ion effect with suitable example.
Answer
A weak electrolyte dissociates partially in aqueous solution to produce cations and anions. Equilibrium exists between ions thus formed and the undissociated molecules.
$BA \rightleftharpoons B ^{+}+ A ^{-}$
For such an equilibrium, the dissociation constant $K$ is defined as
$K =\frac{\left[ B ^{+}\right] \times\left[ A ^{-}\right]}{[ BA ]}$
$K$ is constant for the weak electrolyte at a given temperature.
Now, if another electrolyte BC or DA is added to the solution $B A$, having a common ion either $B ^{+}$ or $A ^{-}$, then the concentration of either $B ^{+}$or $A ^{-}$is increased. However, as $K$ is always constant, the increase in the concentration of any one of the ions shifts the equilibrium to left. In other words, the dissociation of BA is suppressed. This is called common ion effect. For example, the dissociation of a weak acid $CH _3 COOH$ is suppressed by adding $CH _3 COONa$ having common ion  $CH _3 COO^{-}$.
$\begin{aligned}
& CH _3 COOH \rightleftharpoons CH _3 COO ^{-}+ H ^{+} \\
& CH _3 COONa \rightarrow CH _3 COO ^{-}+ Na ^{+}
\end{aligned}$
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Question 173 Marks
At $298 K$, the solubility of silver sulphate is $1.85 \times 10^{-2} mol dm ^{-3}$. Calculate the solubility product of silver sulphate.
Answer
Given : Silver sulphate dissociates as follows :
$\begin{aligned}
\underset{(solid)}{Ag _2 SO _4} \rightleftharpoons 2 Ag _{( aq )}^{+}+ SO _{4( aq )}^{2-} \\
K_{ sp }=\left[ Ag ^{+}\right]^2\left[ SO _{4( aq )}^{2-}\right]
\end{aligned}$
Solubility of $Ag _2 SO _4=1.85 \times 10^{-2} mol dm ^{-3}$
$K _{ sp }=$ Solubility product of $Ag _2 SO _4=$ ?
$\left[ Ag ^{+}\right]=$Concentration of $Ag ^{+}$ion
$=2 \times 1.85 \times 10^{-2}=3.70 \times 10^{-2} mol dm ^{-3}$
[latex $] \backslash$ mathrm $\{S O\}_{-}\{4\} \wedge\{2-\}[/$ latex $]=$ Concentration of $SO _4^{2-}$
$\begin{aligned}
& =1.85 \times 10^{-2} mol dm ^{-3} \\
& \therefore K _{\text {sp }}=\left(3.70 \times 10^{-2}\right)^2 \times\left(1.85 \times 10^{-2}\right) \\
& =13.69 \times 10^{-4} \times 1.85 \times 10^{-2} \\
& =25.33 \times 10^{-6} \\
& =2.533 \times 10^{-5}
\end{aligned}$
Solubility product of $Ag _2 SO _4$
$=2.533 \times 10^{-5}$
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Question 183 Marks
The solubility product of magnesium hydroxide is $1.4 \times 10^{-11}$. Calculate the solubility of magnesium hydroxide.
Answer
Given : $K _{\text {sp }}=1.4 \times 10^{-11} ; S =$ ?
Magnesium hydroxide dissociates as shown below:
$\begin{aligned}
& Mg ( OH )_2 \rightleftharpoons Mg ^{2+}+2( OH )^{-} \\
& K _{ sp }=\left[ Mg ^{2+}\right]\left[ OH ^{-}\right]^2
\end{aligned}$
Let the solubility of $Mg ( OH )_2$ be $S mol dm { }^{-3}$.
$\therefore\left[ Mg ^{2+}\right]=$ Concentration of $Mg ^{2+}$ ions
$= S mol dm ^{-3}$
$\therefore\left[ OH ^{-}\right]=$Concentration of $OH ^{-}$ions
$=2 S mol dm ^{-3}$$\begin{aligned}
& \therefore K _{\text {sp }}= S \times(25)^2=4 S ^3 \\
& K _{ sp }=1.4 \times 10^{-11} \\
& \therefore 1.4 \times 10^{-11}=4 S^3 \\
& 1.4 \times 10^{-11}=4 S^3 \text {. } \\
& S=\sqrt[3]{\frac{1.4 \times 10^{-11}}{4}}=\sqrt[3]{0.35 \times 10^{-11}} \\
& =\sqrt[3]{3.5 \times 10^{-12}} \quad \\
& S=1.518 \times 10^{-4} mol dm ^{-3} \\
\end{aligned}$
Solubility of $Mg ( OH )_2$
$=1.518 \times 10^{-4} mol dm ^{-3}$
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Question 193 Marks
The solubility of lead sulphate is $3.03 \times 10^{-5} kg / dm ^3$. Calculate its solubility product. [Molecular mass of $\left.PbSO _4=303\right]$
Answer
Given : $S=3.03 \times 10^{-5} kg dm ^{-3}, K _{ sp }=$ ?
