Calculate the `pH` of the following half cells solutions and also the molarity of acids: `{: (Pt_(H_(2))),(atm):}|{:H_(2)SO_(4):} " " E = 0.05`
(b) `{: (Pt_(H_(2))),(atm):}|{:HCl:} " " E = 0.1 V`

Explanation

(a) `H_(2) rarr 2H^(+)+2e`
`E_(OP_(H//H^+)) = E_(OP_(H//H^+))^(@)-(0.059)/(2)"log"([H^(+)]^(2))/P_(H_(2))`
`0.05 = 0-(0.059)/(2)"log"([H^(+)]^(2))/P_(H_(2))`
`:. -log[H^(+)] = 0.8475`
`:. pH = 0.8475`
`:. [H^(+)] = 0.1420M`
`:.` Molarity of `H_(2)SO_(4) = 1//2[H^(+)]`
`:. = 1//2 xx 0.1420 = 0.071 M`
(b) Solve accordingly : `pH = 1.6949`
`[H^(+)] = 0.020 M`
`:.` Molarity of `HCl = [H^(+)] = 0.020 M`
No change in calculation if any strong acid producing `H^(+)` is given.