Question

1. What is the pH at the equivalence point for the titration of 50 mL of...

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

Homework Answers

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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