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

Hint: 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
\[\begin{array}{*{20}{l}}
  {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.