Question

At $298\, K$, the solubility of silver chloride in water is $1434 \times 10^{-3} \,g\, L ^{-1}$. The value of $-\log K _{sp}$ for silver chloride is ____.
(Given: Mass of $Ag$ is $1079\, g\, mol ^{-1}$and mass of $Cl$ is $355 \,g \,mol ^{-1}$ )

Hint:

The solubility product constant \( K_{\text{sp}} \) can be calculated from the square of the molar solubility for simple salts like AgCl.

Answer 1

Correct Answer: 10

The dissociation of silver chloride in water is:

\[ \text{AgCl(s)} \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq) \]

The solubility \( S \) of AgCl is given by:

\[ S = 1.434 \times 10^{-3} \, \text{g L}^{-1} \]

The molar solubility of AgCl, \( S \), can be calculated as:

\[ S = \frac{1.434 \times 10^{-3}}{143.4 \times 10^{-3}} \, \text{mol L}^{-1} = 1 \times 10^{-5} \, \text{mol L}^{-1} \]

The solubility product \( K_{\text{sp}} \) is:

\[ K_{\text{sp}} = S^2 = (1 \times 10^{-5})^2 = 10^{-10} \]

Thus, \( -\log K_{\text{sp}} = 10 \).