Solubility of a binary sparingly soluble salt | vedclass

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

Question

(A) $1$. Convert solubility from $g \ dm^{-3}$ to $mol \ dm^{-3}$ (molarity,$S$): $S = \frac{1.12 	imes 10^{-4} \ g \ dm^{-3}}{112 \ g \ mol^{-1}} =

Vedclass · Accepted Answer

$1 	imes 10^{-12}$

Vedclass · Answer

(A) $1$. Convert solubility from $g \ dm^{-3}$ to $mol \ dm^{-3}$ (molarity,$S$): $S = \frac{1.12 	imes 10^{-4} \ g \ dm^{-3}}{112 \ g \ mol^{-1}} = 1.0 	imes 10^{-6} \ mol \ dm^{-3}$. $2$. For a binary salt $(AB)$,the dissociation is $AB ightleftharpoons A^+ + B^-$. $3$. The solubility product expression is $K_{sp} = [A^+][B^-] = S 	imes S = S^2$. $4$. Substitute the value of $S$: $K_{sp} = (1.0 	imes 10^{-6})^2 = 1.0 	imes 10^{-12} \ mol^2 \ dm^{-6}$.

Solubility of a binary sparingly soluble salt is $1.12 \times 10^{-4} \ g \ dm^{-3}$. Calculate its solubility product $(K_{sp})$ if the molar mass of the salt is $112 \ g \ mol^{-1}$.

Similar Questions

The solubility of sparingly soluble salt $AB_2$ is $1.0 \times 10^{-4} \ mol \ dm^{-3}$. What is its solubility product?

The solubility product of $PbCl_2$ at $20 \, ^oC$ is $1.5 \times 10^{-4}$. Calculate the solubility.

What is the molar solubility of $Fe(OH)_2$ $(K_{sp} = 8.0 \times 10^{-16})$ at $pH$ $13.0$ ?

The solubility product of $M(OH)_2$ is $10^{-14}$. What will be the concentration of $M^{+2}$ in a $0.1 \, M$ solution of $NH_4OH$,if $NH_4OH$ is $10\%$ ionized?

Assign $A$,$B$,$C$,$D$ from the given type of reaction.
$CaSO_3 \downarrow + SO_2 + H_2O \longrightarrow Ca(HSO_3)_2$

Vedclass Test Series

Exam Paper Generator

Online Exam Module