If 59.0 mL of 0.40 M HI is mixed with 100. mL of 0.10 M KOH what is the resulting pH?

Answer 1

HI + KOH ==> KI + H2O
59.0 mL x 0.40 M = 23.6 millimoles HI.
100 mL x 0.10M = 10 mmol KOH.
That's an excess of HI by 23.6-10 = 13.6 mmoles HI
M = mmoles/mL = 13.6mmol/159 mL = ?M and convert to pH.

Answer 2

1.07