Question

200 ml of aqueous solution of HCl(pH=2) is mixed with 300 ml of an aqueous solution of NaOH(pH=12).what is the pH of the resulting solution?

Answer 1

Let's find the concentrations of each,

`[HCl] = 10^-2"M"`
`[NaOH] = 10^-2"M"`

The key here is understanding that strong acids and bases completely dissociate into their ions. Thus, for hydrochloric acid the anti-log will do, and for sodium hydroxide we'd need the anti-log of pOH,

`14 - 12 = 2 = "pOH"`

From here it's a basic stoichiometry problem,

`NaOH(aq) + HCl(aq) to NaCl(aq) + H_2O(l)`

We're left with a relatively neutral salt and some strong base left, which will dictate the new pH,

`(0.001"mol")/(0.500"L") = 0.002"M"`
`therefore "pH" = 14+log[OH^-] approx 11.30`

I'm open to feedback if I made a mistake!