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-

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Arrhenius Acid

Equimolar sol...

Question

Equimolar solutions of $H F, H C O O H$ and $H C N$ 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?

H F > H C N > H C O O H

No worries! We‘ve got your back. Try BYJU‘S free classes today!

H F > H C O O H > H C N

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

H C N > H F > H C O O H

No worries! We‘ve got your back. Try BYJU‘S free classes today!

H C O O H > H C N > H F

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $H F > H C O O H > H C N$
solution which has higher $K_{a}$ value will have more acidic strength .
So, $6.8 \times 10^{- 4} > 1.8 \times 10^{- 4} > 4.8 \times 10^{- 8}$
$H F > H C O O H > H C N$

Suggest Corrections

Similar questions

The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10^–4, 1.8 × 10^–4 and 4.8 × 10^–9 respectively. Calculate the ionization constants of the corresponding conjugate base.

Q. The ionization constant of

H F, H C O O H

and

H C N

298 K

are

6.8 \times 10^{- 4}, 1.8 \times 10^{- 4} and 4.8 \times 10^{- 9}

respectively. The ionization constants of the corresponding conjugate bases, respectively are:

Q. The correct descending order of the heat liberated (in

k J

) during the neutralisation of the acids

C H_{3} C O O H (W), H F (X), H C O O H (Y)

and

H C N (Z)

under identical conditions is:

(

K_{a}

C H_{3} C O O H = 1.8 \times 10^{- 5}, H C O O H = 1.8 \times 10^{- 4}, H C N = 4.9 \times 10^{- 10}

and

H F = 3.2 \times 10^{- 4}

)

The heats of neutralisation of $C H_{3} C O O H$ , HCOOH, HCN and $H_{2} S$ are -12.8, -13.2, -3.2 and -3.9 kcal per equivalent. Arrange these acids in increasing order of strength.

Enthalpy of hydrogenation is the enthalpy change when one mole of unsaturated compound is converted to saturated compound. $Δ_{H} h y d r o g e n a t i o n$ helps to calculate resonance energy.

The heat of neutralization of $C H_{3} C O O H, H C O O H, H C N$ and $H_{2} S$ are -12.8, -13.2, -3.2 and -3.9 Kcal per equivalent. Arrange these acids in increasing order of strength.

Join BYJU'S Learning Program

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?

The correct option is C HF>HCOOH>HCNsolution which has higher Ka value will have more acidic strength . So, 6.8×10−4>1.8×10−4>4.8×10−8 HF>HCOOH>HCN

Equimolar solutions of $H F, H C O O H$ and $H C N$ 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?

The correct option is C $H F > H C O O H > H C N$
solution which has higher $K_{a}$ value will have more acidic strength .
So, $6.8 \times 10^{- 4} > 1.8 \times 10^{- 4} > 4.8 \times 10^{- 8}$
$H F > H C O O H > H C N$