A titration is performed by adding 0.791 M KOH to 40 mL of 0.127 M HC 3 H 5 O 2
ID: 957751 • Letter: A
Question
A titration is performed by adding 0.791 M KOH to 40 mL of 0.127 M HC3H5O2.
a) Calculate the pH before addition of any KOH.
b) Calculate the pH after the addition of 1.28, 3.21 and 5.42 mL of the base.(Show your work in detail for one of the volumes.)
c) Calculate the volume of base needed to reach the equivalence point.
d) Calculate the pH at the equivalence point.
e) Calculate the pH after adding 5.00 mL of KOH past the equivalence point.
f) Plot pH (y axis) versus volume of KOH added (x axis) for each calculation above. Sketch the titration curve.
Explanation / Answer
a) pH before addition of any KOH.
pKa of HC3H5O2 = 4.89
pH = 1/2 [pKa - logC]
pH = 1/2 [4.89 - log0.127]
pH = 2.89
b) the pH after the addition of 1.28
millimoles of HC3H5O2 = 0.127 x 40 = 5.08
millimoles of KOH = 0.791 x 1.28 = 1.01
HC3H5O2 + KOH ---------------------->C3H5O2K + H2O
5.08 1.01 0 0
4.07 0.0 1.01
pH = pKa + log [C3H5O2K][HC3H5O2]
pH = 4.89 + log (1.01 / 4.07)
pH = 4.28
note : solve similary at remaining volume of KOH
c) Calculate the volume of base needed to reach the equivalence point.
at the equivalence point.
millimoles of acid = millimoles of base
5.08 = 0.791 x V
V = 6.42 mL
volume of base needed to reach equivalence point = 6.42 mL
d) Calculate the pH at the equivalence point.
here only salt present
salt concentration = C = 5.08 / (40 +6.42) = 0.109 M
pH = 7 + 1/2 [pKa + logC] = 7 + 1/2 [4.89 + log 0.109 ]
pH = 8.96
e) Calculate the pH after adding 5.00 mL of KOH past the equivalence point.
volume of KOH = 5 + 6.42 = 11.42 mL of base
millimoles of base = 11.42 x 0.791 = 9.03
base millimoles remaining after reaction = 9.03 - 5.08 = 3.95
base concentration = 3.95 / (11.42 + 40) = 0.0769 M
[OH-] = 0.0769 M
pOH = -log [OH-] = -log (0.0769 )
pOH = 1.11
pH + pOH = 14
pH = 12.89
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