The dissociation constants for acetic acid and HCN at 25° C are 1.5 × 10-5 and 4.5 × 10-10 respectively. The equilibrium constant for the equilibrium CN -+ CH 3 COOH leftharpoons HCN + CH 3 COO - would be :

Question

The dissociation constants for acetic acid and HCN at 25° C are 1.5 × 10-5 and 4.5 × 10-10 respectively. The equilibrium constant for the equilibrium CN -+ CH 3 COOH leftharpoons HCN + CH 3 COO - would be : (A)  3.0 × 105  (B)  3.0 × 10-5  (C)  3.0 × 10-4  (D)  3.0 × 104

Tardigrade Admin · Accepted Answer

CH 3 COOH Leftrightarrow CH 3 COO -+ H + ; K a=1.5 × 10-5 H ++ CN - HCN ; (1/ K a)=(1/4.5 × 10-10) K a for CN -+ CH 3 COOH Leftrightarrow CH 3 COO -+ HCN is (1.5 × 10-5/4.5 × 10-10)=(1/3) × 105=3 × 104

Q. The dissociation constants for acetic acid and $H CN$ at $2 5^{\circ} C$ are $1.5 \times 1 0^{- 5}$ and $4.5 \times 1 0^{- 10}$ respectively. The equilibrium constant for the equilibrium
$C N^{-} + C H_{3} COO H ⇌ H CN + C H_{3} CO O^{-}$ would be :

$3.0 \times 1 0^{5}$

$3.0 \times 1 0^{- 5}$

$3.0 \times 1 0^{- 4}$

$3.0 \times 1 0^{4}$

Q. The dissociation constants for acetic acid and HCN at 25∘C are 1.5×10−5 and 4.5×10−10 respectively. The equilibrium constant for the equilibrium CN−+CH3​COOH⇌HCN+CH3​COO− would be :

3.0×105

3.0×10−5

3.0×10−4

3.0×104

CH3​COOH⇔CH3​COO−+H+;Ka​=1.5×10−5 H++CN−HCN;Ka​1​=4.5×10−101​ Ka​ for CN−+CH3​COOH⇔CH3​COO−+HCN is 4.5×10−101.5×10−5​=31​×105=3×104

Q. The dissociation constants for acetic acid and $H CN$ at $2 5^{\circ} C$ are $1.5 \times 1 0^{- 5}$ and $4.5 \times 1 0^{- 10}$ respectively. The equilibrium constant for the equilibrium
$C N^{-} + C H_{3} COO H ⇌ H CN + C H_{3} CO O^{-}$ would be :

$3.0 \times 1 0^{5}$

$3.0 \times 1 0^{- 5}$

$3.0 \times 1 0^{- 4}$

$3.0 \times 1 0^{4}$