Question

Part B

A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 13.0 mL of KOH.

Express your answer numerically

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 21.0 mL of HNO3.

Express your answer numerically.

Part D

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 25.0 mL of NaOH.

Express your answer numerically.

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Answer 1

B)

Given:

M(HBr) = 0.15 M

V(HBr) = 50 mL

M(KOH) = 0.25 M

V(KOH) = 13 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 * 13 mL = 3.25 mmol

We have:

mol(HBr) = 7.5 mmol

mol(KOH) = 3.25 mmol

3.25 mmol of both will react

remaining mol of HBr = 4.25 mmol

Total volume = 63.0 mL

[H+]= mol of acid remaining / volume

[H+] = 4.25 mmol/63.0 mL

= 6.746*10^-2 M

use:

pH = -log [H+]

= -log (6.746*10^-2)

= 1.171

Answer: 1.17

C)

Given:

M(HNO3) = 0.5 M

V(HNO3) = 21 mL

M(NH3) = 0.2 M

V(NH3) = 75 mL

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

mol(HNO3) = 0.5 M * 21 mL = 10.5 mmol

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

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

We have:

mol(HNO3) = 10.5 mmol

mol(NH3) = 15 mmol

10.5 mmol of both will react

excess NH3 remaining = 4.5 mmol

Volume of Solution = 21 + 75 = 96 mL

[NH3] = 4.5 mmol/96 mL = 0.0469 M

[NH4+] = 10.5 mmol/96 mL = 0.1094 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 {0.1094/4.688*10^-2}

= 5.113

use:

PH = 14 - pOH

= 14 - 5.1127

= 8.8873

Answer: 8.89

D)

Given:

M(CH3COOH) = 0.35 M

V(CH3COOH) = 52 mL

M(NaOH) = 0.4 M

V(NaOH) = 25 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 * 25 mL = 10 mmol

We have:

mol(CH3COOH) = 18.2 mmol

mol(NaOH) = 10 mmol

10 mmol of both will react

excess CH3COOH remaining = 8.2 mmol

Volume of Solution = 52 + 25 = 77 mL

[CH3COOH] = 8.2 mmol/77 mL = 0.1065M

[CH3COO-] = 10/77 = 0.1299M

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 {0.1299/0.1065}

= 4.831

Answer: 4.83