Posted by Tracey on Friday, March 16, 2012 at 1:30pm.
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.
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