The solubility of a sparingly soluble salt AB | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Step $1$: Calculate molar solubility $(S)$ in $	ext{mol/dm}^3$ by dividing solubility in $	ext{g/dm}^3$ by molar mass.$S = \frac{18.78 	imes 10

Vedclass · Accepted Answer

$4 	imes 10^{-15}$

Vedclass · Answer

(B) Step $1$: Calculate molar solubility $(S)$ in $	ext{mol/dm}^3$ by dividing solubility in $	ext{g/dm}^3$ by molar mass.$S = \frac{18.78 	imes 10^{-4} 	ext{ g/dm}^3}{187.8 	ext{ g/mol}} = 10^{-5} 	ext{ mol/dm}^3$.Step $2$: The dissociation of $AB_2$ is $AB_2(s) ightleftharpoons A^{2+}(aq) + 2B^-(aq)$.Step $3$: The solubility product expression is $K_{sp} = [A^{2+}][B^-]^2 = (S)(2S)^2 = 4S^3$.Step $4$: Substitute $S = 10^{-5}$ into the expression: $K_{sp} = 4(10^{-5})^3 = 4 	imes 10^{-15}$.

The solubility of a sparingly soluble salt $AB_2$ is $18.78 \times 10^{-4} \text{ g/dm}^3$. What is its solubility product? (Molar mass of $AB_2 = 187.8 \text{ g mol}^{-1}$)

Similar Questions

$AgCl$ is soluble in:

The solubility of $BaSO_4$ in water is $2.33 \times 10^{-3} \ g/L$. Its solubility product is....... (Molar mass of $BaSO_4 = 233 \ g/mol$)

The solubility of $AgCl$ in $0.1 \text{ M } NaCl$ is $S \text{ mol/L}$. If the solubility product of $AgCl$ is $1.8 \times 10^{-10}$, then $S$ is approximately:

The solubility of $Ag_2C_2O_4$ is $2 \times 10^{-4} \ mol \ L^{-1}$ at $298 \ K$. What is its solubility product?

Calculate the equilibrium concentration of $Pb^{2+}$ ions in a solution of $PbS$ containing $1 \times 10^{-11} \ mol \ dm^{-3}$ of sulphide ions. (Given $K_{sp}$ for $PbS = 8.0 \times 10^{-28}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module