The solubility of CdSO_{4} in water is 8.0 ti | vedclass

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

Question

(A) In pure water,the solubility product constant $K_{sp}$ is given by $K_{sp} = S^{2} = (8.0 	imes 10^{-4})^{2} = 64 	imes 10^{-8}$.In $0.01 \ M \

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(A) In pure water,the solubility product constant $K_{sp}$ is given by $K_{sp} = S^{2} = (8.0 	imes 10^{-4})^{2} = 64 	imes 10^{-8}$.In $0.01 \ M \ H_{2}SO_{4}$,the dissociation is $H_{2}SO_{4} ightarrow 2H^{+} + SO_{4}^{2-}$.The concentration of $SO_{4}^{2-}$ from $H_{2}SO_{4}$ is $0.01 \ M$.Let the solubility of $CdSO_{4}$ in this solution be $x \ mol \ L^{-1}$.$CdSO_{4} ightleftharpoons Cd^{2+} + SO_{4}^{2-}$.The equilibrium concentrations are $[Cd^{2+}] = x$ and $[SO_{4}^{2-}] = (x + 0.01) \ M$.Since $x \ll 0.01$,we can approximate $(x + 0.01) \approx 0.01$.$K_{sp} = [Cd^{2+}][SO_{4}^{2-}] = x(0.01) = 64 	imes 10^{-8}$.$x = \frac{64 	imes 10^{-8}}{0.01} = 64 	imes 10^{-6} \ mol \ L^{-1}$.Thus,the value is $64$.

Salt	$K_{sp}$ at $298 K$
$AgBr$	$5.0 \times 10^{-13}$
$Zn(OH)_2$	$1.0 \times 10^{-15}$
$Hg_2Cl_2$	$1.3 \times 10^{-18}$

The solubility of $CdSO_{4}$ in water is $8.0 \times 10^{-4} \ mol \ L^{-1}$. Its solubility in $0.01 \ M \ H_{2}SO_{4}$ solution is $..... \ \times 10^{-6} \ mol \ L^{-1}$.
(Round off to the Nearest integer) (Assume that solubility is much less than $0.01 \ M$)

Similar Questions

Calculate the molar solubility of $Ni(OH)_2$ in $0.10 \ M \ NaOH$. The solubility product of $Ni(OH)_2$ is $2 \times 10^{-15}$.

Explain the difference between ionic product and solubility product. Explain the condition for the precipitation of a sparingly soluble salt.

$A$ precipitate of calcium oxalate $(CaC_2O_4)$ will not dissolve in which of the following?

The correct order of solubility of the given salts in water at $298 K$ is:
Salt $K_{sp}$ at $298 K$
$AgBr$ $5.0 \times 10^{-13}$
$Zn(OH)_2$ $1.0 \times 10^{-15}$
$Hg_2Cl_2$ $1.3 \times 10^{-18}$

Solid $Ba(NO_3)_2$ is gradually dissolved in $1 \times 10^{-4} \ M \ Na_2CO_3$ solution. At what concentration of $Ba^{+2}$ will the precipitation of $BaCO_3$ start? (Given: $K_{sp}$ of $BaCO_3 = 5.1 \times 10^{-9}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The solubility of $CdSO_{4}$ in water is $8.0 \times 10^{-4} \ mol \ L^{-1}$. Its solubility in $0.01 \ M \ H_{2}SO_{4}$ solution is $..... \ \times 10^{-6} \ mol \ L^{-1}$.(Round off to the Nearest integer) (Assume that solubility is much less than $0.01 \ M$)