Lead sulphate dissociates as
$PbSO _4 \rightleftharpoons Pb ^{2+}+ SO _4^{2-}$
$($ solid $) \quad(aq) \quad(aq)$
Molecular weight of $PbSO _4=303$
$=303 \times 10^{-3} kg$
The solubility of $PbSO _4$ is $3.03 \times 10^{-5} kg / dm ^3$.
Solubility in $mol dm ^{-3}$
Weight of $PbSO _4$ per $dm ^3$ Molecular weight $3.03 \times 10^{-5}$
$\begin{aligned}
& =\frac{\text { Weight of } PbSO _4 \text { per } dm ^3}{\text { Molecular weight }} \\
& =\frac{3.03 \times 10^{-5}}{303 \times 10^{-3}}=1.0 \times 10^{-4} mol dm ^{-3} \\
& K_{\text {sp }}=\left[ Pb ^{2+}\right]\left[ SO _4^{2-}\right] \\
& {\left[ Pb ^{2+}\right]=\left[ SO _4^{2-}\right]=1.0 \times 10^{-4} mol dm ^{-3}} \\
& \begin{aligned}
\therefore K_{\text {sp }} & =\left(1.0 \times 10^{-4}\right) \times\left(1.0 \times 10^{-4}\right) \\
= & 1.0 \times 10^{-8} \quad \text {  }
\end{aligned}
\end{aligned}$
Solubility product $= 1 . 0 \times 1 0 ^{-8}$.
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Question 203 Marks
Write expression for solubility and solubility product of following sparingly soluble salts :
$(1) \ AgBr\ (2) \ PbI _2\ (3) \ Al ( OH )_3$
Answer
In general, for a sparingly soluble salt $A_x B_y$,
$K _{ sp }= x ^{ x } \cdot y ^{ y } \cdot( S )^{ x + y }$
$(1)$ For binary electrolyte $(BA$ type$)$ :
$ AgBr _{( s )} {\rightleftharpoons} \underset{x=1}{ Ag _{( aq )}^{+}}+\underset{y=1}{ Br _{( aq )}^{-}}$
$ \therefore K_{ sp } =x^x \cdot y^y \cdot(S)^{x+y}$
$ =1^1 \times 1^1 \times( S )^{1+1}$
$ =S^2$
$\therefore S=\sqrt{K_{\text {sp }}} mol dm ^{-3}$
$\text { (2) } PbI _{2(s)} \rightleftharpoons \underset{x=1}{ Pb _{\text {(aq) }}^{2+}}+ \underset{y=2}{2 I _{( aq )}^{-}}$
$\therefore K_{ sp }=x^x \cdot y^y \cdot(S)^{x+y}$
$=1^1 \times 2^2 \times( S )^{1+2}$
$=4 S^3$
$\therefore S=\left(\frac{K_{ sp }}{4}\right)^{\frac{1}{3}} \ mol \ dm ^{-3}$
$(3) Al ( OH )_{3(s)} \rightleftharpoons Al _{( aq )}^{3+}+3 OH _{( aq )}^{-}$
$K_{ sp }= {\left[ Al ^{3+}\right] \times\left[ OH ^{-}\right]^3 }$
$\therefore K_{ sp } =x^x \cdot y^y .(S)^{x+y}$
$ =1^1 \times 3^3 \times(S)^{1+3}$
$ =27 S^4$
$ \therefore S=\left(\frac{K_{ sp }}{27}\right)^{\frac{1}{4}}\  mol \ dm ^3$
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Question 213 Marks
$0.2 dm ^3$ acidic buffer solution contains $1.18 g$ acetic acid and $2.46 g$ sodium acetate. If $K _{ a }$ for acetic acid is $1.8 \times 10^{-5}$ at $25^{\circ} C$, find $pH$ of the solution.
Answer
Given : $W_{ CH _3 COOH }=1.18 g$;
$W_{ CH _3 COONa }=2.46 g;$
$V=0.2 dm ^3, K_{ aCH _3 COOH }=1.8 \times 10^{-5} ; pH =$ ?
