Question

What is the $P^H$ of a solution obtained by dissolving $0.0005$ mole of the strong electrolyte, calcium hydroxide, $Ca{\left(OH\right)}_2$ to form $100ml$ of a saturated solution (aqueous)?
($K_w=1.0\times {10}^{-14}$ ${mole}^2{litre}^{-2}$)

• A. $9.8$
• B. $11.7$
• C. $12.0$
• D. $3.0$

Answer 1

The correct option is C $12.0$

$\left[{OH}^{-}\right]$ = Normality of base

$=\frac{0.0005\times 2\times 1000}{100}$

$={10}^{-2}$

$p^{OH}={-log10}^{-2}=2$

$p^H=14-2=12$