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

Explanation

`[HCl] = 10^(-8) M " " rArr [H^(+)] = 10^(-8) M`
`rArr pH = 8` which is absolutely wrong since the solution must be acidic as it is an acidic solution i.e., `pH lt 7`
The catch here is that since HCl has been added in water (which also dissociates ) so we need to consider `[H^(+)]` from dissociation of `H_(2)O` as well . In the earlier illustrations , we didn't consider this because `[H^(+)]` from acid comes out be to be much higher than that furnished by water but in this case we can not neglect the water contribution .
`{:(H_(2)O,hArr,H^(+),+,OH^(-)),(,,x,,x):}`
`rArr K_(w) = [H^(+)][OH^(-)] = 10^(-14)`
`[H^(+)]_("total") = [H^(+)]_("acid") + [H^(+)]_("water") = 10^(-8) + x`
`rArr K_(w) = (x + 10 ^(-8)) x = 10^(-14)`
` rArr x^(2) + 10^(-8) x - 10^(-14) = 0 `
Solve the above quadratic equation to get :
` x= 9.52 xx 10^(-8) M`
Now, `[H^(+)]_("total") = 10^(-8) + 9.52 xx 10^(-8) = 1.05 xx 10^(-7) M`
`rArr pH =- log_(10) [1.05 xx 10^(-7)] = 6.978 `