Question

Half of the formic acid solution is neutralised on addition of a KOH solution to it. If ${K_a}\left( {HCOOH} \right) = 2 \times {10^{ - 4}}$ then pH of the solution is: $\left( {log2 = 0.3010} \right)$
A.3.6990
B.10.3010
C.3.85
D.4.3010

Answer 1

Since, the ${K_a}$ of formic acid (HCOOH) is given as $2 \times {10^{ - 4}}$. It is given in the question that $50\%$ of the acid is neutralised on addition of a base, therefore concentration of base will be equal to $\dfrac{1}{2}$ and also that of acid is equal to $\dfrac{1}{2}$. Putting the buffer formula, $pH = p{k_a} + log\left( {\dfrac{{base}}{{acid}}} \right)$, we will calculate the pH of the solution.

Complete step by step answer:
The equation formed is:
$HCOOH + KOH \to HCOOK + {H_2}0$
Given in the question is:
${K_a}\left( {HCOOH} \right) = 2 \times {10^{ - 4}}$
When we take the initial concentration of HCOOH as C, then after reaction it becomes $\dfrac{C}{2}$ and $\dfrac{C}{2}$ concentration of HCOOK is also produced. But since $HCO{O^ - }$ is conjugate base of HCOOH, a buffer solution is formed. So, we apply the buffer formula

{pH = p{k_a} + log\left( {\dfrac{{base}}{{acid}}} \right)} \\\ \Rightarrow {pH = p{k_a} + log\left( {\dfrac{{HCOOK}}{{HCOOH}}} \right)} \\\ \Rightarrow {p{k_a} = - log10{K_a}} \\\ \Rightarrow {p{K_a} = - log(2 \times {{10}^{ - 4}})} \\\ \Rightarrow p{K_a} = - log2-log{10^{ - 4}} \\\ \Rightarrow p{K_a} = - 0.3010 + 4 \\\ \Rightarrow p{K_a} = 3.6990 \\\ \Rightarrow p{K_a} \approx 3.7 \\\ \end{array}$$ At half neutralisation point, in this case when half of formic acid (HCOOH) solution is neutralised on addition of KOH solution,

pH = pKa \\
\Rightarrow pH = 3.6990 \\
\Rightarrow pH \approx {\text{ }}3.7 \\

**Therefore, the correct answer is option (A).** **Note:** Since $$50\% $$ of the acid and base is neutralised, therefore concentration of base is equal to $$\dfrac{1}{2}$$ and that of acid is equal to $$\dfrac{1}{2}$$. We know that the Formic acid (HCOOH) is a weak acid and Potassium Hydroxide (KOH) is a strong base.