What should be the $pH$ of solution to dissolve the ${\rm{Cr}}{\left( {{\rm{OH}}} \right)_3}{\rm{\;precipitate}}$ [Given, $\left[ {{\rm{C}}{{\rm{r}}^{3 + }}} \right] = 1.0\;{\rm{mol/L}},\;{K_{{\rm{sp}}}} = 6 \times {10^{ - 31}}$

JEE CHEMISTRY MCQ: Equilibrium

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Application +4 | -1

What should be the

pH

of solution to dissolve the

\mathrm{Cr}{\left( {\mathrm{OH}} \right)_3}\mathrm{\;precipitate}

[Given,

\left[ {\mathrm{C}{\mathrm{r}^{3 + }}} \right] = 1.0\;\mathrm{mol/L},\;{K_{\mathrm{sp}}} = 6 \times {10^{ - 31}}

View answer and explanation

Correct option: C

Option C (5) is correct because the pH must be greater than or equal to 4.253 to dissolve the precipitate, making pH 5 the appropriate choice.

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\mathrm{K}_1

\mathrm{K}_2

\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\quad\ldots

\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\quad\ldots

K{a_1},\,\,\,K{a_2}\,\,\,\mathrm{and}\,\,\,K{a_3}

\begin{array}{l} {H_2}S\leftrightharpoons{H^ + } + H{S^ - }\\ H{S^ - }\leftrightharpoons{H^ + } + {S^{2 - }}\\ {H_2}S\leftrightharpoons2{H^ + } + {S^{2 - }} \end{array}

K{a_1},\,\,K{a_2}\,,K{a_3}

\mathrm{pH}

0.1 \mathrm{~M}-\mathrm{N}_2 \mathrm{H}_4

\mathrm{N}_2 \mathrm{H}_4,

K_{{b}_1}=3.6 \times 10^{-6}, K_{{b}_2}=6.4 \times 10^{-12}, \log 2=0.3

\log 3=0.48

{\text{Ba}}{{\text{O}}_2}(s) \rightleftharpoons {\text{BaO}}(s) + {{\text{O}}_2}(g);\Delta H = - {\text{ve}}.

\text{O}_2