Homework Answers
HBr + KOH --------> KBr + H2O
millimoles of HBr = 21.7 x 0.269 = 5.8373
1) initially only HBr present
HBr is strong acid so dissociates completely
[HBr] = [H+] = 0.269 M
pH = - log [H+] = - log [0.269]
pH = 0.570
2) millimoles of KOH addded = 13.6 x 0.215 = 2.924
millimoles of HBr left = 5.8373 - 2.924 = 2.9133
total volume = 21.7 + 13.6 = 35.3 mL
[HBr] = 2.9133 / 35.3 = 0.082 M
pH = - log [H+]
pH = - log [0.082]
pH = 1.09
3) as it is strong acid / strong base titration
at equivalence point
pH = 7.0
4) millimoles KOH added = 35.0 x 0.215 = 7.525
millimoles of KOH left = 7.525 - 5.8373 = 1.6877
total volume = 35.0 + 21.7 = 56.7 mL
[KOH] = 1.6877 / 56.7 = 0.0297 M
as KOH is strong base [KOH] = [OH-] = 0.0297 M
pOH = - log [OH-]
pOH = - log [0.0297]
pOH = 1.53
pH = 14 - 1.53
pH = 12.47
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