Question

100ml of 0.20M weak acid $HA$ is completely neutralized by $0.20M$ $NaOH$. ${K_b}$for ${A^ - }$ is ${10^{ - 5}}$,the pH at the equivalence point is
A. $11$
B. $7$
C. $10$
D. None of these

Answer 1

As we all know that when an electrolyte is added into a solvent, it gets dissociated into its constituent ions and there exists an equilibrium between ionised and unionised molecules. When in the solution of a weak electrolyte, a strong electrolyte is added then the degree of dissociation increases.

Formula used: ${K_b} = \dfrac{{[HA][O{H^ - }]}}{{[{A^ - }]}}$, $pH = - \log [O{H^ - }]$ and $pH + pOH = 14$

Complete answer:
As we have already learnt that when we add a strong electrolyte in the solution of a weak electrolyte then the degree of dissociation will increase and when an electrolyte is added into a solvent, it gets dissociated into its constituent ions. In the same way, weak acid $HA$ will dissociate into ${H^ + }$ and ${A^ - }$ and their dissociation constant will be given as the ratio of their concentrations.
$HA \rightleftharpoons {H^ + } + {A^ - }$
${K_a} = \dfrac{{[{H^ + }][{A^ - }]}}{{[HA]}}$
And when we talk about the dissociation constant of base that tells about complete dissociation of a base, it is given as:
${A^ - } + HOH \rightleftharpoons HA + O{H^ - }$
${K_b} = \dfrac{{[HA][O{H^ - }]}}{{[{A^ - }]}}$
Using these conditions, we can find out the concentration of hydroxide ions in the solution as the weak acid will be completely neutralized by the strong base, so the solution will be basic. So, for complete neutralization, the volume of $NaOH$required can be calculated using the formula:
${M_1}{V_1} = {M_2}{V_2}$ where ${M_1}$ is molarity of $HA$, ${V_1}$ is its volume, ${M_2}$ is molarity of $NaOH$ and ${V_2}$ is its volume.
$\Rightarrow {V_2} = \dfrac{{100 \times 0.2}}{{0.2}} \\\ \Rightarrow {V_2} = 100ml$
The concentration of ${A^ - }$ in the solution will be:
$\Rightarrow {A^ - } = \dfrac{{100 \times 0.2}}{{100 + 100}} \\\ \Rightarrow {A^ - } = 0.1M$
Now, at equilibrium:
${A^ - } + HOH \rightleftharpoons HA + O{H^ - }$
At initial conc. $0.1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;0\;\;\;\;\;\;\;\; 0$
After t time $\;\;\;\;0.1 - x\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x\;\;\;\;\;\;\;\;\; x$
${K_b} = \dfrac{{[HA][O{H^ - }]}}{{[{A^ - }]}}$
$
\Rightarrow {10^ - }^5 = \dfrac{{x \times x}}{{0.1 - x}} \\

As we know that $$0.1 - x \simeq 0.10$$ so,x = {10^{ - 3}}$. Hence, the concentration of hydroxide ion is:$[O{H^ - }]$$ = x = {10^{ - 3}}$and we can now find$pOH$as:$
pOH = - \log [O{H^ - }] \\
\Rightarrow pOH = - \log [{10^{ - 3}}] \\
\Rightarrow pOH = 3
$Now the pH of the solution can be calculated as:$
pH + pOH = 14 \\
\Rightarrow pH = 14 - pOH \\
\Rightarrow pH = 14 - 3 \\
\Rightarrow pH = 11
$

**Therefore, the correct answer is (A).

Note:**
If the dissociation constant is high means the solution is more acidic then the $p{K_a}$ will be lower and can be given as a negative logarithm of acidic dissociation constant and when the dissociation constant of base is higher meaning the solution is more basic then, its $p{K_b}$ will be low. Ostwald dilution law is applicable to weak electrolytes and the degree of dissociation of weak electrolytes increases with increase in dilution.