A 25.0 mL sample of 0.150 M chloroacetic acid is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point? The of chloroacetic acid is 1.4 ×10-3.
The answer is: pH = 7.86. See solution below:
Let HA represent chloroacetic acid.
HA + NaOH => NaA + H2O
Volume of NaOH required = volume x molarity of HA/molarity of NaOH
= 25.0 x 0.150/0.150 = 25.0 mL
Initial molarity of A- = initial molarity of NaA
= moles of HA/total volume of solution
= (25/1000 x 0.150)/0.050 = 0.075 M
A-+H2O<=> HA + OH-
I0.0750 0
C-a+a +a
E0.075-aa a
Kb of A- = [HA][OH-]/[A-]
= a2/(0.075 - a) = Kw/Kb of HA
= 10-14/1.4 x 10-3 = 7.143 x 10-12
a2 + 7.143 x 10-12a - 5.357 x 10-13 = 0
The positive root of the quadratic equation is:
a = 7.319 x 10-7
[OH-] = a = 7.319 x 10-7 M
pOH = -log[OH-] = 6.14
pH = 14 - pOH = 7.86
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