Question

Part A

A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 25.0 mL of HNO3.

Part B

A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 21.0 mL of NaOH.

Answer 1

(A) Initial moles of NH₃ = 75/1000 X 0.200 = 0.015 mol

Moles of HNO₃ added = 25/1000 X 0.500 =0.0125 mol

NH₃ + HNO₃ --------> NH₄⁺ +NO₃^-

Moles of NH₃ left =0.015 - 0.0125 = 0.0025 mol

Moles of NH₄⁺ =0.0025 mol

K_a(NH₄⁺) = K_w / K_b(NH₃)

=10^-14 / (1.8 X 10^-5) = 5.556 X 10^-10

Henderson- Hasselbalch equation;

pH = pK_a + log ([NH₃] / [NH₄⁺])

= - log K_a + log(moles of NH₃ / molesof NH₄⁺) since volume is the same for both

= - log( 5.556 X 10^-10) + log(0.0025 / 0.0125)

pH = 8.56

(B)

(A) Initial moles of NH₃ = 52/1000 X 0.35 = 0.0182 mol

Moles of HNO₃ added = 21/1000 X 0.400 =0.0084 mol

CH₃COOH + NaOH --------> CH₃COO^- + Na⁺

Moles of CH₃COOH left =0.0182 - 0.0084 = 0.0098 mol

Moles of CH₃COO^- =0.0084 mol

Henderson- Hasselbalch equation;

pH = pK_a + log ([CH₃COO^-] / [CH₃COOH])

= - log K_a + log(moles of CH₃COO^- / molesof CH₃COOH) since volume is the same for both

= - log( 1.8 X 10^-5) + log(0.0084 / 0.0098)

pH = 4.678

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