The freezing point of a 0.1 M formic acid aqueous solution is − 0.204 6 ∘ C . Find the equilibrium constant of the reaction: H CO O − ( a q . ) + H 2 O ( l ) ⇌ H COO H ( a q . ) + O H − ( a q . ) Gi

Question

The freezing point of a 0.1 M formic acid aqueous solution is − 0.204 6 ∘ C . Find the equilibrium constant of the reaction: H CO O − ( a q . ) + H 2 ​ O ( l ) ⇌ H COO H ( a q . ) + O H − ( a q . ) Given: K f ​ ( H 2 ​ O ) = 1.86 K ⋅ k g ⋅ m o l − 1 . Assume the solution to be very dilute.

TestHub · Accepted Answer

Question Explanation: Calculate the hydrolysis constant using freezing point depression.

Concept: Colligative properties and Ionic Equilibrium.

Solution:

First, use the depression in freezing point to find the Van't Hoff factor ( $i$ ) and degree of dissociation ( $α$ ) of the acid. Then, calculate the acid dissociation constant ( $K_{a}$ ). Finally, determine the hydrolysis constant ( $K_{h}$ ) for the conjugate base using the relation $K_{h} = K_{w} / K_{a}$ .

Calculation:

1. Find $α$ :

$\displaylines T_{f} = i \times K_{f} \times m = 0.2046 = (1 + α) \times 1.86 \times 0.1 (1 + α) = 1.1 α = 0.1$

2. Find $K_{a} = \frac{C α ^{2}}{1 - α} = \frac{0.1 \times ( 0.1 ) ^{2}}{1 - 0.1} = \frac{1 0 ^{- 3}}{0.9} = 1.11 \times 1 0^{- 3}$

3. Find $K_{h} = \frac{K _{w}}{K _{a}} = \frac{1 0 ^{- 14}}{1.11 \times 1 0 ^{- 3}} = 9 \times 1 0^{- 12} M$ :

Final Answer: $9 \times 1 0^{- 12} M^{- 1}$

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