The solubility of a salt A_2B_3 is 10^{-3} M | vedclass

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

Question

(A) For a salt of the type $A_2B_3$,the dissociation is given by: $A_2B_3(s) \rightleftharpoons 2A^{3+}(aq) + 3B^{2-}(aq)$.If $S$ is the solubility,th

Vedclass · Accepted Answer

$1.08 	imes 10^{-13}$

Vedclass · Answer

(A) For a salt of the type $A_2B_3$,the dissociation is given by: $A_2B_3(s) ightleftharpoons 2A^{3+}(aq) + 3B^{2-}(aq)$.If $S$ is the solubility,then $[A^{3+}] = 2S$ and $[B^{2-}] = 3S$.The solubility product $K_{sp}$ is given by: $K_{sp} = [A^{3+}]^2 [B^{2-}]^3$.$K_{sp} = (2S)^2 	imes (3S)^3 = 4S^2 	imes 27S^3 = 108S^5$.Given $S = 10^{-3} \ M$,substituting the value:$K_{sp} = 108 	imes (10^{-3})^5 = 108 	imes 10^{-15} = 1.08 	imes 10^{-13}$.

The solubility of a salt $A_2B_3$ is $10^{-3} \ M$. What will be its solubility product?

Similar Questions

In which of the following pairs of insoluble chlorides,one dissolves in concentrated $HCl$ and the other is not soluble?

Find the solubility of $PbI_2$ if its solubility product is $7.0 \times 10^{-9}$.

The molar solubility of $PbI_2$ in $0.2 \ M \ Pb(NO_3)_2$ solution in terms of solubility product of $PbI_2$ is

Solubility product of $CaC_2O_4$ at a given temperature in pure water is $4 \times 10^{-9} \ (mol \ L^{-1})^2$. Solubility of $CaC_2O_4$ at the same temperature is

The $K_{sp}$ for bismuth sulphide $(Bi_{2}S_{3})$ is $1.08 \times 10^{-73}$. The solubility of $Bi_{2}S_{3}$ in $mol \ L^{-1}$ at $298 \ K$ is ......

Vedclass Test Series

Exam Paper Generator

Online Exam Module