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. 

Answers (1)
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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