M S. A titration is performed on 50.0 ml. of 0.10 M benzoic acid (CallsCO) with
ID: 1010616 • Letter: M
Question
M S. A titration is performed on 50.0 ml. of 0.10 M benzoic acid (CallsCO) with 0.10 NaOH a. What is the pH at the starting point? b. What volume of NaOH solution is required to reach the equivalence point? c. What would be the plH after every 10.0 mL of NaOll were added up to and including the equivalence point? equivalence point. midpoint, and the equivalence point. d. What is the pH at the midpoint of the titration? e. Determine the pH after 10.0 additional mL, of base has been added beyond the f. Sketch the pH curve (or graph using Excel, etc). Label the buffer zone, theExplanation / Answer
a)
benzoic acid Ka = 6.3 x 10^-5
pKa = -log Ka = -log (6.3 x 10^-5)
pKa = 4.20
C = 0.1 M
initial pH = 1/2 [pKa - logC]
pH = 1/2 [4.20 -log 0.1]
pH = 2.60
b)
at equivalence point : millimoles of acid = millimoles of base
0.1 x 50 = 0.1 x V
V = 50 mL
volume of NaOH needed = 50 mL
c) after 10 mL of NaOH added
millimoles of base = 10 x 0.1 = 1
millimoles of acid = 50 x 0.1 = 5
C6H5COOH + NaOH -----------------> 6H5COONa + H2O
5 1 0 0 -----------------> initial
4 0 1 1 ------------------> equilibrium
pH = pKa + log [C6H5COONa / C6H5COOH]
pH = 4.20 + log (1/4)
pH = 3.60
after 20 mL NaOH added
millimoles of base = 20 x 0.1 = 2
millimoles of acid = 50 x 0.1 = 5
C6H5COOH + NaOH -----------------> 6H5COONa + H2O
5 2 0 0 -----------------> initial
3 0 2 2 ------------------> equilibrium
pH = pKa + log [C6H5COONa / C6H5COOH]
pH = 4.20 + log (2 / 3)
pH = 4.02
similarly solve for 30 mL , 40 mL , 50 mL
d) pH at mid point
at midpoint : pH = pKa
pH = 4.20
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