What is the value of K_{sp} for a saturated s | vedclass

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

Question

(D) For a saturated solution of $Ba(OH)_2$,the dissociation is $Ba(OH)_2 \rightleftharpoons Ba^{2+} + 2OH^-$. Given $pH = 12$,then $pOH = 14 - 12 = 2$

Vedclass · Accepted Answer

$5 	imes 10^{-7}$

Vedclass · Answer

(D) For a saturated solution of $Ba(OH)_2$,the dissociation is $Ba(OH)_2 ightleftharpoons Ba^{2+} + 2OH^-$. Given $pH = 12$,then $pOH = 14 - 12 = 2$. Therefore,$[OH^-] = 10^{-pOH} = 10^{-2} \ M$. From the stoichiometry,$[Ba^{2+}] = \frac{1}{2} [OH^-] = \frac{1}{2} 	imes 10^{-2} = 0.5 	imes 10^{-2} \ M$. The solubility product constant $K_{sp}$ is given by $K_{sp} = [Ba^{2+}][OH^-]^2$. Substituting the values: $K_{sp} = (0.5 	imes 10^{-2}) 	imes (10^{-2})^2 = 0.5 	imes 10^{-2} 	imes 10^{-4} = 0.5 	imes 10^{-6} = 5 	imes 10^{-7}$. Thus,the correct option is $D$.

What is the value of $K_{sp}$ for a saturated solution of $Ba(OH)_2$ having $pH = 12$?

Similar Questions

If the concentration of $OH^{-}$ ions in the reaction $Fe(OH)_{3(s)} \rightleftharpoons Fe^{3+}_{(aq)} + 3OH^{-}_{(aq)}$ is decreased by $1/4$ times,then the equilibrium concentration of $Fe^{3+}$ will increase by $......$ times.

The solubility of $CaF_2$ is $s$ moles/litre. Then its solubility product is ....

$A$ precipitate of $CaF_{2}$ $(K_{sp} = 1.7 \times 10^{-10})$ will be obtained when equal volumes of the following are mixed:

The solubility of $PbF_2$ (formula weight $= 245$) is $0.46 \ g/L$. What is the solubility product?

The solubility product of $AgCl$ at $25 \, ^\circ C$ is $5 \times 10^{-13}$. Its solubility is = .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module