Solubility product of a salt AB is 1 × {10^{ - 8}} in a solution in which concentration of A is {10

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6-2.Equilibrium-II (Ionic Equilibrium)

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Solubility product of a salt $AB $ is $1 \times {10^{ - 8}}$ in a solution in which concentration of $A$ is ${10^{ - 3}}\,M$. The salt will precipitate when the concentration of $B$ becomes more than

A

${10^{ - 4}}\,M$

B

${10^{ - 7}}\,M$

C

${10^{ - 6}}\,M$

D

${10^{ - 5}}\,M$

Solution

(d) $[B] = \frac{{{K_{sp}}AB}}{{[A]}} = \frac{{1 \times {{10}^{ – 8}}}}{{{{10}^{ – 3}}}} = 1 \times {10^{ – 5}}\,M$

Where ionic product $>$ ${K_{sp}},\,ppt$ formed

$\therefore $ $B$ should be more then ${10^{ – 5}}\,M$.

Standard 11

Chemistry

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