$n_{ CH _3 COOH }=\frac{W_{ CH _3 COOH }}{M_{ CH _3 COOH }}=\frac{1.20}{60}=0.02 mol$
$n_{ CH _3 COONa }=\frac{W_{ CH _3 COONa }}{M_{ CH _3 COONa }}=\frac{2.46}{82}=0.03 mol$
$\left[ CH _3 COOH \right]=\frac{n_{ CH _3 COOH }}{V}=\frac{0.02}{V} M$;
$\left[ CH _3 COONa \right]=\frac{n_{ CH _3 COONa }}{V}=\frac{0.03}{V}$
$\begin{aligned}
p K_{ a } & =-\log _{10} K_{ a } \\
& =-\log _{10} 1.8 \times 10^{-5} \\
& =-(\overline{5} .2553) \\
& =4.7447
\end{aligned}$

$pH = p K_{ a }+\log _{10} \frac{[\text { Salt }]}{[\text { Acid }]}$
$\begin{aligned}
& =4.7447+\log _{10} \frac{\frac{0.03}{V}}{\frac{0.02}{V}} \\
& =4.7447+0.1761 \\
& =4.9208 \quad
\end{aligned}$

$pH =4.9208$
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Question 223 Marks
Explain a buffer action of a basic buffer.
Answer
Mechanism of action of a basic buffer :
(1) A basic buffer solution is a solution containing a weak base and its salt with a strong acid. The weak base dissociates feebly, but the salt dissociates completely. Moreover, due to the presence of the common ion, largely supplied by the salt, the dissociation of the base is further suppressed.$\left( NH _4 OH + NH _4 Cl \right):$
$\begin{aligned}
& NH _4 Cl _{( aq )} \rightarrow NH _{4( aq )}^{+}+ Cl _{( aq )}^{-} \text {(Complete) }
\end{aligned}$

(2) When a small quantity of a strong acid is added to the solution, the hydrogen ions combine with the base producing corresponding cations and water. Thus, the addition of an acid does not change the $pH$ of the buffer.
$H _{( aq )}^{+}+ NH _4 OH _{( aq )} \rightleftharpoons NH _{4( aq )}^{+}+ H _2 O _{( l )}$
This removal of added $H ^{+}$is called reserved basicity.

(3) When a small quantity of a strong base is added, the hydroxide ions combine with $NH _4^{+}$ions to form undissociated $NH _4 OH$. As a result, the hydrogen or hydroxyl ion concentration does not change. Thus, the $pH$ of the solution does not change.
$OH _{( aq )}^{-}+ NH _4^{+} \rightleftharpoons NH _4 OH _{( aq )}$
This removal of added $OH ^{-}$is called reserved acidity.
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Question 233 Marks
Explain a buffer action of an acidic buffer.
Answer
Mechanism of action of an acidic buffer:
(1) An acidic buffer is a mixture of a weak acid and its salt with a strong base. The weak acid dissociates feebly, but the salt dissociates almost completely. Moreover, due to the common ions, largely supplied by the salt, dissociation of the weak acid is further suppressed.
$\left( CH _3 COOH + CH _3 COONa \right):$
$CH _3 COONa _{( aq )} \rightarrow CH _3 COO _{( aq )}+ Na _{( aq )}^{+}$(Complete)

(2) When a small quantity of strong acid $\left( H ^{+}\right)$is added to this mixture, hydrogen ions combine with acetate ions to form undissociated acetic acid. Thus, addition of an acid does not change the pH of the buffer.
$H _{( aq )}^{+}+ CH _3 COO _{( aq )}^{-} \rightleftharpoons CH _3- COOH _{( aq )}$
This removal of added $H ^{+}$is called reserved basicity.

(3) When a small quantity of a strong base $\left( OH ^{-}\right)$is added, the hydroxide ions react with the acid producing the corresponding anions and water. Thus, the concentrations of $H ^{+}$and $OH ^{-}$in the solution do not change and the pH remains constant.
$CH _3 COOH _{( aq )}+ OH _{( aq )}^{-} \rightleftharpoons CH _3 COO _{( aq )}^{-}+ H _2 O _{( l )}$
This removal of added $OH ^{-}$is called reserved acidity.
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Question 243 Marks
What are the types of buffer solutions ?
Answer
These are two types of buffer solutions :
(A) Acidic buffer solution : it is a solution containing a weak acid e.g. $\left( CH _3 COOH \right)$ and its salt of a strong base. e.g. ( $CH _3 COONa$ ).
