10)
concentration of H2A = 0.100 M
H2A -----------> HA- + H+
0.100 0 0
0.100-x x x
Ka1 = x^2 / 0.100 - x
2.46 x 10^-4 = x^2 / 0.100 - x
x = 4.84 x 10^-3
[H+] = 4.84 x 10^-3 M
pH = 2.315
b)
mmoles of H2A = 25 x 0.100 = 2.5
mmoles of NaOH = 10 x 0.200 = 2.0
H2A + NaOH ---------> HA- + H2O
2.5 2.0 0 0
0.5 0 2.0
pH = pKa1 + log [HA- / H2A]
= 3.61 + log [2.0 / 0.5]
pH = 4.211
c)
mmoles of NaOH = 12.5 x 0.2 = 2.5
H2A + NaOH ---------> HA- + H2O
2.5 2.5 0 0
0 0 2.5
pH = pKa1 + pKa2 / 2
= 3.609 + 7.234 / 2
pH = 5.422
d)
mmoles of NaOH = 20 x 0.2 = 4.0
H2A + NaOH ---------> HA- + H2O
2.5 4.0 0 0
0 1.5 2.5
HA- + NaOH -----------> A2- + H2O
2.5 1.5 0 0
1.0 0 1.5
pH = pKa2 + log [A2- / HA-]
= 7.234 + log [1.5 / 1]
pH = 7.410
e)
mmoles of NaOH = 25 x 0.2 = 5
H2A + NaOH ---------> HA- + H2O
2.5 5.0 0 0
0 2.5 2.5
HA- + NaOH -----------> A2- + H2O
2.5 2.5 0 0
0 0 2.5
concentration of A2- = 2.5 / 25 + 25 = 0.05 M
A2- + H2O ---------> HA- + OH-
0.05 0 0
0.05-x x x
Kb1 = x^2 / 0.05 - x
1.715 x 10^-7 = x^2 / 0.05 - x
x = 9.26 x 10^5
[OH-] = 9.26 x 10^5
pOH = 4.03
pH = 9.966
f)
mmoles of NaOH = 40 x 0.2 = 8
pH = 12.664
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