Question

1. What is the pH at the equivalence point for the titration of 50 mL of an aqueous solution of 0.10 M C6H5COOH (benzoic acid) with a 0.2 M NaOH solution at 25°C? (Ka for benzoic acid is 6.28 x 10−5

2. Which of the following sparingly soluble salts would be more soluble in acidic aqueous solutions at

25°C?

i. AgCl

ii. Mg(OH)2

iii. AgF

iv. AuI3

Answer 1

1)
ofr equivalence point,
moles of C6H5COOH = mol of NaOH
0.10 M * 50 ml = 0.2 M * V(NaOH)
V(NaOH) = 25 mL

at equivalence point, whole of C6H5COOH will be converted to C6H5COO-
mol of C6H5COONa formed = 0.10 M * 50 mL = 5 mmol

total volume = 50 mL + 25 mL = 75 mL

[C6H5COO-] = number of mol / volume
= 5 mmol / 75 mL
= 0.067 M

Kb of C6H5COO- = 10^-14 / (6.28*10^-5)
= 1.59*10^-10

C6H5COO- + H2O <---------> C6H5COOH + OH-
0.067 0 0 (initial)
0.067-x x x (at equilibrium)

Kb = x*x / (0.067-x)
since Kb is small, x will be small and it be ignored as compared to 0.067
1.59*10^-10 = x^2 / (0.067)
x = 3.27*10^-6 M

[OH-] = x = 3.27*10^-6 M
pOH = -log [OH-] = 5.5

pH = 14 - pOH
= 14 - 5.5
= 8.5

Answer: 8.5

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