Question

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 titrated with 0.400 mol L−1 NaOH. Calculate the pH after the addition of 17.0 mL of NaOH. Express your answer numerically.

Answer 1

B)
Given:
M(HBr) = 0.15 M
V(HBr) = 50 mL
M(KOH) = 0.25 M
V(KOH) = 14 mL


mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.15 M * 50 mL = 7.5 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.25 M * 14 mL = 3.5 mmol


We have:
mol(HBr) = 7.5 mmol
mol(KOH) = 3.5 mmol
3.5 mmol of both will react
remaining mol of HBr = 4 mmol
Total volume = 64.0 mL

[H+]= mol of acid remaining / volume
[H+] = 4 mmol/64.0 mL
= 6.25*10^-2 M


use:
pH = -log [H+]
= -log (6.25*10^-2)
= 1.2041
Answer: 1.20

C)
Given:
M(HNO3) = 0.5 M
V(HNO3) = 28 mL
M(NH3) = 0.2 M
V(NH3) = 75 mL


mol(HNO3) = M(HNO3) * V(HNO3)
mol(HNO3) = 0.5 M * 28 mL = 14 mmol

mol(NH3) = M(NH3) * V(NH3)
mol(NH3) = 0.2 M * 75 mL = 15 mmol



We have:
mol(HNO3) = 14 mmol
mol(NH3) = 15 mmol

14 mmol of both will react
excess NH3 remaining = 1 mmol
Volume of Solution = 28 + 75 = 103 mL
[NH3] = 1 mmol/103 mL = 0.0097 M
[NH4+] = 14 mmol/103 mL = 0.1359 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 {0.1359/9.709*10^-3}
= 5.891

use:
PH = 14 - pOH
= 14 - 5.8909
= 8.1091

Answer: 8.11

D)
Given:
M(CH3COOH) = 0.35 M
V(CH3COOH) = 52 mL
M(NaOH) = 0.4 M
V(NaOH) = 17 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 * 17 mL = 6.8 mmol


We have:
mol(CH3COOH) = 18.2 mmol
mol(NaOH) = 6.8 mmol

6.8 mmol of both will react

excess CH3COOH remaining = 11.4 mmol
Volume of Solution = 52 + 17 = 69 mL
[CH3COOH] = 11.4 mmol/69 mL = 0.1652M

[CH3COO-] = 6.8/69 = 0.0986M

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 {9.855*10^-2/0.1652}
= 4.52


Answer: 4.52