Equimolar solutions of HF, HCOOH and HCN at 298 K have the values of Ka as 6.8× 10- 4,1.8× 10- 4 and 4.8× 10- 9 respectively. What will be the order of their acidic strength?

Question

Equimolar solutions of HF, HCOOH and HCN at 298 K have the values of Ka as 6.8× 10- 4,1.8× 10- 4 and 4.8× 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

Tardigrade Admin · Accepted Answer

Solution which has higher Ka value will have more acidic strength . So, Ka underset6.8 × 1 0- 4H F> underset1.8 × 1 0- 4H C O O H> underset4.8 × 1 0- 9 H C N

Q. Equimolar solutions of HF, HCOOH and HCN at 298 K have the values of $K_{a}$ as $6.8 \times 1 0^{- 4}, 1.8 \times 1 0^{- 4}$ and $4.8 \times 1 0^{- 9}$ respectively. What will be the order of their acidic strength?

HF > HCN > HCOOH

HF > HCOOH > HCN

HCN > HF > HCOOH

HCOOH > HCN > HF

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

So, $K_{a} 6.8 \times 1 0^{- 4} H F > 1.8 \times 1 0^{- 4} H COO H > 4.8 \times 1 0^{- 9} H CN$

Q. Equimolar solutions of HF, HCOOH and HCN at 298 K have the values of Ka​ as 6.8×10−4,1.8×10−4 and 4.8×10−9 respectively. What will be the order of their acidic strength?