Question

What should be the pH after 33 ml of 0.1 M NaOH was added to 40 ml of 0.05 M HCl?

Answer 1

12.25

Explanation:

`33ml(0.10M NaOH) + 40ml(0.05M HCl)` => 1:1 Rxn Ratio

=> `0.033(0.10)"mole" NaOH + 0.040(0.05)"mole" HCl`

=> `0.0033 "mole" NaOH + 0.0020 "mole" HCl`

=>`(0.0033 - 0.0020) "mole". "of". NaOH` in excess

=> `0.0013` mole `NaOH` remains in excess in 73 ml of solution.

=> `[NaOH]_(excess)` = `[`OH^-]`= ((0.0013"mole"OH^-)/(0.073LiterSolution))`

= `0.017 M (OH^-)`

=> `pOH = -log[OH^-] = -log(0.017) = -(-1.75) = 1.75`

=> `pH = 14 - pOH = 14 - 1.75 = 12.25`