At 25^{circ} C, the solubility product of a s | vedclass

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

Question

(B) For a salt of $MX_{2}$ type, the dissociation is given by: $MX_{2} \rightleftharpoons M^{2+} + 2X^-$. If $s$ is the solubility in $mol / L$, then

Vedclass · Accepted Answer

$2 	imes 10^{-3}$

Vedclass · Answer

(B) For a salt of $MX_{2}$ type, the dissociation is given by: $MX_{2} ightleftharpoons M^{2+} + 2X^-$. If $s$ is the solubility in $mol / L$, then $[M^{2+}] = s$ and $[X^-] = 2s$. The solubility product $K_{sp}$ is given by: $K_{sp} = [M^{2+}][X^-]^2 = (s)(2s)^2 = 4s^3$. Given $K_{sp} = 3.2 	imes 10^{-8}$. So, $4s^3 = 3.2 	imes 10^{-8}$. $s^3 = \frac{3.2 	imes 10^{-8}}{4} = 0.8 	imes 10^{-8} = 8 	imes 10^{-9}$. Taking the cube root, $s = \sqrt[3]{8 	imes 10^{-9}} = 2 	imes 10^{-3} \ mol / L$.

At $25^{\circ} C$, the solubility product of a salt of $MX_{2}$ type is $3.2 \times 10^{-8}$ in water. The solubility (in $mol / L$) of $MX_{2}$ in water at the same temperature will be:

Similar Questions

The solubility product $({K_{sp}})$ of $BaCO_3$ is $1.5 \times 10^{-9}$. At what concentration of $Ba^{2+}$ ions will precipitation begin when solid $Ba(NO_3)_2$ is added to a $10^{-4} \ M$ solution of $Na_2CO_3$?

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

Evaporation of $1 \ L$ of a saturated solution of $CaCO_3$ leaves a residue of $7.0 \ g$. What is the solubility product $(K_{sp})$ for $CaCO_3$?

The solubility of $Al(OH)_3$ in a buffer solution at $pH = 8.0$ is [$K_{sp}$ of $Al(OH)_3 = 2.7 \times 10^{-27}$]

Assign $A$,$B$,$C$,$D$ from the given type of reaction.
$CaSO_3 \downarrow + SO_2 + H_2O \longrightarrow Ca(HSO_3)_2$

Vedclass Test Series

Exam Paper Generator

Online Exam Module