Question

A solution prepared by dissolving 2.8 gm of lime. $CaO$ in enough water to make 1 00 L of lime water $\left(Ca\right(OH{)}_2(aq.))$. If solubility of $Ca(OH{)}_2$ in water is 1 48 gm. The $pH$ of the solution obtained will be:
[log 2 = 0.3, Atomic masses are $Ca=40,O=16,H=11$]

• A. 12.3
• B. 1.3
• C. 12.6
• D. 13

Answer 1

Answer: (D)

13

Moles of $CaO=\frac{2.8}{56}=0.05$ moles.

$CaO+H_{2}O\rightleftharpoons C{a}^{2+}+2O{H}^{-}$

$\therefore$ Moles of $O{H}^{-}=2\times$moles of $CaO=2\times 0.05=0.1$

$\left[O{H}^{-}\right]=\frac{0.1}{1}=0.1M$

$pOH=-\log \left[O{H}^{-}\right]=1$

$pH=14-pOH=13$