# 7.49     Calculate the pH of the following solutions: (b)      0.3 g of $Ca(OH)_2$ dissolved in water to give 500 mL of solution.

M manish

The calcium hydroxide ion dissociates into-
$Ca(OH)_{2}\rightarrow Ca^{2+}+2OH^-$
Molecular weight of $Ca(OH)_{2}$ = 74
the concentration of $[Ca(OH)_{2}]$ = $\frac{0.3\times 1000}{74\times500} = 0.0081 M$

$\therefore [OH^-]= [Ca(OH)_2] = 0.0081 M$

We know that,
$[H^+] = \frac{K_w}{[OH^-]} = \frac{10^{-14}}{0.0162}$
= $61.7 \times 10^{-14}$

Thus $P^H = - \log [H^+]= -\log (61.7 \times 10^{-14})$
= 14 - 1.79  = 12.21

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