What is the pH of , `(a) 5 xx 10^(-8) `M HCl and (b) `5 xx 10^(-10) ` M HCl .

Explanation

(a) Here if we consider contribution of `H^(+)` ions only from HCl , i.e., `[H^(+)] = 5 xx 10^(-8) , " then " pH gt 7`, which is not possible , because pH of acidic solution, no matter how dilute it is , cannot be `gt ` 7. Here we have to take into consideration contribution of `H^(+)` ion from water also.
Let contribution of `H^(+)` ion from water is x mol `L^(-1)` in presence of `5 xx 10^(-8)` M HCl , then :
`{:(H_(2)O ,hArr,H^(+),+,OH^(-)),("From water ",,x,,x),("From HCl",,5xx10^(-8),,),("At equilibrium",,(5 xx 10^(-8)+x),,):}`
`K_(w) = [H^(+)][OH^(-)] = (5 xx 10^(-8) + x) x = 10^(-14) rArr x = 0.78 xx 10^(-7)`
`[H^(+)]_("Total") = 5 xx 10^(-8) + 0.78 xx 10^(-7) = 1.28 xx 10^(-7)`
`pH = - log[H^(+)]_("Total") = - log(1.28 xx 10^(-7)) = 6.89`
HCl is so dilute its contribution to `[H^(+)]` is negligible therefore `pH = 7`.