Chapter 6: Equilibrium

Ans: Degree of ionization, $\alpha  = 0.132$

Concentration, ${\text{c}} = 0.1{\text{M}}$

Thus, the concentration of ${{\text{H}}_3}{{\text{O}}^ + } = $c. $\alpha $

$ \Rightarrow 0.1 \times 0.132 = 0.0132{\text{}}$  

$pH= {\text{}} - \log \left[ {{H^ + }} \right]{\text{}} = {\text{}} - \log (0.0132){\text{}} = 1.879:1.88{\text{}}$

Now

${K_a}$ = $C{\alpha ^2}{\text{}}$ = $0.1 \times{(0.132)^2}{\text{}}$

${\text{}}{K_a} = .0017{\text{}}p$

${K_a} = 2.75$

Ans:

${{{\text{K}}_{\text{b}}} = 4.27 \times {{10}^{ - 10}}} $ 

${{\text{c}} = 0.001{\text{M}}} $ 

${{\text{pH}} = ?\quad } $ 

${\alpha  = ?} $ 

${{k_a} = C{\alpha ^2}} $ 

${4.27 \times {{10}^{ - 10}} = 0.001 \times {\alpha ^2}} $ 

${4270 \times {{10}^{ - 10}} = {\alpha ^2}} $ 

${65.34 \times {{10}^{ - 5}} = \alpha  = 6.53 \times {{10}^{ - 4}}} $ 

$Then,[anion] =$\text{c}\alpha =0.01\times 65.34\times 10^{-5}$ 

${ = 0.65 \times {{10}^{ - 5}}} $ 

${pOH}{ =  - \log \left( {.065 \times {{10}^{ - 5}}} \right)} $ 

${ = 6.187} $

 $pH = 7.813$

Now,

${{K_a} \times {K_b} = {K_w}} $ 

${\therefore 4.27 \times {{10}^{ - 10}} \times {K_a} = {K_w}} $ 

${{K_a} = \dfrac{{{{10}^{ - 14}}}}{{4.27 \times {{10}^{ - 10}}}}} $ 

${ = 2.34 \times {{10}^{ - 5}}}$ 

Thus, the ionization constant of the conjugate acid of aniline is $2.34 \times {10^{ - 5}}$.

Ans:

${{\text{c}} = 0.05{\text{M}}}$

${p{K_a} = 4.74}$ 

$p{K_a} =  - \log {K_a}$ 

${{K_a} = 1.82 \times {{10}^{ - 5}}} $

$ c = 0.05M$  

$p{K_{a} = 4.74} $

$ pK_{a} =  - \log( K_{a} )$  

${K_{a}} = 1.82 \times 10^{-5} $

${{K_a} = c{\alpha ^2}\quad \alpha  = \sqrt {\dfrac{{{K_a}}}{c}} }$

${\alpha  = \sqrt {\dfrac{{1.82 \times {{10}^{ - 5}}}}{{5 \times {{10}^{ - 2}}}}}  = 1.908 \times {{10}^{ - 2}}}$ 

$ K_a = c\alpha ^2 \quad \alpha  = \sqrt {\dfrac{{K_a}}{{c}}} $ 

${\alpha  = \sqrt {\dfrac{{1.82 \times {{10}^{ - 5}}}}{{5 \times {{10}^{ - 2}}}}}  = 1.908 \times {{10}^{ - 2}}} $

When HCl is added to the solution, the concentration of ${{\text{H}}^ + }$ ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease. Case I: When $0.01{\text{MHCl}}$ is taken.

Let $x$ be the amount of acetic acid dissociated after the addition of HCl.

${{\text{C}}{{\text{H}}_3}{\text{COOH}}}{{{\text{H}}^ + } + }{{\text{C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - }} $ 

${{\text{Initial conc}}{\text{. }}}{0.05{\text{M}}} {{\text{ Afterdissociation }}}{0.05 - x}{0.01 + x}x $

As the dissociation of a very small amount of acetic acid will take place, the values i.e., $0.05 - {\text{x}}$ and $0.01 + {\text{x}}$ can be taken as $0.05$ and $0.01$ respectively.

