What is the solubility product of ${\rm{Ca}}{{\rm{F}}_2},$ if its saturated solution contains $0.017\;{\rm{g}}$ of ${\rm{Ca}}{{\rm{F}}_2},$ per litre?

JEE CHEMISTRY MCQ: Equilibrium

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Application +4 | -1

What is the solubility product of

\mathrm{Ca}{\mathrm{F}_2},

if its saturated solution contains

0.017\;\mathrm{g}

\mathrm{Ca}{\mathrm{F}_2},

per litre?

View answer and explanation

Correct option: B

4.14 \times {10^{ - 11}}

Option B (

4.14 \times {10^{ - 11}}

) is correct because it is calculated using the solubility product expression,

{K_{sp}} = 4{s^3}

, where s is the molar solubility of

\mathrm{Ca}{\mathrm{F}_2}

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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