Question

Be sure to answer all parts. Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (K2 1.54 x 10, with 0.1000 M NaOH solution after the following additions of titrant. (а) 15.00 mL: pH (b) 20.60 mL: pH (с) 26.00 mL: pH

Answer 1

a)when 15.0 mL of NaOH is added

Given:

M(C3H7COOH) = 0.1 M

V(C3H7COOH) = 20 mL

M(NaOH) = 0.1 M

V(NaOH) = 15 mL

mol(C3H7COOH) = M(C3H7COOH) * V(C3H7COOH)

mol(C3H7COOH) = 0.1 M * 20 mL = 2 mmol

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

mol(NaOH) = 0.1 M * 15 mL = 1.5 mmol

We have:

mol(C3H7COOH) = 2 mmol

mol(NaOH) = 1.5 mmol

1.5 mmol of both will react

excess C3H7COOH remaining = 0.5 mmol

Volume of Solution = 20 + 15 = 35 mL

[C3H7COOH] = 0.5 mmol/35 mL = 0.0143M

[C3H7COO-] = 1.5/35 = 0.0429M

They form acidic buffer

acid is C3H7COOH

conjugate base is C3H7COO-

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 {4.286*10^-2/1.429*10^-2}

= 5.29

Answer: 5.29

2)when 20.6 mL of NaOH is added

Given:

M(C3H7COOH) = 0.1 M

V(C3H7COOH) = 20 mL

M(NaOH) = 0.1 M

V(NaOH) = 20.6 mL

mol(C3H7COOH) = M(C3H7COOH) * V(C3H7COOH)

mol(C3H7COOH) = 0.1 M * 20 mL = 2 mmol

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

mol(NaOH) = 0.1 M * 20.6 mL = 2.06 mmol

We have:

mol(C3H7COOH) = 2 mmol

mol(NaOH) = 2.06 mmol

2 mmol of both will react

excess NaOH remaining = 0.06 mmol

Volume of Solution = 20 + 20.6 = 40.6 mL

[OH-] = 0.06 mmol/40.6 mL = 0.0015 M

use:

pOH = -log [OH-]

= -log (1.478*10^-3)

= 2.8304

use:

PH = 14 - pOH

= 14 - 2.8304

= 11.1696

Answer: 11.17

c)when 26.0 mL of NaOH is added

Given:

M(C3H7COOH) = 0.1 M

V(C3H7COOH) = 20 mL

M(NaOH) = 0.1 M

V(NaOH) = 26 mL

mol(C3H7COOH) = M(C3H7COOH) * V(C3H7COOH)

mol(C3H7COOH) = 0.1 M * 20 mL = 2 mmol

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

mol(NaOH) = 0.1 M * 26 mL = 2.6 mmol

We have:

mol(C3H7COOH) = 2 mmol

mol(NaOH) = 2.6 mmol

2 mmol of both will react

excess NaOH remaining = 0.6 mmol

Volume of Solution = 20 + 26 = 46 mL

[OH-] = 0.6 mmol/46 mL = 0.013 M

use:

pOH = -log [OH-]

= -log (1.304*10^-2)

= 1.8846

use:

PH = 14 - pOH

= 14 - 1.8846

= 12.1154

Answer: 12.12