Question
100 ~mL of 2 M of formic acid ( p K_a=3.74 ) is neutralise by NaOH , at the equivalence point pH is
BITSAT
Chemistry
Ionic Equilibrium
2022
BITSAT 2022
BITSAT Chemistry Question (2022) — Solution
Options
- A. 7
- B. 6
- C. 9.5
- D. 8.87
Answer
D. 8.87
Step-by-step solution
Sodium formate is present at the equivalence point. It is the salt of weak acid + strong base. So, final solution will be basic in nature. As we know pH =7+ p K_a 2 + C 2 where, C is concentration of salt. Total volume of solution =100+100=200 ~mL Concentration of salt (C)=2 100 200 =1 M pH =7+ 3.74 2 + [1] 2 pH =7+1.87+0 pH =8.87
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