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

1 Answer
Jul 3, 2017

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