The $$pH$$  of neutral water at $$25{\,^ \circ }C$$  is 7.0. As the temperature increases, ionisation of water increases, however, the concentration of $${H^ + }$$  ions and $$O{H^ - }$$  ions are equal. What will be the $$pH$$  of pure water at $$60{\,^ \circ }C?$$

Solution :
The $$pH$$  of neutral water at $$25{\,^ \circ }C$$  is 7.0.
$$\therefore \left[ {{H^ + }} \right] = \left[ {O{H^ - }} \right] = {10^{ - 7}}$$         $$\left( {\because pH = - \log \left[ {{H^ + }} \right]} \right)$$
$${\text{Now}},\,{K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]$$      $$ = {\left( {1 \times {{10}^{ - 7}}} \right)^2} = 1 \times {10^{ - 14}}$$
As the temperature increases, ionisation of water increases, thus $$\left[ {{H^ + }} \right]$$  and $$\left[ {O{H^ - }} \right]$$  increases equally. Now
$${K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] > 1 \times {10^{ - 14}}$$         $$\left( {\because \left[ {{H^ + }} \right] = \left[ {O{H^ - }} \right]} \right)$$
$$\eqalign{ & {\text{or}}\,\,{\left[ {{H^ + }} \right]^2} > 1 \times {10^{ - 14}} \cr & \therefore \left[ {{H^ + }} \right] > 1 \times {10^{ - 7}}\,\,{\text{and}}\,\,pH < 7 \cr} $$