a) It is a triprotic acid. n = 3. At half equivalence point, pH = pKa1 = 3 Ka1 =
ID: 699759 • Letter: A
Question
a)
It is a triprotic acid. n = 3.
At half equivalence point,
pH = pKa1 = 3
Ka1 = 10-3
At one and a half equivalence point,
pH = pKa2 = 7
Ka2 = 10-7
At two and a half equivalence point,
pH = pKa3 = 9
Ka3 = 10-9
b)
Equilibrium concentration can be calculated from given fractions at required pH
Initial [H3A]0 = 0.002 M
Mole balance equation:
[H3A] + [H2A-] + [HA2-] + [A3-] = 0.002
pH
[H3A]
[H2A-]
[HA2-]
[A3-]
2
0.001
0.001
0
0
5
0.0
0.002
0
0
7
0
0.001
0.001
0
12
0
0
0
1
c)
Buffer capacity is maximum near pH = 7 when [acidic species] = [basic species]
At pH = 7 and pH = 9 buffer capacity is maximum.
pH
[H3A]
[H2A-]
[HA2-]
[A3-]
2
0.001
0.001
0
0
5
0.0
0.002
0
0
7
0
0.001
0.001
0
12
0
0
0
1
Explanation / Answer
The graph below is the fractional composition diagram for a polyprotic acid (HA). Given the initial concentration of the acid is 0.002 M, answer the following questions: (a) Find n. Identify and estimate the acid dissociation constants (b) Calculate the equilibrium concentrations of all acidic or basic species at pH 2, pH 5, pH 7, and at pH 12. (c) Identify two pH values where the buffer capacities of the acid are maximunm 0.8 0.4- 0.2 0 0 2 4 6 10 12 14
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