Question

The mean ionic activity coefficient for a 0.01 M aqueous solution of Ca$_3$(PO$_4$)$_2$ is .......... (Round off to three decimal places) (Given: \( \log_{10} \gamma_{\pm} = -0.509 Z_{\pm}^2 \sqrt{T} \))

Hint:

The activity coefficient depends on temperature and ionic strength. Be sure to account for the temperature and the charges of the ions when calculating.

Answer 1

Correct Answer: 0.063 - 0.069

Given: 0.01 M aqueous solution of Ca₃(PO₄)₂

Dissociation: $$\text{Ca}_3(\text{PO}_4)_2 \rightarrow 3\text{Ca}^{2+} + 2\text{PO}_4^{3-}$$

Ion concentrations:

• [Ca²⁺] = 3 × 0.01 = 0.03 M
• [PO₄³⁻] = 2 × 0.01 = 0.02 M

Ionic strength (I): $$I = \frac{1}{2}\sum c_i z_i^2$$ $$I = \frac{1}{2}[0.03 \times (2)^2 + 0.02 \times (3)^2]$$ $$I = \frac{1}{2}[0.03 \times 4 + 0.02 \times 9]$$ $$I = \frac{1}{2}[0.12 + 0.18]$$ $$I = \frac{1}{2}[0.30] = 0.15$$

Mean ionic activity coefficient using Debye-Hückel equation: $$\log_{10} \gamma_{\pm} = -0.509 \times z_+ |z_-| \sqrt{I}$$

For Ca₃(PO₄)₂: z₊ = 2, z₋ = -3, so |z₋| = 3

$$\log_{10} \gamma_{\pm} = -0.509 \times 2 \times 3 \times \sqrt{0.15}$$ $$\log_{10} \gamma_{\pm} = -3.054 \times 0.3873$$ $$\log_{10} \gamma_{\pm} = -1.183$$

$$\gamma_{\pm} = 10^{-1.183} = 0.0656$$

Answer: 0.066 (rounded to three decimal places)