Question

Equimolar solutions of HF, HCOOH and HCN at 298 K have the values of $K_{a}$ as $6.8\times 10^{- 4},1.8\times 10^{- 4}$ and $4.8\times 10^{- 9}$ respectively. What will be the order of their acidic strength?

• A. HF > HCN > HCOOH
• B. HF > HCOOH > HCN
• C. HCN > HF > HCOOH
• D. HCOOH > HCN > HF

Answer 1

Answer: (B)

Solution which has higher Ka value will have more acidic strength .

So, $K_{a}\underset{6.8 \times 1 0^{- 4}}{H F}>\underset{1.8 \times 1 0^{- 4}}{H C O O H}>\underset{4.8 \times 1 0^{- 9 }}{H C N}$