The $$pH$$ of a 10^-8 molar solution of $$HCl$$  in water is

Solution :
TIPS/Formulae :
(i) $$pH$$ of acid cannot be more than 7.
(ii) While calculating $$pH$$ in such case, consider contribution $${\left[ {{H^ + }} \right]}$$  from water also.
Molar conc. of $$HCl = {10^{ - 8}}.$$   (given)
∴ $$pH = 8.$$  But this cannot be possible as pH of an acidic solution can not be more than 7. So we have to consider $${\left[ {{H^ + }} \right]}$$  coming from $${{\text{H}}_2}O.$$
$$\eqalign{ & {\text{Total}}\,\left[ {{H^ + }} \right] = {\left[ {{H^ + }} \right]_{HCl}} + {\left[ {{H^ + }} \right]_{{H_2}O}} \cr & {\text{Ionisation of}}\,{H_2}O:\,{H_2}O \rightleftharpoons {H^ + } + O{H^ - } \cr & {K_w} = {10^{ - 14}} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right] \cr & {\text{Let }}x{\text{ be the cone}}{\text{. of}}\,\left[ {{H^ + }} \right]\,{\text{from}}\,{H_2}O \cr & {\text{or}}\,\left[ {{H^ + }} \right] = x = \left[ {O{H^ - }} \right]\,\,\,\,\,\,\,\,\,\left[ {\because \left[ {{H^ + }} \right] = {{\left[ {OH} \right]}^ - }{\text{in water}}} \right] \cr & \therefore \,\,{10^{ - 14}} = \left( {x + {{10}^{ - 8}}} \right)\left( x \right)\,{\text{or}}\,x = 9.5\, \times {10^{ - 8}}M \cr & [{\text{For quadratic equation }}x{\text{ = }}\frac{{ - b \pm \sqrt {4ac} }}{{2a}}] \cr & \therefore \,\,{\text{Total}}\,\left[ {{H^ + }} \right] = {10^{ - 8}} + 9.5 \times {10^{ - 8}} = 10.5 \times {10^{ - 8}}\,{\text{or}}\,{\text{pH}} \cr & = - {\text{log}}\left( {10.5 \times {{10}^{ - 8}}} \right) = 6.98 \cr & {\text{It is between 6 and 7}}{\text{.}} \cr} $$