Question

100 mL of 2M formic acid (\(pK_a\) = 3.74) is neutralized by NaOH. At the equivalence point, the pH is:

• A. 7
• B. 6
• C. 9.5
• D. 8.87

Hint:

At the equivalence point of a weak acid-strong base titration, the pH is determined by the hydrolysis of the conjugate base of the weak acid.

Answer 1

The Correct Option is D

At the equivalence point, formic acid (HCOOH) is neutralized by NaOH, forming its conjugate base, formate \({HCOO}^-\). The pH at the equivalence point depends on the hydrolysis of the formate ion, which acts as a weak base. 
Step 1: The neutralization reaction is: \[ {HCOOH} + {OH}^- \rightarrow {HCOO}^- + {H}_2 {O} \] 
Step 2: The concentration of formate ion \({HCOO}^-\) at the equivalence point is the same as the initial concentration of formic acid (since the reaction is 1:1), i.e., 2M. 
Step 3: The hydrolysis of formate ion \({HCOO}^-\) in water: \[ {HCOO}^- + {H}_2 {O} \rightleftharpoons {HCOOH} + {OH}^- \] 
Step 4: The \( K_b \) of formate can be calculated from the relation: \[ K_b = \frac{K_w}{K_a} = \frac{1 \times 10^{-14}}{1.7 \times 10^{-4}} \approx 5.88 \times 10^{-11} \] 
Step 5: For a 2M solution of formate ion, the concentration of \({OH}^-\) ions produced can be found using the equilibrium expression for \( K_b \).
Step 6: Solving the equilibrium gives the pOH, and the pH is then calculated as: \[ {pH} = 14 - {pOH} \approx 8.87 \]