Calculate the pH of `10^(-8)` M HCl solution .

Explanation

If we use the relation, `pH=-log [H_3O^+]` , we get pH equl to 8. But this is not correct because an acidic solution cannot have pH greater than 7. It may be noted that in very dilute acidic solution, when `H^+` concentrations from acid and water are comparable,the concentration of `H^+` from water cannot be neglected. Therefore,
`[H^+]_"total"=[H^+]_"acid"+[H^+]_"water"`
Since HCl is a strong acid and is completely ionized
`[H^+]_"HCl"=1.0xx10^(-8)`
The concentration of `H^+` from ionization of water is equal to the `[OH^-]` from water ,
`[H^+]_(H_2O)=[OH^-]_(H_2O)=x` say
`[H^+]_"total"=1.0xx10^(-8)+x`
But `[H^+][OH^-]=1.0xx10^(-14)`
`(1.0xx10^(-8)+x)(x)=1.0xx10^(-14)`
`x^2+10^(-8)x-10^(-14)=0`
Solving for x, we get `x=9.5xx10^(-8)`
`therefore [H^+]=1.0xx10^(-8)+9.5xx10^(-8)=10.5xx10^(-8)`
`=1.05xx10^(-7)`
`pH=-log [H^+]=-log (1.05xx10^(-7))`=6.98