$pH$ of an acidic buffer is given by following Henderson Hasselbalch equation,
$pH = p K_{ a }+\log _{10} \frac{[\text { Salt }]}{[\text { Acid }]}$
where $pK _{ a }=-\log _{10} K _{ a }$
and $K _{ a }$ is the dissociation constant of weak acid.

(B) Basic buffer solutions : It is a solution containing a weak base (e.g. $NH _4 OH$ ) and its salt of strong acid, (e.g. $NH _4 Cl$ ).
pOH of a basic buffer is given by Henderson Hassebalch equation,
$pOH = p K_{ b }+\log _{10} \frac{[\text { Salt }]}{[\text { Base }]}$
where $pK _{ b }=-\log 10 K _{ b }$
and $K_b$ is the dissociation constant of a weak base.
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Question 263 Marks
A salt of strong acid and strong base does not undergo hydrolysis. Explain.
###
An aqueous solution of sodium chloride is neutral. Explain.
Answer
(1) Sodium chloride is a salt of strong acid $HCl$ and strong base $NaOH$.
(2) In water, it reacts forming $HCl$ and $NaOH$.
(3) As both are strong, they dissociate almost completely to liberate $H ^{+}$and $OH ^{-}$ions, respectively.
(4) $H ^{+}$and $OH ^{-}$ions combine together to form weakly dissociating $H _2 O$. As there are no free $H ^{+}$ ions and $OH ^{-}$ions, the solution is neutral and the salt does not undergo hydrolysis.
$NaCl + H _2 O \rightleftharpoons NaOH + HCl$
$\begin{aligned}
& Na ^{+}+ Cl ^{-}+ H _2 O \rightleftharpoons Na ^{+}+ OH ^{-}+ H ^{+}+ Cl ^{-} \\
& H _2 O \rightleftharpoons H ^{+}+ OH ^{-}
\end{aligned}$
Since the solution contains equal number of $H ^{+}$and $OH ^{-}$ions, it is neutral.
Hence the salt of strong acid and strong base does not undergo hydrolysis.
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Question 273 Marks
$NH _4 OH$ is $4.3 \%$ ionised at $298 K$ in $0.01 M$ solution. Calculate the ionization constant and $pH$ of $NH _4 OH$.
Answer
Given : Per cent dissociation $=4.3$,
$C =0.01 M , K _{ b }=?, pH =?$
The degree of dissociation and dissociation constant of $NH _4 OH$ are related to each other by the formula :
$K_b=\alpha^2 C$
$K _{ b }=$ Dissociation constant of $NH _4 OH =$ ?
$\alpha=$ Degree of dissociation of $NH _4 OH =4.3 \%$
$=4.3 \times 10^{-2}$
$C =$ Molar concentration of $NH _4 OH =0.01 M$
$\begin{aligned}
& \therefore K _{ b }=\left(4.3 \times 10^{-2}\right)^2 \times 0.01 \\
& =18.49 \times 10^{-4} \times 10^{-2} \\
& \therefore K _{ b }=1.849 \times 10^{-5}
\end{aligned}$
>Since $NH _4 OH$ is a monoacidic base,
$\begin{aligned}
& {\left[ OH ^{-}\right]=\alpha C} \\
& =4.3 \times 10^{-2} \times 0.01 \\
& =4.3 \times 10^{-4} mol dm ^{-3} \\
& pOH =-\log _{10}[ OH -] \\
& =-\log _{10} 4.3 \times 10^{-4} \\
& =-[0.6335-4]=3.3665 \\
& pH + pOH =14 \\
& pH =14-3.3665=10.6335 \\
& K _{ b }=1.849 \times 10^{-5}, pH =10.6335 \\
&
\end{aligned}$
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Question 283 Marks
The $pH$ of a $0.2 M$ solution of ammonia is $10.78$ . Calculate $(i) \ OH ^{-}$ions concentration $(ii)$ the degree of dissociation $(iii)$ the dissociation constant.
