Equilibrium - Result Question 81
####85. If $pH$ of a saturated solution of $Ba(OH)_2$ is 12 , the value of its $K _{(s p)}$ is :
[2010]
(a) $4.00 \times 10^{-6} M^{3}$
(b) $4.00 \times 10^{-7} M^{3}$
(c) $5.00 \times 10^{-6} M^{3}$
(d) $5.00 \times 10^{-7} M^{3}$
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Solution:
(d) $Ba(OH)_2(s) \longrightarrow Ba^{2+}(aq)+2 OH^{-}$(aq) $pH=12$ or $pOH=2$
$[OH^{-}]=10^{-2} M$
$Ba(OH)_2 \longrightarrow Ba^{2+}+2 OH^{-}$
$0.5 \times 10^{-2} \quad 10^{-2}$
$[\therefore.$ Concentration of $Ba^{2+}$ is half of $.OH^{-}]$
$ \begin{aligned} K _{s p} & =[Ba^{2+}][OH^{-}]^{2} \\ & =[0.5 \times 10^{-2}][1 \times 10^{-2}]^{2} \\ & =0.5 \times 10^{-6}=5 \times 10^{-7} M^{3} \end{aligned} $
