Question

What is the pH of a solution prepared by mixing 20.0 mL of 0.6 M acetic acid (Ka = 1.8 × 10-5) with 10 mL of 1.0 M NaOH? Report your answer to two decimal places. What is the pH of a solution prepared by mixing 20.0 mL of 0.6 M acetic acid (Ka = 1.8 × 10-5) with 20 mL of the 1.0 M NaOH in Part A? Report your answer to two decimal places.

Answer 1

1)

Given:

M(CH3COOH) = 0.6 M

V(CH3COOH) = 20 mL

M(NaOH) = 1 M

V(NaOH) = 10 mL

mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)

mol(CH3COOH) = 0.6 M * 20 mL = 12 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 1 M * 10 mL = 10 mmol

We have:

mol(CH3COOH) = 12 mmol

mol(NaOH) = 10 mmol

10 mmol of both will react

excess CH3COOH remaining = 2 mmol

Volume of Solution = 20 + 10 = 30 mL

[CH3COOH] = 2 mmol/30 mL = 0.0667M

[CH3COO-] = 10/30 = 0.3333M

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 {0.3333/6.667*10^-2}

= 5.444

Answer: 5.44

2)

Given:

M(CH3COOH) = 0.6 M

V(CH3COOH) = 20 mL

M(NaOH) = 1 M

V(NaOH) = 20 mL

mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)

mol(CH3COOH) = 0.6 M * 20 mL = 12 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 1 M * 20 mL = 20 mmol

We have:

mol(CH3COOH) = 12 mmol

mol(NaOH) = 20 mmol

12 mmol of both will react

excess NaOH remaining = 8 mmol

Volume of Solution = 20 + 20 = 40 mL

[OH-] = 8 mmol/40 mL = 0.2 M

use:

pOH = -log [OH-]

= -log (0.2)

= 0.699

use:

PH = 14 - pOH

= 14 - 0.699

= 13.301

Answer: 13.30