Question

The solubility of silver acetate in pure water at $25^\circ C$ $8.35{\text{g}}{\text{.litr}}{{\text{e}}^{ - 1}}$ and ${\text{61}}{\text{.8g}}{\text{.litr}}{{\text{e}}^{ - 1}}$ in an acid solution of ${\text{pH = 3}}$ . Calculate:
i.${{\text{K}}_{sp}}$ of silver acetate and
ii.Dissociation constant of acetic acid.

Answer 1

Hint: The solubility product constant is an equilibrium constant which describes the relation between the solid and its dissolved ions. The solubility product constant depends upon temperature, it changes with the change in temperature of environment or surroundings.

Complete step by step answer:
(i)
We have given two conditions about the solubility of silver acetate, we have to calculate the for ${{\text{K}}_{sp}}$ both of them.
So in the first condition we have given the solubility of silver acetate in pure water.
The reaction is:
${\text{AgOAc}} \to {\text{A}}{{\text{g}}^ + }{\text{ + Ac}}{{\text{O}}^ - }$
                      $s$ $s$
Where $s$ is the solubility which is equal to $\left( {\dfrac{{8.35g}}{{1L}} \times \dfrac{{1mole}}{{166.91g}}} \right)$ $5.033 \times {10^{ - 2}}mol/L$ . Here the given solubility given in gram per liter is converted into mole per liter,
Therefore,
${{\text{K}}_{sp}} = \left[ {A{g^ + }} \right]\left[ {Ac{O^ - }} \right]$
${{\text{K}}_{sp}} = {{\text{s}}^2}$
${{\text{K}}_{sp}} = {(5.033 \times {10^{ - 2}})^2} = 2.50 \times {10^{ - 3}}$
Hence the ${{\text{K}}_{sp}}$ value for the silver acetate in pure water is $2.50 \times {10^{ - 3}}$.
Now in second condition we have provided the solubility of silver acetate in acid solution of ${\text{pH = 3}}$.
The ${\text{pH = 3}}$therefore the concentration of ${{\text{H}}_3}{{\text{O}}^ + }$ is $1 \times {10^{ - 3}}.$
The reaction is:

$
  {\text{AgOAc + }}{{\text{H}}_3}{{\text{O}}^ + } \to {\text{A}}{{\text{g}}^ + }{\text{ + HOAc + }}{{\text{H}}_2}{\text{O}} \\
  {\text{1}} \times {\text{1}}{{\text{0}}^{ - 3}}{\text{ s s}} \\
  \; \\
 $
Where $s$ is the solubility which is equal to $\left( {\dfrac{{61.8g}}{{1L}} \times \dfrac{{1mole}}{{166.91g}}} \right)$ $3.702 \times {10^{ - 1}}mol/L$ . Here the given solubility given in gram per liter is converted into mole per liter.
Therefore,
${{\text{K}}_{sp}} = \dfrac{{\left[ {A{g^ + }} \right]\left[ {HOAc} \right]}}{{\left[ {Ac{O^ - }} \right]}}$
${{\text{K}}_{sp}} = \dfrac{{{s^2}}}{{1 \times {{10}^{ - 3}}}} = \dfrac{{{{(3.702 \times {{10}^{ - 1}})}^2}}}{{1 \times {{10}^{ - 3}}}}$
${{\text{K}}_{sp}} = 1.371 \times {10^2}$
Hence the ${{\text{K}}_{sp}}$ value for the silver acetate in pure water is $1.371 \times {10^2}$.
(ii)
Now we have calculated the dissociation constant of acetic acid.
Here the dissociation constant we have to calculate is ${K_a}$. Which is basically the equilibrium constant of dissociation reaction of acetic acid.
We have two chemical equations
${\text{AgOAc}} \to {\text{A}}{{\text{g}}^ + }{\text{ + Ac}}{{\text{O}}^ - }$ ${K_1} = 2.503 \times {10^{ - 3}}$
${\text{AgOAc + }}{{\text{H}}_3}{{\text{O}}^ + } \to {\text{A}}{{\text{g}}^ + }{\text{ + HOAc + }}{{\text{H}}_2}{\text{O}}$ ${K_2} = 1.37 \times {10^2}$
When we subtract these two equations we get the equation for dissociation of acetic acid. When we subtract these equations their equilibrium constants are divided to get the equilibrium constant for resultant reaction.
So,
${\text{HOAc + }}{{\text{H}}_2}{\text{O}} \to {{\text{H}}_3}{{\text{O}}^ + } + {\text{Ac}}{{\text{O}}^ - }$ ${\text{K = }}\dfrac{{{{\text{K}}_1}}}{{{{\text{K}}_2}}} = \dfrac{{2.503 \times {{10}^{ - 3}}}}{{1.37 \times {{10}^2}}} = 1.82 \times {10^{ - 5}}$
Therefore ${K_a} = 1.82 \times {10^{ - 5}}$


Note:
The dissociation constant is basically an equilibrium constant for the reaction in which a complex dissociates into ions or its substituents. The dissociation constant is also known as ionization constant. It is defined for a complex not for the whole reaction.