The solubility of BaCl_2 is 4 times 10^{-6} | vedclass

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

Question

(C) For $BaCl_2$,the dissociation is: $BaCl_2 ightleftharpoons Ba^{2+} + 2Cl^-$. Given solubility $s = 4 	imes 10^{-6} \ M$. $K_{sp} = [Ba^{2+}][Cl

Vedclass · Accepted Answer

$256 	imes 10^{-18}$

Vedclass · Answer

(C) For $BaCl_2$,the dissociation is: $BaCl_2 ightleftharpoons Ba^{2+} + 2Cl^-$. Given solubility $s = 4 	imes 10^{-6} \ M$. $K_{sp} = [Ba^{2+}][Cl^-]^2 = (s)(2s)^2 = 4s^3$. $K_{sp} = 4 	imes (4 	imes 10^{-6})^3 = 4 	imes 64 	imes 10^{-18} = 256 	imes 10^{-18}$. In the presence of $10^{-2} \ M \ Ba(OH)_2$,the common ion is $Ba^{2+}$. $[Ba^{2+}] = 10^{-2} \ M$ (from $Ba(OH)_2$) and $[Cl^-] = 2s'$. $K_{sp} = [Ba^{2+}][Cl^-]^2 = (10^{-2})(2s')^2 = 4 	imes 10^{-2} 	imes (s')^2$. $256 	imes 10^{-18} = 4 	imes 10^{-2} 	imes (s')^2$. $(s')^2 = 64 	imes 10^{-16}$. $s' = 8 	imes 10^{-8} \ M$.

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 $BaCl_2$ is $4 \times 10^{-6} \ M$. Calculate its $K_{sp}$. Also,find its new solubility in the presence of $10^{-2} \ M \ Ba(OH)_2$.

Similar Questions

For the equilibrium reaction: $Fe(OH)_3(s) \rightleftharpoons Fe^{3+}(aq) + 3OH^-(aq)$,if the concentration of $OH^-$ is reduced to $\frac{1}{4}$ of its initial value,by what factor will the concentration of $Fe^{3+}$ increase at equilibrium?

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

$A$ $2.901 \ L$ saturated solution is formed by $0.08 \ g$ of $CaF_2$ at $298 \ K$. Calculate the $K_{sp}$ of $CaF_2$. (Molar mass of $CaF_2$ is $78 \ g \ mol^{-1}$)

The solubility product of $AgI$ at $25\,\text{°C}$ is $1.0 \times 10^{-16}\,\text{mol}^2\,\text{L}^{-2}$. The solubility of $AgI$ in $10^{-4}\,\text{N}$ solution of $KI$ at $25\,\text{°C}$ is approximately (in $\text{mol L}^{-1}$)

The $pH$ of a saturated solution of $Ca(OH)_{2}$ is $9$. The solubility product $(K_{sp})$ of $Ca(OH)_{2}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module