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

Explanation

Note. At the first instance, it appears that as `[H^(+)]=10^(-8)`, therefore, pH should be 8. But pH cannot be 8 but should be less than 7 because the solution is acidic. The reason is that from `H_(2)O, [H^(+)]=106(-7)` M which cannot be neglected in comparison to `10^(-8)`M.
The pH can be calculated as follows :
From, acid, `[H^(+)]=[OH^(-)]=x "mol" L^(-1) ` (say) (as in presence of acid, `[H^(+)]` from `H_(2)O ~=10^(-7)M`)
Thus, in the solution, now we have `[H^(+)]=(10^(-8)+x)M and [OH^(-)]=x M`
But `[H^(+)] [OH^(-)]=K_(w) = 10^(-14)`
`:. (10^(-8) +x)(x) = 10^(-14) or x^(2) + 10^(-8) x - 10^(-14) = 0`
`x=(-10^(-8)pmsqrt(10^(-16)+4xx10^(-14)))/(2) = (-10^(-8)+sqrt(401)xx10^(-8))/(2)` (Taking +ve value only)
`=(19.025xx10^(-8))/(2) = 9.5125xx10^(-8)`
i.e, `[OH^(-)]=9.5125xx10^(-8)M`
`:. pOH = -log (9.5125xx10^(-8))=8-0.9783~=7.02 " " :. "" pH=14-7.02=6.98`
Alternatively, in a simplified manner, we have
From acid, `[H^(+)]=10^(-8)M, "From" H_(2)O , [H^(+)]=10^(-7)M`
`:. "Total" [H^(+)]=10^(-8)+10^(-7) = 10^(-8) (1+10)=11xx10^(-8)M`
`:. pH = -log [H^(+)]=-log(11xx10^(-8))=-[log 11 + log 10^(-8)]`
`=-[1.0414-8]=6.9586~~ 6.96`