Question

The solubility of electrolytes ${MX}_1,{MX}_2$ and ${MX}_{3}is1\times {10}^{-3}$ moles per litre. Hence their respective solubility products are:

• A. ${10}^{-6},4\times {10}^{-9},27\times {10}^{-12}$
• B. ${10}^{-9},4\times {10}^{-9},32\times {10}^{-12}$
• C. ${10}^{-9},8\times {10}^{-8},32\times {10}^{-12}$
• D. none of these

Answer 1

Answer: (A)

${10}^{-6},4\times {10}^{-9},27\times {10}^{-12}$

Given solubilities:

$M{X}_1=1\times {10}^{-3}$

$M{X}_2={10}^{-3}$

$M{X}_3={10}^{-3}$

$M{X}_1\rightleftharpoons M^{\oplus }+X^{⊝}$

$K_{sp}=s^2$

$\Rightarrow K_{sp}={10}^{-6}$

$M{X}_2\rightleftharpoons M^{2+}+2X^{⊝}$

$K_{sp}=4s^3=4\times {10}^{-9}$

$\Rightarrow K_{sp}=4\times {10}^{-9}$

$M{X}_3\rightleftharpoons M^{2+}+3S^{⊝}$

$K_{sp}=27s^4$ $s={10}^{-3}$

$\Rightarrow K_{sp}=27\times {10}^{-12}$