Considering acetic acid dissociates in water, its dissociation constant is . If 5 mL of acetic acid is dissolved in 1 litre water, the solution will freeze at °C, provided pure water freezes at 0 °C.

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Solution

Q.
Considering acetic acid dissociates in water, its dissociation constant is $6.25 \times 10^{- 5}$ . If 5 mL of acetic acid is dissolved in 1 litre water, the solution will freeze at $- x \times 10^{- 2}$ °C, provided pure water freezes at 0 °C. $x$ = ________ .(Nearest integer)

Given:    ${(K_{f})}_{water}$ = 1.86 K kg ${mol}^{- 1}$ .

density of acetic acid is 1.2 g ${mol}^{- 1}$ .

molar mass of water = 18 g ${mol}^{- 1}$ .

molar mass of acetic acid = 60 g ${mol}^{- 1}$ .

density of water = 1 g ${cm}^{- 3}$

  Acetic acid dissociates as

   ${CH}_{3} COOH ⇌ {CH}_{3} {COO}^{⊝} + H^{\oplus}$ [2024]

Ans.
(19)

$W_{{CH}_{3} COOH} = density \times volume = 1.2 \frac{g}{mL} \times 5 mL = 6 g$

$W_{H_{2} O} = density \times volume = 1 \frac{g}{mL} \times 1000 mL = 1000 g$

   $m = \frac{W_{{CH}_{3} COOH} \times 1000}{M_{{CH}_{3} COOH} \times W_{water} (in g)} = \frac{6 \times 1000}{60 \times 1000 (in g)} m = 0.1 m$

${CH}_{3} COOH ⇌ {CH}_{3} {COO}^{-} + H^{+}$

   $K_{a} = \frac{C α^{2}}{1 - α} \approx C α^{2}$

   $α = \sqrt{\frac{K_{a}}{C}} = \sqrt{\frac{6.25 \times 10^{- 5}}{0.1}} = 0.025$

   $i = 1 + (n - 1) α = 1 + (2 - 1) \times 0.025 = 1.025$

$Δ T_{f} = i m k_{f} = 1.025 \times 0.1 \times {1.86}^{\circ} C = {0.19}^{\circ} C$

$T_{f} ° - T_{f} = {0.19}^{\circ} C$

$0 - T_{f} = {0.19}^{\circ} C$

$T_{f} = - {0.19}^{\circ} C = 19 \times {10^{- 2}}^{\circ} C$

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