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

1) 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 23.0 mL of HNO3.

2)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.

Solutions

Expert Solution

Part.1

Initial moles of NH₃ = 0.075 L x 0.200 M = 0.015 mol

Moles of HNO₃ added = 0.023 L x 0.500 M = 0.0115 mol

NH₃ + HNO₃ ----------> NH₄⁺ + NO₃^-

Moles of NH₃ left = 0.015 - 0.0115 = 0.0035 mol

Moles of NH₄⁺ = 0.0115 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 0.0035/0.0115

= -log(5.556 x 10^-10) + log 0.0035/0.0115

= 9.26 + log 0.3043

= 9.26 - 0.5167

pH = 8.74

Part.2

Initial moles of CH₃COOH = 0.052 L x 0.35 M = 0.0182 mol

Moles of NaOH added = 0.021 L x 0.400 M = 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 Ka + log([CH₃COO^-]/[CH₃COOH])

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

= 4.74 + log 0.8571

= 4.74 - 0.06697

pH = 4.67

queen_honey_blossom answered 1 year ago

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