Question

A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 16.0 mL of KOH.

Answer 1

moles HBr = 0.0500 L x 0.15 M=0.0075

moles KOH = 0.0160 L x 0.25 M=0.0040

moles H+ in excess = 0.0075 - 0.0040= 0.0035

total volume = 0.070 L

[H+]= 0.0035/0.070=0.050 M

pH =1.55