Ionic Equilibrium

Let $\,\,\,s = \,\,\,$ solubility

$Ag{\rm I}{O_3}\,\rightleftharpoons\,\mathop {A{g^ + }}\limits_s \,\,\mathop {{\rm I}{O_3}^ - }\limits_s $

${K_{sp}} = \left[ {A{g^ + }} \right]\left[ {{\rm I}{O_3}^ - } \right]$

$ = s \times s = {s^2}$

Given $\,\,\,{K_{sp}} = 1 \times {10^{ - 8}}$

$\therefore$ $\,\,\,s = \sqrt {{K_{sp}}} = \sqrt {1 \times {{10}^{ - 8}}} $

$ = 1.0 \times {10^4}\,\,mol/lit$

$ = 1.0 \times {10^{ - 4}} \times 283\,g/lit$

(as Molecular mass of $Ag\,{\rm I}{O_3} = 283$ )

$ = {{1.0 \times {{10}^{ - 4}} \times 283 \times 100} \over {1000}}gm/100ml$

$ = 2.83 \times {10^{ - 3}}\,gm/100\,ml$

Conjugate acid-base pair differ by only one proton.

$O{H^ - } \to {H^ + } + {O^2}^ - \,\,$ Conjugate base of $O{H^ - }$ is ${O^{2 - }}$

$M{X_2}\,\rightleftharpoons\,\,\mathop {{M^{ 2 + }}}\limits_{s\,\,\,\,\,\,\,\,} \,\, + \,\,\mathop {2{X^ - }}\limits_{2s} $

Where $s$ is the solubility of $M{X_2}$

then ${K_{sp}} = $ $s \times {\left( {2s} \right)^2}$ = $4{s^3}$;

$ \Rightarrow $$ 4 \times {10^{ - 12}} = 4{s^3};$

$ \Rightarrow $ $s = 1 \times {10^{ - 4}}$

$\left[ {{M^{ 2+ }}} \right] = s $ $ = 1 \times 10^ {- 4}$

$pH = - \log \left[ {{H^ + }} \right] = \log {1 \over {\left[ {{H^ + }} \right]}};$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5.4 = \log {1 \over {\left[ {{H^ + }} \right]}}$

On solving, $\left[ {{H^ + }} \right] = 3.98 \times {10^{ - 6}}$

$M{X_4}\rightleftharpoons\,\mathop {{M^{4 + }}}\limits_{S\,\,\,\,\,\,\,} + \mathop {4{X^ - }}\limits_{4S} $

${K_{sp}} = \left[ s \right]{\left[ {4s} \right]^4} = 256\,{s^5}$

$\therefore$ $\,\,\,\,\,s = {\left( {{{{K_{sp}}} \over {256}}} \right)^{1/5}}$

NOTE : Conjugate acid-base differ by ${H^ + }$

$\mathop {{H_2}PO_4^ - }\limits_{Acid} \mathop \to \limits^{ - H} \,\mathop {\,HPO{{_4^ - }^ - }}\limits_{conjugate\,\,base} $

$pH$ of an acidic solution should be less than $7.$ The reason is that from ${H_2}O.\left[ {{H^ + }} \right] = {10^{ - 7}}M$ which cannot be neglected in comparison to ${10^{ - 8}}M.$ The $pH$ can be calculated as.

from acid, $\,\,\,\left[ {{H^ + }} \right] = {10^{ - 8}}M.$

from ${H_2}O,\,\,\left[ {{H^ + }} \right] = {10^{ - 7}}M$

$\therefore$ $\,\,\,$ Total $\left[ {{H^ + }} \right] = {10^{ - 8}} + {10^{ - 7}}$

$ = {10^{ - 8}}\left( {1 + 10} \right) = 11 \times {10^{ - 8}}$

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

$ = - \log 11 \times {10^{ - 8}}$

$ = - \left[ {\log 11 + 8\log \,10} \right]$

$ = - \left[ {1.0414 - 8} \right] = 6.9586$

$A{B_2}\rightleftharpoons\,{A^{ + 2}} + 2{B^ - }$

$\left[ A \right] = 1.0 \times {10^{ - 5}},\,\,$

$\left[ B \right] = \left[ {2.0 \times {{10}^{ - 5}}} \right],$

${K_{sp}} = {\left[ B \right]^2}\left[ A \right]$

$ = {\left[ {2 \times {{10}^{ - 5}}} \right]^2}\left[ {1.0 \times {{10}^{ - 5}}} \right]$

$ = 4 \times {10^{ - 15}}$

The rain water after thunderstorm contains dissolved acid and therefore the $pH$ is less than rain water without thunderstorm.

${\left( {HS{O_4}} \right)^ - }\,\,\,$ can accept and donate a proton

${\left( {HS{O_4}} \right)^ - } + {H^ + } \to {H_2}S{O_4}\,\,\,$ (acting as base)

${\left( {HS{O_4}} \right)^ - } - {H^ + } \to {H_2}S{O_4}^{2 - }.\,\,\,$ (acting as acid)

$Mg{\left( {OH} \right)_2} \to \mathop {\left[ {M{g^{2 + }}} \right]}\limits_x + 2\mathop {\left[ {O{H^ - }} \right]}\limits_{2x} $

${K_{sp}} = \left[ {Mg} \right]{\left[ {OH} \right]^2}$

$ = \left[ x \right]{\left[ {2x} \right]^2}$

$ = x.4{x^2} = 4{x^3}.$

NOTE : A buffer is a solution of weak acid and its salt with strong base and vice versa.

$HCL$ is strong acid and $NaCL$ is its salt with strong base.

$pH$ is less than $7$ due to $HCL$.