JEE CHEMISTRY MCQ: Equilibrium

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Analyzing +4 | -1

What is the solubility product of

{Al}({OH})_3

in water? Given:

{Al}({OH})_4^{-}(\mathrm{aq}) \rightleftharpoons {Al}^{3+}(\mathrm{aq})+4 {OH}^{-}(\mathrm{aq}); K=1.3 \times 10^{-34}

{Al}({OH})_3(\mathrm{s})+{OH}^{-}(\mathrm{aq}) \rightleftharpoons {Al}({OH})_4^{-}(\mathrm{aq}) ; K=38.5

View answer and explanation

Correct option: B

5 \times 10^{-33}

The correct answer is B (

5 \times 10^{-33}

) because the solubility product is calculated by multiplying the equilibrium constants of the given reactions.

More Questions from Equilibrium

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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