If solubility product of AB _{2} is 3.20 × 10^{-11}, M ^{3} , then solubility of AB

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6-2.Equilibrium-II (Ionic Equilibrium)

medium

If the solubility product of $AB _{2}$ is $3.20 \times 10^{-11}\, M ^{3}$,

then the solubility of $AB _{2}$ in pure water is......... $\times 10^{-4}$ $mol$ $L ^{-1}$.

[Assuming that neither kind of ion reacts with water]

$8$

$4$

$2$

$10$

(JEE MAIN-2020)

Solution

$\begin{array}{*{20}{c}} {A{B_2}\left( S \right) \rightleftharpoons A_{\left( {aq.} \right)}^{ + 2} + 2B_{\left( {aq.} \right)}^ – \,:\,{K_{sp}}} \\ {\,\,\,\,\,\,\,\,\,\,S\,\,\,\,\,\,\,\,\,\,\,\,2s} \end{array}$

$K _{ SP }= S ^{1} \times(2 s )^{2}=4 s ^{3}$

$3.2 \times 10^{-11}=4 \times S ^{3}$

$S =2 \times 10^{-4} M / L$

Standard 11

Chemistry

Concentration of $H _2 SO _4$ and $Na _2 SO _4$ in a solution is $1 M$ and $1.8 \times 10^{-2} M$, respectively. Molar solubility of $PbSO _4$ in the same solution is $X \times 10^{- Y } M$ (expressed in scientific notation). The value of $Y$ is. . . .

[Given: Solubility product of $PbSO _4\left(K_{s p}\right)=1.6 \times 10^{-8}$. For $H _2 SO _4, K_{a l}$ is very large and $\left.K_{a 2}=1.2 \times 10^{-2}\right]$

hard

(IIT-2022)

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medium

(KVPY-2014)

A litre of solution is saturated with $AgCl$. To this solution if $1.0 \times {10^{ – 4}}$ mole of solid $NaCl$ is added, what will be the $[A{g^ + }]$, assuming no volume change

medium

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hard

(AIEEE-2012)

The solubility of $CaC{O_3}$ in water is $3.05 \times {10^{ – 4}}\,moles/litre.$ Its solubility product will be

medium

If the solubility product of $AB _{2}$ is $3.20 \times 10^{-11}\, M ^{3}$, then the solubility of $AB _{2}$ in pure water is......... $\times 10^{-4}$ $mol$ $L ^{-1}$. [Assuming that neither kind of ion reacts with water]

Solution

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If the solubility product of $AB _{2}$ is $3.20 \times 10^{-11}\, M ^{3}$,

then the solubility of $AB _{2}$ in pure water is......... $\times 10^{-4}$ $mol$ $L ^{-1}$.

[Assuming that neither kind of ion reacts with water]