The solubility of PbF_2 in water at 25,^{circ | vedclass

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

Question

(A) The solubility of $PbF_2$ in water is $S = 10^{-3}\, M$. The solubility product constant $K_{sp}$ is given by $K_{sp} = [Pb^{2+}][F^-]^2 = (S)(2S)

Vedclass · Accepted Answer

$1.6 	imes 10^{-6}\, M$

Vedclass · Answer

(A) The solubility of $PbF_2$ in water is $S = 10^{-3}\, M$. The solubility product constant $K_{sp}$ is given by $K_{sp} = [Pb^{2+}][F^-]^2 = (S)(2S)^2 = 4S^3$. $K_{sp} = 4 	imes (10^{-3})^3 = 4 	imes 10^{-9}$. In $0.05\, M\, NaF$ solution,$NaF$ dissociates completely to provide $[F^-] = 0.05\, M$. Let the solubility of $PbF_2$ in this solution be $S'\, M$. Then $[Pb^{2+}] = S'$ and $[F^-] = (2S' + 0.05)\, M$. $K_{sp} = S'(2S' + 0.05)^2 = 4 	imes 10^{-9}$. Since $S'$ is very small,we assume $2S' \ll 0.05$,so $(2S' + 0.05) \approx 0.05$. $S' 	imes (0.05)^2 = 4 	imes 10^{-9}$. $S' 	imes 2.5 	imes 10^{-3} = 4 	imes 10^{-9}$. $S' = \frac{4 	imes 10^{-9}}{2.5 	imes 10^{-3}} = 1.6 	imes 10^{-6}\, M$.

$S.No.$	$Formula$	$K_{sp}$
$1$	$PQ$	$4.0 \times 10^{-20}$
$2$	$PQ_2$	$3.2 \times 10^{-14}$
$3$	$PQ_3$	$2.7 \times 10^{-35}$

The solubility of $PbF_2$ in water at $25\,^{\circ}C$ is $\sim 10^{-3}\, M$. What is its solubility in $0.05\, M\, NaF$ solution? Assume the latter to be fully ionised.

Similar Questions

The solubility products of three sparingly soluble salts are given below. What is the correct decreasing order of their molar solubility?
$S.No.$ $Formula$ $K_{sp}$
$1$ $PQ$ $4.0 \times 10^{-20}$
$2$ $PQ_2$ $3.2 \times 10^{-14}$
$3$ $PQ_3$ $2.7 \times 10^{-35}$

Assign $A$,$B$,$C$,$D$ from the given type of reaction.
$HgS \downarrow + HNO_3 \text{ (Conc.)} \longrightarrow \text{No dissolution}$

Solubility of $CaF_2$ in a solution of $Ca(NO_3)_2$ will be represented by the concentration term?

The $K_{sp}$ of an electrolyte $AB$ is $1 \times 10^{-10}$. If $[A^{+}] = 10^{-5} \ M$,which concentration of $B^{-}$ will not result in the precipitation of $AB$?

The solubility product of $AgCl$ is $1.5625 \times 10^{-10}$ at $25\ ^oC$. Its solubility in grams per litre will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module