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6-2.Equilibrium-II (Ionic Equilibrium)
hard
Upto what $pH$ must a solution containing a precipitate of $Cr{(OH)_3}$ be adjusted so that all of precipitate dissolves (When $C{r^{3 + }} = 0.1\,mol/\,l,\,{K_{sp}} = 6 \times {10^{ - 31}})$
A
Upto $4.4$
B
Upto $4.1$
C
Upto $4.2$
D
Upto $4.0$
Solution
(d) ${K_{sp}} = [C{r^{3 + }}]\,{[O{H^ – }]^3}$
${[OH]^{ – 3}} = {K_{sp/C{r^{3 + }}}} = \frac{{6 \times {{10}^{31}}}}{{1 \times {{10}^{ – 1}}}} = 6 \times {10^{ – 30}}$
${[OH]^ – } = 1.8 \times {10^{ – 10}}$
$pOH = (\log 1.8 + \log {10^{10}})$$ = 10 + 0.25 + 1$$ = 11.25$
$pH = 14 – 11.25$$ = 2.27$
Standard 11
Chemistry
