The freezing point of a 0.08 molal aqueous solution of $N a H S O_{4}$ is $- 0.372 ° C$ . The dissociation constant for the following reaction is ( $K_{f}$ for $H_{2} O = 1.86 K kg mo l^{- 1}$ )
$H S O_{4}^{-} ⇌ H^{+} + S O_{4}^{2 -}$

Question

The freezing point of a 0.08 molal aqueous solution of

N a H S O_{4}

is

- 0.372 ° C

. The dissociation constant for the following reaction is (

K_{f}

for

H_{2} O = 1.86 K kg mo l^{- 1}

)

H S O_{4}^{-} ⇌ H^{+} + S O_{4}^{2 -}

Aicharya · Accepted Answer

N a H S O 4 → N a + 0.08 + H S O 4 - 0.08 H S O 4 - 0.08 ( 1 - α ) ⇌ H + 0.08 α + S O 4 2 - 0.08 α i = 0.08 + 0.08 1 - α + 0.08 α + 0.08 α 0.08 = 2 + α Δ T = i × K f × m 0.372 = i × 1.86 × 0.08 i = 2.5 So 2 + α = 2.5 α = 0.5 Dissociation constant, K = 0.08 α × 0.08 α 0.08 1 - α = 0.08 × 0.5 × 0.08 × 0.5 0.08 × 0.5 = 0.04

The freezing point of a 0.08 molal aqueous solution of N a H S O 4 is - 0.372 ° C . The dissociation…

Solution