Question

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 28.0 mL of HNO3 .

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 33.0 mL of NaOH .

Please Explain Steps

Answer 1

A)

Given:

M(HNO3) = 0.500 M

V(HNO3) = 28.0 mL

M(NH3) = 0.200 M

V(NH3) = 75.0 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.500 M * 28.0 mL = 14.0 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.200 M * 75.0 mL = 15.0 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)

= 4.7447

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 4.7447+ log {0.1359/0.0097}

= 5.89

use:

PH = 14 - pOH

= 14 - 5.89

= 8.11

PH = 8.11

B)

Given:

M(CH3COOH) = 0.35 M

V(CH3COOH) = 52 mL

M(NaOH) = 0.4 M

V(NaOH) = 33 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 * 33 mL = 13.2 mmol

We have:

mol(CH3COOH) = 18.2 mmol

mol(NaOH) = 13.2 mmol

13.2 mmol of both will react

excess CH3COOH remaining = 5 mmol

Volume of Solution = 52 + 33 = 85 mL

[CH3COOH] = 5 mmol/85 mL = 0.06M

[CH3COO-] = 13.2/85 = 0.16M

They form acidic buffer

acid is CH3COOH

conjugate base is CH3COO-

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.7447

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.7447+ log {0.1553/0.0588}

= 5.1663

PH = 5.17