Answer
Given: $pH =10.78, C =0.02 M ,\left[ OH ^{-}\right]=? K _{ b }=?$
As $NH _3$ is a base,
$pOH =14- pH$
$ pH =10.78$
$ \therefore pOH =14-10.78=3.22$
$ pOH =-\log _{10}\left[ OH ^{-}\right]$
$ \therefore 3.22=-\log _{10}\left[ OH ^{-}\right]$
$ \therefore-3.22=\log _{10}\left[ OH ^{-}\right]$
$ {\left[ OH ^{-}\right]=\operatorname{antilog}(-3.22)}$
$ \therefore\left[ OH ^{-}\right]=\operatorname{antilog}(\overline{4} .78)$
$ =6.026 \times 10^{-4} M$
As $NH _4 OH$ is a monoacidic base, $\left[ OH ^{-}\right]$
${\left[ OH ^{-}\right]=\alpha C \therefore \alpha=\frac{\left[ OH ^{-}\right]}{C}}$
$C=0.02\ mol\ dm ^{-3}$
$\therefore \alpha=\frac{6.026 \times 10^{-4}}{0.02}$
$=3.013 \times 10^{-2} $$ \therefore K_{ b } =\alpha^2 C$
$ =\left(3.013 \times 10^{-2}\right)^2 \times 0.02$
$ =9.078 \times 10^{-4} \times 2 \times 10^{-2}$
$ =18.156 \times 10^{-6} \doteqdot 1.816 \times 10^{-5} $
$\left[ OH ^{-}\right]=6.026 \times 10^{-4}\ mol\ dm ^{-3}$,
$\alpha=3.013 \times 10^{-2}$
$K_b=1.816 \times 10^{-5}$
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Question 293 Marks
What are approximate concentrations of $H _3 O ^{+}$and $OH ^{-}$in, (a) pure water or neutral solution, (b) acidic solution and (c) basic solution ? Also mention $pH$ values.
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Question 303 Marks
0.01 mole of a weak base is dissolved in $8 dm ^3$ of water. The dissociation constant of the base is $4.0 \times 10^{-10}$. Calculate the degree of dissociation of the base in the solution.
Answer
Given : $V=8 dm ^3 ; n =0.01$ mole; $K _{ b }=4 \times 10^{-10}$
$\alpha=?$
The degree of dissociation and dissociation constant of a weak base are related to each other by the following formula :
$K _{ b }=\alpha^2 COR \alpha=\sqrt{\frac{K_b}{C}}$
$K _{ b }=$ Dissociation constant of the base $=4.0 \times 10^{-10}$
$\alpha=$ Degree of dissociation of the base $=$ ?
$C=$ Molar concentration of the base
$\begin{aligned}
& =0.01 \text { mole in } 8 dm ^3 \\
& =\frac{0.01}{8} mol dm ^{-3} \\
& \therefore \alpha=\sqrt{\frac{4.0 \times 10^{-10}}{0.01 / 8}} \\
& =\sqrt{32 \times 10^{-8}} \\
& =4 \sqrt{2} \times 10^{-4}=4 \times 1.414 \times 10^{-4} \\
& =5.656 \times 10^{-4} \text {  } \\
\end{aligned}$
Degree of dissociation of the base $=5.656 \times 10^{-4}$
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Question 313 Marks
Explain : (A) $BF _3$ is a Lewis acid, (B) $NH _3$ is a Lewis base.
Answer
(A) According to Lewis theory, an acid is a substance which can accept a pair of electrons.
In $BF _3$ molecule, the octet of $B$ is incomplete, hence it needs two electrons or a pair of electrons to complete its octet. Hence $BF _3$ acts as a Lewis acid.
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(B) According to Lewis theory, a base is a substance which can donate a pair of electrons. In $NH _3$ molecule, nitrogen atom has one lone pair of elctrons to donate.
Image
The reaction between BF3 and NH3 can be represented as,

Image
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Question 323 Marks
For each of the following reactions, identify the Lowry-Bronsted conjugate acid-base pairs:
(1) $CH _3 COOH _{( aq )}+ NH _3 \rightleftharpoons N H _{4( aq )}^{+}+$
$CH _3 COO _{( aq )}^{-}$
(2) $CN _{(\text {aq) }}^{-}+ H _2 O _{(i) } \rightleftharpoons OH _{( aq )}^{-}+ H C N _{( aq )}$ 
(3) $HPO _{4( aq )}^{2-}+ H _2 O _{( i )} \rightleftharpoons OH _{( aq )}^{-}+ H _2 PO _{4( aq )}^{-}$
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Question 333 Marks
Explain neutralisation reaction according to Arrhenius theory.
Answer
Neutralisation reaction : According to Arrhenius theory neutralisation is a reaction between an acid and a base in their aqueous solutions produciijg salt and unionised water.
$HCl _{( aq )}+ NaOH _{( aq )} \rightarrow NaCl _{( aq )}+ H _2 O _{( l )}$
Since strong acid, strong base and salt dissociate completely, the above reaction is represented as,
Image
Hence, according to Arrhenius theory neutralisation reaction is defined as a reaction between $H ^{+}$ ions and $OH ^{-}$ions forming unionised water molecules.
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