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 -, 0.04, 0.02, 0.01, 0.2

MARKS App · 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

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