[tex] \text{Given, Concentration of HCl}= 10 {}^{ - 6} M[/tex]
[tex] \text{After Dilution conc. of HCl =} \frac{10 {}^{ - 6} }{100} = {10}^{ - 8}M [/tex]
[tex] \therefore \text{pH = - log[H}^{ + } ][/tex]
[tex] \therefore \text{ pH }= - \log [{} {10}^{ - 8}] = 8[/tex]
[tex] \text{But} \: \text{This} \: \text{Is} \: \text{Not True Because} \\ \text{ An Acidic Solution Cannot have} \text{pH Greater than 7.} \\ \text{ In this case} [ {H}^{ + } ] \text{ Ions Of Water Molecule} \\ \: \text {Cannot Be Neglected.}[/tex]
[tex] {[H}^{ + } ] = {[H}^{ + } ] _{HCl} +{ [H + }^{ } ] _{H_2 O}[/tex]
[tex][H {}^{ + } ][OH {}^{ - } ] = 10 {}^{ - 14} \\ [H {}^{ + } ] = {10}^{ - 7} \\ [H {}^{ + } ] \text{total} = {10}^{ - 8} + {10}^{ - 7} \\ = {10}^{ - 8} (1 + 10) \\ = 11 \times {10}^{ - 8} [/tex]
[tex] \text{Now from, pH} = - \text{log}[H {}^{ + } ] \\ = \text{ - log}(11 \times {10}^{ - 8} ) \\ = - \text{ log}11 - \text{ log} {10}^{ - 8} \\ = 6.957[/tex]
