(A) Initial moles of NH3 = 75/1000 x 0.200 = 0.015 mol
Moles of HNO3 added = 17/1000 x 0.500 = 0.0085 mol
NH3 + HNO3 => NH4+ + NO3-
Moles of NH3 left = 0.015 - 0.0085 = 0.0065 mol
Moles of NH4+ = 0.0085 mol
Ka(NH4+) = Kw/Kb(NH3)
= 10-14/1.8 x 10-5 = 5.556 x 10-10
Henderson-Hasselbalch equation:
pH = pKa + log([NH3]/[NH4+])
= -log Ka + log(moles of NH3/moles of NH4+) since volume is the same for both
= -log(5.556 x 10-10) + log(0.0065/0.0085)
= 9.14
(B) Initial moles of CH3COOH = 52/1000 x 0.35 = 0.0182 mol
Moles of NaOH added = 33/1000 x 0.400 = 0.0132 mol
CH3COOH + NaOH => CH3COO- + Na+
Moles of CH3COOH left = 0.0182 - 0.0132 = 0.005 mol
Moles of CH3COO- = 0.0132 mol
Henderson-Hasselbalch equation:
pH = pKa + log([CH3COO-]/[CH3COOH])
= -log Ka + log(moles of CH3COO-/moles of CH3COOH) since volume is the same for both
= -log(1.8 x 10-5) + log(0.0132/0.005)
= 5.17
A) A 75.0mL- volume of 0.200M NH3 (Kb= 1.8*10^-50 is titrated with 0.500HNO3. Calculate the pH after the addition of 1...
Titrations 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 19.0 mL of HNO3. Express your answer numerically. 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 23.0 mL of NaOH. Express your answer numerically.
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 11.0mL of HNO3.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 31.0mL of NaOH.
Part C A 75.0-mL volume of 0.200 M NH3 (Kb = 1.8 x 10-5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 15.0 mL of HNO3 . Express your answer numerically. Part D A 52.0-mL volume of 0.35 M CH3COOH (Ka = 1.8 x 10-5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 19.0 mL of NaOH. Express your answer numerically.
A: 48.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 24.0 mL of KOH at 25 ∘C B: Consider the titration of 50.0 mL of 0.20 M NH3 (Kb=1.8×10−5) with 0.20 M HNO3. Calculate the pH after addition of 50.0 mL of the titrant at 25 ∘C. C: A 30.0-mL volume of 0.50 M CH3COOH (Ka=1.8×10−5) was titrated with 0.50 M NaOH. Calculate the pH after addition of 30.0 mL of NaOH at 25...
1) A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 16.0 mL of KOH.Express your answer numerically. pH=_______ 2) A 75.0-mL volume of 0.200 M NH3 (Kb = 1.8 x10-5) is titrated with 0.500 M HNO3. Calculate the pH after the addition of 13.0 mL of HNO3.Express your answer numerically. pH=_______ 3) A 52.0-mL volume of 0.35 M CH3COOH (Ka = 1.8 x10-5 ) is titrated with 0.40 M NaOH. Calculate...
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 28.0 mL of HNO3. Express your answer numerically.
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 13.0 mL of H
A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 15.0 mL of KOH. Express your answer numerically. pH = SubmitHintsMy AnswersGive UpReview Part Part C 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 27.0 mL of HNO3. Express your answer numerically. pH = SubmitHintsMy AnswersGive UpReview Part Part D A 52.0-mL volume of 0.35 M CH3COOH...
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 27.0 mL of HNO3. Express your answer numerically. A 52.0-mL volume of 0.35 MM CH3COOH (Ka=1.8×10−5Ka=1.8×10−5) is titrated with 0.40 M NaOH. Calculate the pH after the addition of 15.0 mL of NaOH. and Imagine that you are in chemistry lab and need to make 1.00 LL of a solution with a pH of 2.80. You have in front...
A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 12.0 mL of KOH. Express your answer numerically. 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 13.0 mL of HNO3. Express your answer numerically. 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 33.0 mL...