Sample of water has hardness $$77.5\,ppm\,C{a^{2 + }}.$$    If this is passed through an ion exchange column where $$C{a^{2 + }}$$  is replaced by $${H^ + },$$ what is the $$pH$$  of water after it has been so treated?

Solution :
$$C{a^{2 + }} + 2RH \to {R_2}Ca + 2{H^ + }$$
$${10^6}\,mL\,\,{\text{of}}\,\,{H_2}O\,\,{\text{has}}\,\,{\text{77}}{\text{.5}}\,{\text{g}}\,C{a^{2 + }}$$       $$ = \frac{{77.5}}{{40}}\,mole\,\,C{a^{2 + }}$$
$$ \Rightarrow 2 \times \frac{{77.5}}{{40}}\,mole\,{H^ + } = $$      $$3.875\,mole\,{H^ + }$$
$$\therefore {10^3}\,mL\,\,{\text{of}}\,\,{H_2}O\,\,{\text{has}}\,\,{H^ + } = $$       $$3.875 \times {10^{ - 3}}\,M$$
$$\therefore pH = - \log \left[ {{H^ + }} \right] = 2.41$$