The solubility of a sparingly soluble salt A_ | vedclass

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

Question

(B) The molar solubility $S$ is given by $S = \frac{x}{M} \ mol/L$. The dissociation of the salt is $A_3B_2 \rightleftharpoons 3A^{+2} + 2B^{-3}$. The

Vedclass · Accepted Answer

$54 \frac{x^4}{M^4}$

Vedclass · Answer

(B) The molar solubility $S$ is given by $S = \frac{x}{M} \ mol/L$. The dissociation of the salt is $A_3B_2 ightleftharpoons 3A^{+2} + 2B^{-3}$. The concentration of $B^{-3}$ ions is $[B^{-3}] = 2S = 2(\frac{x}{M})$. The solubility product $K_{sp}$ is given by $K_{sp} = [A^{+2}]^3 [B^{-3}]^2 = (3S)^3 (2S)^2 = 27S^3 	imes 4S^2 = 108S^5$. Substituting $S = \frac{x}{M}$,we get $K_{sp} = 108(\frac{x}{M})^5$. The ratio $\frac{K_{sp}}{[B^{-3}]} = \frac{108S^5}{2S} = 54S^4$. Substituting $S = \frac{x}{M}$,the ratio is $54(\frac{x}{M})^4$.

The solubility of a sparingly soluble salt $A_3B_2$ (Molar mass $= M \ g/mol$) in water is $x \ g/L$. The ratio of the solubility product of the salt to the molar concentration of $B^{-3}$ ion is:

Similar Questions

$AgCl$ is soluble in:

$A$ saturated solution of $Ag_2SO_4$ is $2.5 \times 10^{-2} \ M$. The value of its solubility product is:

The solubility of $PbCl_2$ at $25 \ ^oC$ is $6.3 \times 10^{-3} \ mol/L$. Its solubility product at that temperature is:

What is the solubility product $(K_{sp})$ of calcium phosphate in pure water? $[S = \text{molar solubility}]$

Which of the following has the maximum solubility? (Given $K_{sp}$ values)

Vedclass Test Series

Exam Paper Generator

Online Exam Module