\(0.5\) molal aqueous solution of a weak acid \((HX)\) is \(20\%\) ionized. If \(K_1\) for water is \(1.86\, \ K\, \ kg\, \ mol^{-1}\) the lowering in freezing point of solution will be:
Answer & explanation
Correct answer: option 2
The correct answer is option 2. 1.12 K.
Ionization: We are given that the solution is \(20\%\) ionized. This translates to a degree of ionization \((\alpha )\) to 0.2.
van't Hoff Factor \((i)\): Since \(HX\) dissociates into \(H^+\) and \(X^-\) ions (\(1:1\) ratio), the van't Hoff factor accounts for these dissociated particles along with the undissociated molecule:
\(i = 1\) (undissociated molecule) \(+\, \ \alpha \) (fraction ionized) \(+\, \ 1\) (ions formed)
\(i = 1 + 0.2 + 1 = 1.2\)
Freezing Point Depression: The equation relates the change in freezing point \((\Delta T_f)\) to molality \((m)\), van't Hoff factor \((i)\), and the molal freezing point depression constant of the solvent \((K_f)\) for water \((1.86\, \ K\, \ kg/mol)\):
\(\Delta T_f = i × K_f × m\)
Calculation: Plugging in the values:
\(\Delta T_f = 1.2 × 1.86 × 0.5\)
or, \(\Delta T_f = 1.12 K\)
Therefore, the solution's freezing point will decrease by 1.12 K.