Number of moles of NaOH = V(NaOH) * M(NaOH)= 0.150 L * 0.1 moles/L = 0.015 moles
Number of moles of formic acid, HCOOH = V(HCOOH) * M(HCOOH) = 0.200 L * 0.1 moles/L = 0.020 moles
Here, the limiting reagent is NaOH
The reaction is represented as:
HCOOH + NaOH ↔HCOONa + H2O
Moles of HCOONa formed = Moles of the limiting reagent, NaOH = 0.015 moles
Moles of HCOOH remaining = 0.020-0.015 = 0.005 moles
Total final volume is given as 1 L
Therefore: [HCOOH] = 0.005 moles/1 L = 0.005 M
[HCOONa] = 0.015/1 = 0.015 M
pKa of HCOOH = 3.74
As per Henderson-Hasselbalch equation
pH = pka + log[HCOONa]/[HCOOH] = 3.74+log[0.015/0.005] = 4.22
Therefore, pH of the final solution = 4.22
