C)
Given:
M(HNO3) = 0.5 M
V(HNO3) = 15 mL
M(NH3) = 0.2 M
V(NH3) = 75 mL
mol(HNO3) = M(HNO3) * V(HNO3)
mol(HNO3) = 0.5 M * 15 mL = 7.5 mmol
mol(NH3) = M(NH3) * V(NH3)
mol(NH3) = 0.2 M * 75 mL = 15 mmol
We have:
mol(HNO3) = 7.5 mmol
mol(NH3) = 15 mmol
7.5 mmol of both will react
excess NH3 remaining = 7.5 mmol
Volume of Solution = 15 + 75 = 90 mL
[NH3] = 7.5 mmol/90 mL = 0.0833 M
[NH4+] = 7.5 mmol/90 mL = 0.0833 M
They form basic buffer
base is NH3
conjugate acid is NH4+
Kb = 1.8*10^-5
pKb = - log (Kb)
= - log(1.8*10^-5)
= 4.745
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 4.745+ log {8.333*10^-2/8.333*10^-2}
= 4.745
use:
PH = 14 - pOH
= 14 - 4.7447
= 9.2553
Answer: 9.26
D)
Given:
M(CH3COOH) = 0.35 M
V(CH3COOH) = 52 mL
M(NaOH) = 0.4 M
V(NaOH) = 19 mL
mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 0.35 M * 52 mL = 18.2 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.4 M * 19 mL = 7.6 mmol
We have:
mol(CH3COOH) = 18.2 mmol
mol(NaOH) = 7.6 mmol
7.6 mmol of both will react
excess CH3COOH remaining = 10.6 mmol
Volume of Solution = 52 + 19 = 71 mL
[CH3COOH] = 10.6 mmol/71 mL = 0.1493M
[CH3COO-] = 7.6/71 = 0.107M
They form acidic buffer
acid is CH3COOH
conjugate base is CH3COO-
Ka = 1.8*10^-5
pKa = - log (Ka)
= - log(1.8*10^-5)
= 4.745
use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.745+ log {0.107/0.1493}
= 4.6
Answer: 4.60
Part C A 75.0-mL volume of 0.200 M NH3 (K) = 1.8 x 10-5) is titrated...
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 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 25.0 mL of NaOH. Express your answer numerically. ► View Available Hint(s) O AQ 0 2 ? pH = 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 28.0 mL of HNO3. Express your answer numerically. ► View Available...
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.
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...
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...
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...
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.
? Part D A 30.0-mL volume of 0.50 M CH3COOH (Ka 1.8 x 10-5) was titrated with 0.50 M NaOH. Calculate the pH after addition of 30.0 mL of NaOH at 25 C Express the pH numerically View Available Hint(s) PH Submit Return to Assigment Provide Feedback
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.
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