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

$A$ solution is saturated with $SrCO_3$ and $SrF_2$. The $[CO_3^{2-}]$ is found to be $1.2 \times 10^{-3} \ M$. The concentration of $F^{-}$ in the solution would be:
Given: $K_{sp}(SrCO_3) = 7.0 \times 10^{-10}$, $K_{sp}(SrF_2) = 7.9 \times 10^{-10}$

If the solubility product $(K_{sp})$ of $Ni(OH)_2$ is $1.9 \times 10^{-15}$, the molar solubility of $Ni(OH)_2$ in $1.0 \ M \ NaOH$ is:

Which of the following pairs of ions would be expected to form a precipitate when their solutions are mixed?

The compound not soluble in acetic acid is

The solubility product of $AgCl$ is $10^{-10} \, M^2$. The minimum volume (in $m^3$) of water required to dissolve $14.35 \, mg$ of $AgCl$ is approximately

Vedclass Test Series

Exam Paper Generator

Online Exam Module