Question

Calculate the pH of the following solutions:

0.3 g of Ca(OH)₂ dissolved in water to give 500 mL of solution.

Answer 1

Strength of Ca(OH)₂ = ${\displaystyle \frac{0.3}{500}\times 1000}$
= 0.6 gL^-1

Molar mass of Ca(OH)₂ = 40 + 32 + 2 = 74

Molarity of Ca(OH)₂ = ${\displaystyle \frac{0.6}{74}}$ = 8.11 x 10^-3M

1 mole of Ca(OH)₂ = ${\displaystyle \frac{0.6}{74}}$ = 8.11 x 10^-3 M

1 mole of Ca(OH)₂ on dissociation produces
two moles of OH^-1 ions.

${\displaystyle \therefore \left[{\text{OH}}^{-}\right]=2\times 8.11\times {10}^{-3}\text{M}}$

= ${\displaystyle 16.22x{10}^{-3}\text{M}}$

But ${\displaystyle \left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]={\text{k}}_{\text{w}}=1\times {10}^{-14}}$

${\displaystyle \therefore \left[{\text{H}}^{+}\right]=\frac{1\times {10}^{-14}}{{\text{OH}}^{-}}=\frac{1x{10}^{-14}}{16.22\times {10}^{-3}}}$

= ${\displaystyle 6.16\times {10}^{-3}}$M

Now pH = ${\displaystyle -\log \left[{\text{H}}^{+}\right]}$

= ${\displaystyle -\log [6.16\times {10}^{-13}]}$

= ${\displaystyle -[\log 6.16+{\log }^{-13}]}$

= ${\displaystyle -[0.7896-13]}$

${\displaystyle =12.2104}$