Question

Kp for the reaction

$CuSO_4 \cdot 3H_2O_{(s)} + 2H_2O_{(g)} \rightarrow CuSO_4 \cdot 5H_2O_{(s)}$ is $1.0 \times 10^4 atm^{-2}$ at certain temperature. What lowest relative humidity of air can be achieved using $CuSO_4.3H_2O$ as drying agent at that temperature? Aqueous tension at the given temperature = 16 torr

• A. 47.5%
• B. 23.75%
• C. 95%
• D. 50%

Answer 1

Answer: (A)

47.5%

Answer 2

The reaction is $CuSO_4 \cdot 3H_2O_{(s)} + 2H_2O_{(g)} \rightleftharpoons CuSO_4 \cdot 5H_2O_{(s)}$. The equilibrium constant $K_p$ for this reaction is given by $K_p = \frac{1}{P_{H_2O}^2}$, where $P_{H_2O}$ is the partial pressure of water vapour at equilibrium. Given $K_p = 1.0 \times 10^4 atm^{-2}$, we have: $1.0 \times 10^4 = \frac{1}{P_{H_2O}^2}$ $P_{H_2O}^2 = \frac{1}{1.0 \times 10^4} atm^2 = 1.0 \times 10^{-4} atm^2$ $P_{H_2O} = \sqrt{1.0 \times 10^{-4}} atm = 1.0 \times 10^{-2} atm$

Convert this partial pressure to torr: $P_{H_2O} = 1.0 \times 10^{-2} atm \times 760 \frac{\text{torr}}{\text{atm}} = 7.6 \text{ torr}$

This partial pressure of water vapour (7.6 torr) is the equilibrium pressure that the drying agent ($CuSO_4 \cdot 3H_2O$ in equilibrium with $CuSO_4 \cdot 5H_2O$) can maintain in the air. This is the lowest partial pressure of water vapour that can be achieved.

The aqueous tension at the given temperature is the saturated vapour pressure of water, which is 16 torr. Relative humidity ($\phi$) is calculated as: $\phi = \frac{\text{Partial pressure of water vapour}}{\text{Saturated vapour pressure of water}} \times 100\%$ $\phi = \frac{P_{H_2O}}{P_{H_2O, \text{saturation}}} \times 100\%$ $\phi = \frac{7.6 \text{ torr}}{16 \text{ torr}} \times 100\%$ $\phi = 0.475 \times 100\% = 47.5\%$

Thus, the lowest relative humidity that can be achieved using $CuSO_4 \cdot 3H_2O$ as a drying agent is 47.5%.