Question

Be sure to answer all parts. Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (K, = 1.54 x 10-5), with 0.1000 M NaOH solution after the following additions of titrant. (a) 13.00 mL: pH = (b) 20.90 mL: pH = 11. 30 (c) 30.00 mL: pH =

Answer 1

1)when 13.0 mL of NaOH is added

Given:

M(CH3CH2CH2COOH) = 0.1 M

V(CH3CH2CH2COOH) = 20 mL

M(NaOH) = 0.1 M

V(NaOH) = 13 mL

mol(CH3CH2CH2COOH) = M(CH3CH2CH2COOH) * V(CH3CH2CH2COOH)

mol(CH3CH2CH2COOH) = 0.1 M * 20 mL = 2 mmol

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

mol(NaOH) = 0.1 M * 13 mL = 1.3 mmol

We have:

mol(CH3CH2CH2COOH) = 2 mmol

mol(NaOH) = 1.3 mmol

1.3 mmol of both will react

excess CH3CH2CH2COOH remaining = 0.7 mmol

Volume of Solution = 20 + 13 = 33 mL

[CH3CH2CH2COOH] = 0.7 mmol/33 mL = 0.0212M

[CH3CH2CH2COO-] = 1.3/33 = 0.0394M

They form acidic buffer

acid is CH3CH2CH2COOH

conjugate base is CH3CH2CH2COO-

Ka = 1.54*10^-5

pKa = - log (Ka)

= - log(1.54*10^-5)

= 4.812

use:

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

= 4.812+ log {3.939*10^-2/2.121*10^-2}

= 5.081

Answer: 5.08

2)when 20.9 mL of NaOH is added

Given:

M(CH3CH2CH2COOH) = 0.1 M

V(CH3CH2CH2COOH) = 20 mL

M(NaOH) = 0.1 M

V(NaOH) = 20.9 mL

mol(CH3CH2CH2COOH) = M(CH3CH2CH2COOH) * V(CH3CH2CH2COOH)

mol(CH3CH2CH2COOH) = 0.1 M * 20 mL = 2 mmol

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

mol(NaOH) = 0.1 M * 20.9 mL = 2.09 mmol

We have:

mol(CH3CH2CH2COOH) = 2 mmol

mol(NaOH) = 2.09 mmol

2 mmol of both will react

excess NaOH remaining = 0.09 mmol

Volume of Solution = 20 + 20.9 = 40.9 mL

[OH-] = 0.09 mmol/40.9 mL = 0.0022 M

use:

pOH = -log [OH-]

= -log (2.2*10^-3)

= 2.6575

use:

PH = 14 - pOH

= 14 - 2.6575

= 11.3425

Answer: 11.34

3)when 30.0 mL of NaOH is added

Given:

M(CH3CH2CH2COOH) = 0.1 M

V(CH3CH2CH2COOH) = 20 mL

M(NaOH) = 0.1 M

V(NaOH) = 30 mL

mol(CH3CH2CH2COOH) = M(CH3CH2CH2COOH) * V(CH3CH2CH2COOH)

mol(CH3CH2CH2COOH) = 0.1 M * 20 mL = 2 mmol

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

mol(NaOH) = 0.1 M * 30 mL = 3 mmol

We have:

mol(CH3CH2CH2COOH) = 2 mmol

mol(NaOH) = 3 mmol

2 mmol of both will react

excess NaOH remaining = 1 mmol

Volume of Solution = 20 + 30 = 50 mL

[OH-] = 1 mmol/50 mL = 0.02 M

use:

pOH = -log [OH-]

= -log (2*10^-2)

= 1.699

use:

PH = 14 - pOH

= 14 - 1.699

= 12.301

Answer: 12.30