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1. ⇒ (MHT CET 2021 20th September Evening Shift )

Solubility of $AgCl$ is $7.2 \times 10^{- 7} mol {dm}^{- 3}$ . What is it's solubility product?

3.6 \times 10^{- 13}

7.2 \times 10^{- 14}

2.59 \times 10^{- 14}

5.18 \times 10^{- 13}

Correct Option is (D)

$\begin{aligned} \begin{aligned} S = 7.2 \times 10^{- 7} mol {dm}^{- 3}, K_{sp} = ? \\ For AgCl, K_{sp} = S^{2} \\ = {(7.2 \times 10^{- 7})}^{2} = 5.18 \times 10^{- 13} \end{aligned} \end{aligned}$

2. ⇒ (MHT CET 2021 20th September Morning Shift )

The solubility of AgCl in it's solution is 1.25 $\times$ 10 $^{- 5}$ mol dm $^{- 3}$ . What is solubility product of AgCl?

1.56 \times 10^{- 10}

3.50 \times 10^{- 6}

1.10 \times 10^{- 5}

2.53 \times 10^{- 3}

Correct Option is (A)

${AgCl}_{(s)} ⇌ {Ag}_{(aq)}^{+} + {Cl}_{(aq)}^{-}$

Here, x = 1, y = 1

$∴$ Ksp = S $^{2}$ = (1.25 $\times$ 10 $^{- 5}$ ) $^{2}$ = 1.56 $\times$ 10 $^{- 10}$

⇒Ionic Equilibrium