${{K_a} = \dfrac{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H^ + }} \right]}}{{\left[ {C{H_3}COOH} \right]}}}$

$\therefore {K_{a}} = \dfrac{{(0.01)x}}{{0.05}}$ 

$x = \dfrac{{1.82 \times 10^{-5} \times 0.05}}{{0.01}}$ 

$x = 1.82 \times 10^{-3} \times 0.05M$

$ {{K_a} = \dfrac{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H^ + }} \right]}}{{\left[ {C{H_3}COOH} \right]}}} $ 

${\therefore {K_a} = \dfrac{{(0.01)x}}{{0.05}}} $ 

${x = \dfrac{{1.82 \times {{10}^{ - 5}} \times 0.05}}{{0.01}}} $ 

${x = 1.82 \times {{10}^{ - 3}} \times 0.05M} $

Now,

$\alpha = \dfrac{{{\text{ Amount of acid dissociated }}}}{{{\text{ Amount of acid taken }}}}$

$ \Rightarrow \dfrac{{1.82 \times {{10}^{ - 5}} \times 0.05}}{{0.05}} = 1.82 \times {10^{ - 3}}$

Case II: When $0.1{\text{MHCl}}$ is taken.

Let the amount of acetic acid dissociated in this case be $X$. As we have done in the first case, the concentrations of various species involved in the reaction are:

${\left[ {{\text{C}}{{\text{H}}_3}{\text{COOH}}} \right] = 0.05 - X;0.05{\text{M}}}$

${\left[ {{\text{C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - }} \right] = X}$

${\left[ {{{\text{H}}^ + }} \right] = 0.1 + X;0.1{\text{M}}}$ 

${{K_a} = \dfrac{{\left[ {{\text{C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - }} \right]\left[ {{{\text{H}}^ + }} \right]}}{{\left[ {{\text{C}}{{\text{H}}_3}{\text{COOH}}} \right]}}}$

$\therefore K_a = \dfrac{{(0.1)x}}{{0.05}} $

${\left[ {{\text{C}}{{\text{H}}_3}{\text{COOH}}} \right] = 0.05 - X;0.05{\text{M}}} {\left[{{\text{C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - }} \right] = X} $ 

${\left[ {{{\text{H}}^ + }} \right] = 0.1 + X;0.1{\text{M}}} $ 

${{K_a} = \dfrac{{\left[ {{\text{C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - }}\right]\left[{{{\text{H}}^+}}\right]}}{{\left[{{\text{C}}{{\text{H}}_3}{\text{COOH}}} \right]}}} $$ \therefore K_a = \dfrac{{(0.1)x}}{{0.05}} $

$x = \dfrac{{1.82 \times {{10}^{ - 5}} \times 0.05}}{{0.1}}$ $x = 1.82 \times {10^{ - 4}} \times 0.05M$ Now,

$\alpha = \dfrac{{{\text{ Amount of acid dissociated }}}}{{{\text{ Amount of acid taken }}}}$

$ \Rightarrow \dfrac{{1.82 \times {{10}^{ - 4}} \times 0.05}}{{0.05}} = 1.82 \times {10^{ - 4}}$

Answer: is also $0.1{\text{M}}$ in ${\text{NaOH}}?$

Ans:${{K_b} = 5.4 \times {{10}^{ - 4}}} $ 

${{\text{c}} = 0.02{\text{M}}} $

${\text{c}} = 0.02{\text{M}}$ Then, $\alpha  = \sqrt {\dfrac{{{K_b}}}{c}} $ $ = \sqrt {\dfrac{{5.4 \times {{10}^{ - 4}}}}{{0.02}}} $ $ = 0.1643$

Now, if $0.1{\text{M}}$ of ${\text{NaOH}}$ is added to the solution, then ${\text{NaOH}}$ (being a strong base) undergoes complete ionization.

NaOH (aq)↔ Na+ (aq) + OH- (aq)

                                   0.1 M                 0.1 M

And,

${{{\left( {{\text{C}}{{\text{H}}_3}} \right)}_2}{\text{NH}} + {{\text{H}}_2}{\text{O}} \leftrightarrow {{\left( {{\text{C}}{{\text{H}}_3}} \right)}_2}{\text{NH}}_2^ +  + {\text{OH}}} $ 

${(0.02 - x)\quad x} $ 

${;0.02{\text{M}}} $ 

${{\text{ Then, }}\left[ {{{\left( {{\text{C}}{{\text{H}}_3}} \right)}_2}{\text{NH}}_2^ + } \right] = x} $ 

${\left[ {{\text{O}}{{\text{H}}^ - }} \right] = x + 0.1;0.1} $

${ \Rightarrow {K_b} = \dfrac{{\left[ {{{\left( {{\text{C}}{{\text{H}}_3}} \right)}_2}{\text{NH}}_2^ + } \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]}}{{\left[ {{{\left( {{\text{C}}{{\text{H}}_3}} \right)}_2}NH} \right]}}} $ 

${5.4 \times {{10}^{ - 4}} = \dfrac{{x \times 0.1}}{{0.02}}} $ 

${x = 0.0054} $

It means that in the presence of $0.1{\text{MNaOH}},0.54\% $ of dimethylamine will get dissociated.

