Question

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.  

Answer 1

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