Solveeit Logo
Login

Question

Question: What is \({{K}_{a}}\) of \(ZnC{{l}_{2}}\)?...

What is Ka{{K}_{a}} of ZnCl2ZnC{{l}_{2}}?

Explanation

Solution

When any acid or a base is added to a solution, it dissociates to some extent. This extent of dissociation when calculated for acids is called dissociation constant for an acid, denoted by Ka{{K}_{a}}. The value for the base dissociation constant multiplied by acid dissociation constant is Ka×Kb=1×10−14{{K}_{a}}\times {{K}_{b}}=1\times {{10}^{-14}}.

Complete answer:
The dissociation of any acid in the solution leads to formation of ions in a particular concentration. The concentration of the ions produced by the acid upon the concentration of the dissociated acid gives us dissociation constant for an acid, denoted by Ka{{K}_{a}}.
ZnCl2ZnC{{l}_{2}} is zinc chloride that acts as an acid, when dissolved in water undergoes a reaction to produce zinc hydroxide base, and hydrochloric acid. The reaction is, ZnCl2+2H2O⇌Zn(OH)2+2HClZnC{{l}_{2}}+2{{H}_{2}}O\rightleftharpoons Zn{{(OH)}_{2}}+2HCl , in ionic form this equation will be, Zn+++2H2O⇌Zn(OH)2+2H+Z{{n}^{++}}+2{{H}_{2}}O\rightleftharpoons Zn{{(OH)}_{2}}+2{{H}^{+}} the dissociation constant for the acid zinc chloride will be calculated as,
Ka=[Zn(OH)2][H+][Zn++]{{K}_{a}}=\dfrac{[Zn{{(OH)}_{2}}][{{H}^{+}}]}{[Z{{n}^{++}}]}
This Ka{{K}_{a}} is called a hydrolysis constant as zinc chloride is subjected to hydrolysis that produces the mentioned ions in the formula. The Ka{{K}_{a}} can also be expressed as assuming that the concentration of the ZnCl2ZnC{{l}_{2}} solution is 0.001 M and the base dissociation constant Kb{{K}_{b}} for zinc hydroxide, we have,
Zn(OH)2⇌Zn+++2OH−Zn{{(OH)}_{2}}\rightleftharpoons Z{{n}^{++}}+2O{{H}^{-}} , so dissociation constant for base will be,
Kb=[Zn++][OH−]2[Zn(OH)2]{{K}_{b}}=\dfrac{[Z{{n}^{++}}]{{[O{{H}^{-}}]}^{2}}}{[Zn{{(OH)}_{2}}]}
We have dissociation constant of water Kw{{K}_{w}}as Kw=[H+][OH−]{{K}_{w}}=[{{H}^{+}}][O{{H}^{-}}]
Therefore the dissociation constant for acid will be, Ka=Kw2Kb{{K}_{a}}=\dfrac{{{K}_{w}}^{2}}{{{K}_{b}}}
Hence, Ka{{K}_{a}} for ZnCl2ZnC{{l}_{2}}is Ka=[Zn(OH)2][H+][Zn++]{{K}_{a}}=\dfrac{[Zn{{(OH)}_{2}}][{{H}^{+}}]}{[Z{{n}^{++}}]}or Ka=Kw2Kb{{K}_{a}}=\dfrac{{{K}_{w}}^{2}}{{{K}_{b}}}

Note:
By putting the values of dissociation constants of water and zinc hydroxide we can calculate the value of Ka{{K}_{a}} for ZnCl2ZnC{{l}_{2}}. For calculating these values it is important to know the concentration of the ionic species present in the solution. Ka{{K}_{a}} and Kb{{K}_{b}}tells us the strength of acids and bases respectively through their dissociation.