Assuming complete dissociation, calculate the `pH` of the following solutions,
a. `0.003 M HCl, b. 0.005 M NaOH`,
c. `0.002 M HBr, d. 0.002 M KOH`

Answer

a) `2.52` , b) `11.70` , c) `2.70` , d) `11.30`

Explanation

(i) `0.003 MHCl`:
`H_(2)O + HCl harr H_(3)O^(+) + Cl^(-)`
Since HCl is completelly ionized
`[H_(3)O^(+)] = [HCl]`
`rArr [H_(3)O^(+)] = 0.003`
Now, `pH = - "log"[H_(3)O^(+)] = -"log" (.003)`
`= 2.52`
Hence, the pH of the solution is 2.52.
(ii) 0.005MNaOH:
`NaOH_((aq)) harr Na_((aq))^(+) + HO_((aq))^(-)`
`[HO^(-)] = [NaOH]`
`rArr [HO^(-)] = .005`
`pOH = -"log"[HO^(-)] = -"log"(.005)`
`pOH = 2.30`
`:. pH = 14-2.30`
`= 11.70`
Hence, the pH of the solution is `11.70`
(ii) `0.002 HBr`:
`HBr + H_(1)O harr H_(3)O^(+) + Br^(-)`
`[H_(3)O^(+)] = [HBr]`
`rArr [H_(3)O^(+)] = .002`
`:. pH = -"log" [H_(3)O^(+)]`
`= - "log" (0.002)`
`= 2.69`
Hence the pH of the solution is `2.69`.
(iv) `0.002 M KOH`:
`KOH_((aq)) harr K_((aq))^(+) OH_((aq))^(-)`
`[OH^(-)] = [KOH]`
`rArr [OH^(-)] = .002`
Now, `pOH = -"log" [OH^(-)]`
`= 2.69`
`:. pH = 14 - 2.69`
`= 11.31`
Hence the pH of the solution `11.31`