Part B A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the
ID: 1016329 • Letter: P
Question
Part B A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 20.0 mL of KOH. Express your answer numerically.
pH=
Part C A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×105) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 11.0 mL of HNO3. Express your answer numerically.
pH=
Part D A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×105) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 33.0 mL of NaOH. Express your answer numerically.
pH=
Explanation / Answer
Part B A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 20.0 mL of KOH. Express your answer numerically.
50 mL is 0.05L x 0.15 M = 0.0075 moles
20 mL is 0.02 L x 0.25 M = 0.005 moles
So we have 0.0025 moles of HBr in excess in 50 + 20 = 70 mL which is 0.0025/0.07 = 0.0357 M
pH = -log H+
pH = -log 0.0357
pH = 1.45
Part C A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×105) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 11.0 mL of HNO3. Express your answer numerically.
0.075L x 0.2 = 0.015 moles
0.011 x 0.5 = 0.0055 moles
So we will have a solution with 0.015-0.0055 = 0.0095 moles NH3 in 0.086 L = 0.0095/0.086 = 0.1104M
And 0.0055 moles NH4+ in 0.086 = 0.0055/0.086 = 0.0639 M
Ka of NH4 is Kw/Kb = 10-14/1.8×105 = 5.55 x 10-10
pKa = -log Ka = 9.25
pH = pKa + log base/acid
pH = 9.25 + log 0.1104/0.0639
pH = 9.25 + 0.237
pH = 9.487
Part D A 52.0-mL volume of 0.35 M CH3COOH (Ka=1.8×105) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 33.0 mL of NaOH. Express your answer numerically.
0.052L x 0.35 M = 0.0182 moles
0.033 x 0.4 M = 0.0132 moles
So CH3COO- is 0.0132 moles in 52 + 33 mL = 0.085L so concentration is 0.0132/0.085 = 0.155 M
CH3COOH = 0.0182-0.0132 = 0.005 in 0.085 L = 0.0588 M
pKa = -log Ka
pKa = -log 1.8 x 10-5 = 4.744
pH = pKa + log base/acid
pH = 4.744 + log 0.155/0.0588
pH = 4.744 + 0.421
pH = 5.165