Ans:

1. Human muscle fluid 6.83

${{\text{pH}} = 6.83{\text{pH}} =  - \log \left[ {{{\text{H}}^ + }} \right]} $ 

${\therefore 6.83 =  - \log \left[ {{{\text{H}}^ + }} \right]} $ 

${\left[ {{{\text{H}}^ + }} \right] = 1.48\quad {{10}^{ - 7}}{\text{M}}} $

1. Human stomach fluid, 1.2:

${{\text{pH}} = 1.2\quad  \times } $ 

${1.2 =  - \log \left[ {{{\text{H}}^ + }} \right]} $ 

${\therefore \left[ {{{\text{H}}^ + }} \right] = 0.063} $

1. Human blood, 7.38:

${\text{pH}} = 7.38 =  - \log \left[ {{{\text{H}}^ + }} \right]$

$\therefore \left[ {{{\text{H}}^ + }} \right] = 4.17\quad {10^{ - 8}}{\text{M}}$

1. Human saliva, 6.4:

${{\text{pH}} = 6.4\quad  \times } $ 

${6.4 =  - \log \left[ {{{\text{H}}^ + }} \right]} $ 

${\left[ {{{\text{H}}^ + }} \right] = 3.98\quad {{10}^{ - 7}}} $

Ans: The hydrogen ion concentration in the given substances can be calculated by using the given relation: ${\text{pH}} =  - \log \left[ {{{\text{H}}^ + }} \right]$

1. ${\text{pH}}$ of milk $ = 6.8$

Since, ${\text{pH}} =  - \log \left[ {{{\text{H}}^ + }} \right]$

$6.8 =  - \log \left[ {{{\text{H}}^ + }} \right]\log $

$\left[ {{{\text{H}}^ + }} \right] =  - 6.8$

$\left[ {{{\text{H}}^ + }} \right] = \operatorname{anitlog} ( - 6.8)$

$ = 1.5\quad {19^{ - 7}}{\text{M}}$

1. ${\text{pH}}$ of black coffee $ = 5.0$

Since, ${\text{pH}} =  \times \log \left[ {{{\text{H}}^ + }} \right]$

$5.0 =  - \log \left[ {{{\text{H}}^ + }} \right]\log $

$\left[ {{{\text{H}}^ + }} \right] =  - 5.0$

$\left[ {{{\text{H}}^ + }} \right] = \operatorname{anitlog} ( - 5.0)$

$ = {10^{ - 5}}{\text{M}}$

1. ${\text{pH}}$ of tomato juice $ = 4.2$

Since, ${\text{pH}} =  - \log \left[ {{{\text{H}}^ + }} \right]$

$4.2 =  - \log \left[ {{{\text{H}}^ + }} \right]\log $

$\left[ {{{\text{H}}^ + }} \right] =  - 4.2$

$\left[ {{{\text{H}}^ + }} \right] = \operatorname{anitlog} ( - 4.2)$

$ = 6.31\quad {10^{ - 5}}{\text{M}}$

1. ${\text{pH}}$ of lemon juice $ = 2.2$ 

Since, ${\text{pH}} =  - $ log $\left[ {{{\text{H}}^ + }} \right]$ 

$2.2 =  - \log \left[ {{{\text{H}}^ + }} \right]$log 

$\left[ {{{\text{H}}^ + }} \right] =  - 2.2$ 

$\left[ {{{\text{H}}^ + }} \right] = \operatorname{anitlog} ( - 2.2)$ 

$\quad  = 6.31\quad {10^{ - 3}}{\text{M}}$

1. ${\text{pH}}$ of egg white $ = 7.8$

Since, ${\text{pH}} =  - \log \left[ {{{\text{H}}^ + }} \right]$

$7.8 =  - \log \left[ {{{\text{H}}^ + }} \right]\log $

$\left[ {{{\text{H}}^ + }} \right] =  - 7.8$

$\left[ {{{\text{H}}^ + }} \right] = \operatorname{anitlog} ( - 7.8)$

$ = 1.58\quad {10^{ - 8}}{\text{M}}$

Ans: $\left[ {KO{H_{aq}}} \right] = \dfrac{{0.561}}{{\dfrac{1}{5}}}{\text{g/L}}$

${\text{}} = 2.805{\text{g/L}}$

$ = 2.805 \times \dfrac{1}{{56.11}}M{\text{}} =.05M$

$KO{H_{aq}} \to K_{(aq)}^ + {\text{}} + OH_{(aq)}^ - {\text{}}$

$\left[ {O{H^ - }} \right] = .05M$

$ = \left[ {{K^ + }} \right]{\text{}}\left[ {{H^ + }} \right]\left[ {{H^ - }} \right] = {K_w}{\text{}}$

