A 25.00 mL aliquot of 0.100 F malonic acid (propanedioic acid) is titrated with
ID: 1056538 • Letter: A
Question
A 25.00 mL aliquot of 0.100 F malonic acid (propanedioic acid) is titrated with 0.100 NaOH.
a) the initial pH (before addition of any NaOH) is?
b)What is the pH at the first half-equivalence point (halfway to the first equivalence point)?
c)What is the pH at the first equivalence point?
d)What is the pH at the second half-equivalence (halfway between the first and second equivalences)?
e)What is the pH at the second equivalence point?
f)What is the pH after adding 75.0 mL of the NaOH to the titration?
Explanation / Answer
pKa1 of malonic acid = 2.83
pKa2 = 5.69
a)
H2M ------------------> H+ + HM-
0.1 0 0
0.1 - x x x
Ka1 = x^2 / 0.1 - x
1.48 x 10^-3 = x^2 / 0.1 - x
x = 0.0114
[H+] = 0.0114 M
pH = -log [H+] = -log (0.0114)
pH = 1.94
b)
at half - equivalence point :
pH = pKa1 = 2.83
pH = 2.83
c)
At equivalence point :
pH = pKa1 + pKa2 / 2
= 2.83 + 5.69 / 2
pH = 4.26
d)
At second half equivalnence point :
pH = pKa2
pH = 5.69
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