Question : pH of a saturated solution of Ca(OH)_{2} is 9. The solubility product (K_{sp}) of Ca(OH)_{2} is :

**(1) 0.5 x 10 ^{-15}**

(2) 0.25 x 10^{-10}

(3) 0.125 x 10^{-15}

(4) 0.5 x 10^{-10}

answer : from all above options the correct answer is “**(1) 0.5 x 10 ^{-15}**”

solution of question is given below –

We know pH + pOH = 14

Therefore , 9 + pOH = 14

**→** pOH = 5

**→** [OH^{–}] = 10^{-5}

The reaction involved is Ca(OH)_{2} ⇌ Ca^{2+ }+ 2OH^{–}

Also , [Ca^{2+}] = 10^{-5}/2

K_{sp} = [Ca^{2+}] [OH^{–}]

= (10^{-5}/2) (10^{-5})^{2} = 0.5 x 10^{-15}

that is why we have marked option (1) for correct answer.