Question

Hydroxyl ion concentration $[O{H^ - }]$ in case of sodium acetate can be expressed as (where ${K_a}$ is dissociation constant of $C{H_3}COOH$ and c is the concentration of sodium acetate :-
A) $[O{H^ - }] = \,{(\,c{K_w}.\,{K_a})^{\dfrac{1}{2}}}$
B) $[O{H^ - }] = \,\,c{K_w}.\,\sqrt {{K_a}}$
C) $[O{H^ - }] = \,\,\dfrac{{c{K_w}^{\dfrac{1}{2}}}}{{{K_a}}}$
D) $[O{H^ - }] = \,\,c.\,{K_a}.\,{K_w}$

Answer 1

The question is from ionic equilibrium, here a salt of strong base and weak acid is given which is sodium acetate. It forms when sodium hydroxide reacts with acetic acid. After salt formation its hydrolysis takes place, use the formula for it and solve further to get the relation. The resulting solution will be basic in nature but the expression contains ${K_a}$ that is the dissociation constant of acetic acid.

Complete step-by-step answer:
Let’s start by discussing the reaction $NaOH\, + \,C{H_3}COOH \to \,C{H_3}COONa$ we get sodium acetate which on hydrolysis gives acetate ion and sodium ion. Among the two, sodium ion is considered as passive ions because these ions do not react with water and if they react they further dissociate.
$C{H_3}COONa\, \rightleftharpoons \,\,C{H_3}CO{O^ - }\, + \,N{a^ + }$
${H_2}O\, \rightleftharpoons \,{H^ + }\, + \,O{H^ - }$
The overall reaction can be written as: $\boxed{\,C{H_3}CO{O^ - }\, + {H_2}O\, \rightleftharpoons \,C{H_3}COOH\, + \,O{H^ - }}$
Question is given for the above reaction plot that acetate ions whose dissociation constant is ${K_a}$ will react with water and give hydroxyl ions. After the reaction we can write many expressions like hydrolysis constant ${K_h}$, concentration of ${H^ + }$ ions, concentration of hydroxyl ions $[O{H^ - }]$, pH etc.
So, if we write an expression of hydroxyl concentration $[O{H^ - }]$ it will be a function of ${K_w}$ , concentration c and dissociation constant ${K_a}$ .
We have the relation as: $\left[ {O{H^ - }} \right]\, = \,\sqrt {{K_h}\,c}$ here ${K_h} = \,\dfrac{{{K_w}}}{{{K_a}}}$ it is the ratio of equilibrium constant for water and acetic acid.
On putting the value of ${K_h}$ in the above equation we will get, $\left[ {O{H^ - }} \right]\, = \,\sqrt {\dfrac{{{K_w}}}{{{K_a}}}\,c}$
Let’s compare the answer with the option above and we will see that option A is the correct answer.

Hence the correct answer is option ‘A’.

Note: The expression of hydrogen ions and hydroxyl ions is different for different salt hydrolysis. There are salt hydrolysis of three types, one is when weak acid and strong base reacts, when weak base like ammonium hydroxide reacts with strong acid like $HCl$ and the last one is when both components are weak, weak acid and weak base.