$\left[ {{H^ + }} \right] = \dfrac{{{K_w}}}{{\left[ {O{H^ - }} \right]}}{\text{}}$

$ = \dfrac{{{{10}^{ - 14}}}}{{0.05}}{\text{}}$

$ = 2 \times {10^{ - 11}}M{\text{}}$

$\therefore pH = 12.70$

Ans:Solubility of ${\text{Sr}}{({\text{OH}})_2} = 19.23\;{\text{g}}/{\text{L}}$

Then, concentration of ${\text{Sr}}{({\text{OH}})_2}$

${ = \dfrac{{19.23}}{{121.63}}M} $

${ = 0.1581M} $

${\operatorname{Sr} {{(OH)}_{2(aq)}} \to Sr_{(aq)}^{2 + } + 2{{\left( {O{H^ - }} \right)}_{(aq)}}} $

${\therefore \left[ {S{r^{2 + }}} \right] = 0.1581M} $

${\left[ {O{H^ - }} \right] = 2 \times 0.1581M = 0.3126M} $

Now,

${{K_w} = \left[ {O{H^ - }} \right]\left[ {{H^ + }} \right]} $ 

${\dfrac{{{{10}^{ - 14}}}}{{0.3126}} = \left[ {{H^ + }} \right]} $ 

${ \Rightarrow \left[ {{H^ + }} \right] = 3.2 \times {{10}^{ - 14}}} $ 

${\therefore pH = 13.495;13.50} $

Answer: and also its ${\text{pH}}$. What will be its degree of ionization if the Ans:is $0.01{\text{M}}$ in HCl also?

Ans: Let the degree of ionization of propanoic acid be $\alpha $. Then, representing propionic acid as HA, we have:

${{\text{HA       }} + \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\text{H}}_2}{\text{O}} \Leftrightarrow {{\text{H}}_3}{{\text{O}}^ + }{\text{     +   }}{{\text{A}}^ - }}$ 

${(.05 - 0.0\alpha ) \approx 0.5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.05\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,.05\alpha \,\,\,\,\,\,\,\,} $

$K_{a}=\dfrac{\left[H_{3} O^{+}\right]\left[A^{-}\right]}{[H A]} $

$=\dfrac{(.05 \alpha)(.05 \alpha)}{0.05}=.05 \alpha^{2} $

$\alpha=\sqrt{\dfrac{K_{a}}{.05}}=1.63 \times 10^{-2}$

Then, $\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = .05\alpha  = .05 \times 1.63 \times {10^{ - 2}} = {K_b}.15 \times {10^{ - 4}}{\text{M}}$

$\therefore {\text{pH}} = 3.09$

In the presence of $0.1{\text{M}}$ of ${\text{HCl}}$, let $\alpha $ ' be the degree of ionization.

Then, $\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = 0.01$

${\left[ {{A^ - }} \right] = 005{\alpha ^\prime }} $ 

${[HA] = .05} $ 

${{K_a} = \dfrac{{0.01 \times .05{\alpha ^\prime }}}{{.05}}} $ 

${1.32 \times {{10}^{ - 5}} = .01 \times {\alpha ^\prime }} $ 

${{\alpha ^\prime } = 1.32 \times {{10}^{ - 5}}} $

Ans: ${{\text{c}} = 0.1{\text{M}}} $ 

${{\text{pH}} = 2.34} $ 

${ - \log \left[ {{H^ + }} \right] = pH} $ 

${ - \log \left[ {{H^ + }} \right] = 2.34} $ 

${\left[ {{H^ + }} \right] = 4.5 \times {{10}^{ - 3}}} $

Also,

$\left[ {{H^ + }} \right] = c\alpha $ $4.5 \times {10^{ - 3}} = 0.1 \times \alpha $ $\dfrac{{4.5 \times {{10}^{ - 3}}}}{{0.1}} = \alpha $ $\alpha  = 4.5 \times {10^{ - 3}} = .045$ Then, 

$=0.1 \times\left(45 \times 10^{-3}\right)^{2} $

$=202.5 \times 10^{-6} $

$=2.02 \times 10^{-4}$