The solubility of Ag2CrO4 is 25 mg/lit. Calculate solubility product, if molecular weight in 331.7 g/mol.

Question

GPT-4o · Answer

To calculate the solubility product ($ K_{sp} $) of silver chromate (Ag$_2$CrO$_4$), we need to perform the following steps: 1. **Convert the solubility from mg/L to moles/L:** Given the solubility of Ag$_2$CrO$_4$ is 25 mg/L and the molecular weight is 331.7 g/mol, we first convert the solubility to molar concentration. $$ \text{Solubility (in moles/L)} = \frac{\text{Solubility (in mg/L)}}{\text{Molecular Weight (in g/mol)}} \times \frac{1 \text{ g}}{1000 \text{ mg}} $$ Substituting the values, $$ \text{Solubility} = \frac{25 \text{ mg/L}}{331.7 \text{ g/mol}} \times \frac{1 \text{ g}}{1000 \text{ mg}} $$ $$ \text{Solubility} = \frac{25}{331.7 \times 1000} \text{ mol/L} $$ $$ \text{Solubility} = 7.54 \times 10^{-5} \text{ mol/L} $$ 2. **Determine the concentrations of Ag$^+$ and CrO$_4^{2-}$ ions:** When Ag$_2$CrO$_4$ dissolves in water, it dissociates according to the following equation: $$ \text{Ag}_2\text{CrO}_4 \, (s) \rightarrow 2\text{Ag}^+ \, (aq) + \text{CrO}_4^{2-} \, (aq) $$ If the solubility of Ag$_2$CrO$_4$ is $ s $ mol/L (which we calculated above as $ 7.54 \times 10^{-5} $ mol/L), then the concentrations of the ions at equilibrium will be: $$ [ \text{CrO}_4^{2-} ] = s = 7.54 \times 10^{-5} \text{ mol/L} $$ $$ [ \text{Ag}^+ ] = 2s = 2 \times 7.54 \times 10^{-5} \text{ mol/L} = 1.508 \times 10^{-4} \text{ mol/L} $$ 3. **Calculate the solubility product ($ K_{sp} $):** The solubility product expression for Ag$_2$CrO$_4$ is: $$ K_{sp} = [ \text{Ag}^+ ]^2 [ \text{CrO}_4^{2-} ] $$ Substituting the equilibrium concentrations, $$ K_{sp} = (1.508 \times 10^{-4} \text{ mol/L})^2 \times (7.54 \times 10^{-5} \text{ mol/L}) $$ $$ K_{sp} = (2.273 \times 10^{-8}) \times 7.54 \times 10^{-5} $$ $$ K_{sp} = 1.714342 \times 10^{-12} $$ $$ K_{sp} \approx 1.71 \times 10^{-12} $$ Thus, the solubility product ($ K_{sp} $) of Ag$_2$CrO$_4$ is approximately $ 1.71 \times 10^{-12} $.