The solubility product of PbCl_2 at 298 K is | vedclass

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

Question

(A) The dissociation of $PbCl_2$ is given by: $PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)$.Let the solubility be $s \ mol/L$.Then,$[Pb^{2+}]

Vedclass · Accepted Answer

$6.3 	imes 10^{-3}$

Vedclass · Answer

(A) The dissociation of $PbCl_2$ is given by: $PbCl_2(s) ightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)$.Let the solubility be $s \ mol/L$.Then,$[Pb^{2+}] = s$ and $[Cl^-] = 2s$.The solubility product expression is: $K_{sp} = [Pb^{2+}][Cl^-]^2$.Substituting the values: $K_{sp} = (s)(2s)^2 = 4s^3$.Given $K_{sp} = 1 	imes 10^{-6}$.$4s^3 = 1 	imes 10^{-6}$.$s^3 = \frac{1 	imes 10^{-6}}{4} = 0.25 	imes 10^{-6} = 250 	imes 10^{-9}$.$s = \sqrt[3]{250 	imes 10^{-9}} \approx 6.3 	imes 10^{-3} \ mol/L$.

The solubility product of $PbCl_2$ at $298 \ K$ is $1 \times 10^{-6}$. Its solubility in $mol/L$ is:

Similar Questions

Find out the solubility of $Ni(OH)_2$ in $0.1 \ M \ NaOH$.
Given that the solubility product constant $(K_{sp})$ of $Ni(OH)_2$ is $2 \times 10^{-15}$.

The solubility product of salt $B_2A$ is $3.2 \times 10^{-11}$ at $298 \ K$. What is the solubility of the salt at the same temperature?

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 $0.01 \ M$)

The solubility of $BaSO_{4}$ in water is $2.42 \times 10^{-3} \ g \ L^{-1}$ at $298 \ K$. The value of solubility product $(K_{sp})$ will be (Given molar mass of $BaSO_{4} = 233 \ g \ mol^{-1}$)

Which of the following provides the best explanation for the solubility of $MnS$ in dilute $HCl$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module