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Chapter 16: Problem 139

(a) Assuming complete dissociation and no ionpair formation, calculate the freezing point of a \(0.50 \mathrm{~m}\) NaI solution. (b) What is the freezing point after the addition of sufficient \(\mathrm{HgI}_{2},\) an insoluble compound, to the solution to react with all the free \(\mathrm{I}^{-}\) ions in solution? Assume volume to remain constant.

Short Answer

Expert verified
The freezing points of the NaI solution are -1.86 °C before the addition of HgI2 and -0.93 °C after the addition of HgI2.

Step by step solution

01

Calculation of the new freezing point

First, we use the given molality and calculate the freezing point depression using the formula \(ΔT = i·K_f·m\), where i is the van't Hoff factor, Kf is the cryoscopic constant for water which is 1.86 ℃/molal, and m is the molality. Since the van't Hoff factor for NaI is 2, the freezing point depression becomes \(ΔT = 2·1.86 ℃/molal·0.50 molal = 1.86 ℃\). Now, the freezing point of the solution will be the difference between the normal freezing point of water and the depression. This means, the freezing point is \(0 ℃ - 1.86 ℃ = -1.86 ℃\).
02

Calculation after addition of HgI2

After HgI2 is added, it reacts with the I- ions to form a compound NaHgI3. This new compound doesn't dissociate in the solute, so the van't Hoff factor becomes 1. Hence, the depression in freezing point is now \(ΔT = 1·1.86 ℃/molal·0.50 molal = 0.93 ℃\). Hence, the new freezing point is \(0 ℃ - 0.93 ℃ = -0.93 ℃\)

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