Question

(A) pH of ${10^{ - 7}}M$$NaOH$ solution exists between $7$ to $7.3$ at ${25^ \circ }C$.
(R) Due to common ion effect ionization of water is suppressed.
Both (R) and (A) are true and reason is the correct explanation of assertion.
Both (R) and (A) are true but the reason is not the correct explanation of assertion.
Assertion (A) is true but reason (R) is false.
Assertion (A) and reason (R) both are false.
Assertion (A) is false but reason (R) is true.

Answer 1

Hint: $NaOH$ is a strong base, it dissociates completely in an aqueous solution to give $N{a^ + }$ and $O{H^ - }$ ions. Water molecules undergo self-ionization to form ${H^ + }$ and $O{H^ - }$ions. Due to the common ion effect of $O{H^ - }$ ions, the degree of dissociation of water decreases in the presence of $NaOH$.

Complete step by step answer:
$NaOH$ is a strong base, it undergoes complete dissociation in water, the dissociation of $NaOH$ is given by the following equation:
According to the given concentration of $NaOH$ at ${25^ \circ }C$, which is ${10^{ - 7}}M$, we get a pH of $7$
$
  [O{H^ - }] = [NaOH] \\
  [O{H^ - }] = {10^{ - 7}} \\
  pOH = - \log [O{H^ - }] = - \log ({10^{ - 7}}) \\
  pOH = 7 \\
   \Rightarrow pH = 14 - 7 = 7 \\
  $
But, the pH was found to exist between $7$ to $7.3$, which is slightly alkaline.
Water is a weak electrolyte, it does not completely dissociate, the dissociation of water is given by the following equation:
${H_2}O\,\,(l)\,\, \rightleftharpoons \,\,{H^ + }\,(aq)\,\, + \,\,\,O{H^ - }\,(aq)$
$O{H^ - }$ ions are formed in both the dissociation of $NaOH$and water. The concentration of $O{H^ - }$ ions increases when $NaOH$ dissociates in an aqueous solution. The $O{H^ - }$ions combine with ${H^ + }$ to form water and shift the above equilibrium to the left. Since the concentration of $O{H^ - }$ is increasing and the ${H^ + }$ ions are decreasing, the solution becomes slightly alkaline and the pH becomes higher than $7$ i.e., between $7$ to $7.3$.
The reason for the slightly alkaline pH is the common ion effect. The common ion effect is the decrease in the degree of dissociation of a weak electrolyte in the presence of a strong electrolyte having a common ion.
Here the common ion is $O{H^ - }$ion and water is the weak electrolyte whose degree of ionization is decreased in the presence of a strong electrolyte which is $NaOH$.

So, the correct option is (A): Both (R) and (A) are true and reason is the correct explanation of assertion.

Note:
Common ion-effect is the consequence of Le Chatelier’s principle. It states that if a system in equilibrium is subjected to a change of concentration, temperature, or pressure, the equilibrium shifts in a direction that tends to undo the effect of the change imposed. So, here when the $O{H^ - }$ ions increases, then according to Le Chatelier’s principle, the effect will be to decrease the concentration of $O{H^ - }$. This is only possible if $O{H^ - }$ interacts with ${H^ + }$ to form water and the equilibrium shifts in the backward direction.