1. Number of moles of HBr = Molarity x Volume (L)
= 0.15 M x 0.05 L = 0.0075 mol
Number of moles of KOH = 0.25 M x 0.014 L = 0.0035 mol
Number of moles of HBr unreacted = 0.0075 - 0.0035 = 0.004 mol
Total volume of solution = 50 + 14 = 64 mL = 0.064 L
Since HBr is a strong acid, 0.004 mol HBr will dissociate into 0.004 mol H+
Hence [H+] = Number of moles / Volume (L)
= 0.004 mol / 0.064 L
= 0.0625 M
Hence pH = - log [H+]
pH = - log (0.0625)
pH = 1.2
2. Number of moles of NH3= Molarity x Volume (L)
= 0.2 M x 0.075 L
= 0.015 mol
Number of moles of HNO3 = 0.5 M x 0.017 L
= 0.0085 mol
HNO3 reacts with NH3 to produce NH4+
Hence mol of NH3 remaining = 0.015 - 0.0085 = 0.0065 mol
Since HNO3 is the limiting reagent, the number of moles of NH4+ is equal to the number of moles of HNO3
Mol of NH4+ produced = 0.0085 mol
Given, Kb of NH3= 1.8 x 10-5
Since Ka x Kb = Kw = 10-14
Hence Ka = Kw / Kb = (10-14) / (1.8 x 10-5)
Ka of NH4+ ds= 5.56 x 10-10
pKa = - log Ka
pKa = 9.25
Using Henderson-Hasselbalch equation we get
pH = pKa + log [NH3] / [NH4+]
Putting the values we get,
pH = 9.25 + log (0.0065) / (0.0085)
pH = 9.14
3. Number of moles of CH3COOH = Molarity x Volume
= 0.35 M x 0.052 L
= 0.0182 mol
Number of moles of NaOH = 0.40 M x 0.033 L
= 0.0132 mol
CH3COOH reacts with NaOH to prodcue CH3COO-
Number of moles of CH3COOH remaining = 0.0182 - 0.0132 = 0.005 mol
Since NaOH is the limiting reagent, hence no. of moles of NaOH = No. of moles of CH3COO-
No. of moles of CH3COO- = 0.0132 mol
Given Ka = 1.8 x 10-5
pKa = - log Ka
pKa = 4.745
Using Henderson-Hasselbalch equation we get
pH = pKa + log [CH3COO-] / [CH3COOH]
Putting the values we get,
pH = 4.745 + log (0.0132) / (0.005)
pH = 5.17
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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...
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 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...
Part B: A 50.0 mL volume of 0.15 mol L−1 HBr is titrated with 0.25 mol L−1 KOH. Calculate the pH after the addition of 14.0 mL of KOH. Express your answer numerically. Part C: A 75.0 mL volume of 0.200 mol L−1 NH3 (Kb=1.8×10−5) is titrated with 0.500 mol L−1 HNO3. Calculate the pH after the addition of 28.0 mL of HNO3. Express your answer numerically. Part D: A 52.0 mL volume of 0.350 mol L−1 CH3COOH (Ka=1.8×10−5) is...
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.
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.
6.)Part B A 64.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 32.0 mL of KOH. Express the pH numerically. 7.) Part C 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. Express the pH numerically. 8.) Part D A 30.0-mL volume of 0.50 M CH3COOH (Ka=1.8×10−5) was titrated with 0.50 M NaOH....
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 15.0 mL of NaOH . Express your answer numerically.
Part B A 72.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 36.0 mL of KOH at 25 ∘C. Express the pH numerically. Part C 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. Express the pH numerically. Part D A 30.0-mL volume of 0.50 M CH3COOH (Ka=1.8×10−5) was titrated with 0.50 M NaOH. Calculate the...
Part B A 96.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 48.0 mL of KOH at 25 ∘C. Express the pH numerically. Part C 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. Express the pH numerically. Part D A 30.0-mL volume of 0.50 M CH3COOH (Ka=1.8×10−5) was titrated with 0.50 M NaOH. Calculate...