Question

You perform a titration using 25.0 mL of a 0.20 M propionic acid (HC3H5O2) as your analyte and 0.25 M NaOH as your titrant. Assume a new 25.0 mL of analyte is used for each step, determine the pH when the following volumes of titrant are used.

(Propionic acid Ka= 1.32 x 10^-5)

a. 15.0 mL titrant added.

b. 25.0 mL titrant added.

Answer 1

a)when 15.0 mL of NaOH is added

Given:

M(HC3H5O2) = 0.2 M

V(HC3H5O2) = 25 mL

M(NaOH) = 0.25 M

V(NaOH) = 15 mL

mol(HC3H5O2) = M(HC3H5O2) * V(HC3H5O2)

mol(HC3H5O2) = 0.2 M * 25 mL = 5 mmol

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

mol(NaOH) = 0.25 M * 15 mL = 3.75 mmol

We have:

mol(HC3H5O2) = 5 mmol

mol(NaOH) = 3.75 mmol

3.75 mmol of both will react

excess HC3H5O2 remaining = 1.25 mmol

Volume of Solution = 25 + 15 = 40 mL

[HC3H5O2] = 1.25 mmol/40 mL = 0.0312M

[C3H5O2-] = 3.75/40 = 0.0938M

They form acidic buffer

acid is HC3H5O2

conjugate base is C3H5O2-

Ka = 1.32*10^-5

pKa = - log (Ka)

= - log(1.32*10^-5)

= 4.879

use:

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

= 4.879+ log {9.375*10^-2/3.125*10^-2}

= 5.357

Answer: 5.36

b)when 25.0 mL of NaOH is added

Given:

M(HC3H5O2) = 0.2 M

V(HC3H5O2) = 25 mL

M(NaOH) = 0.25 M

V(NaOH) = 25 mL

mol(HC3H5O2) = M(HC3H5O2) * V(HC3H5O2)

mol(HC3H5O2) = 0.2 M * 25 mL = 5 mmol

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

mol(NaOH) = 0.25 M * 25 mL = 6.25 mmol

We have:

mol(HC3H5O2) = 5 mmol

mol(NaOH) = 6.25 mmol

5 mmol of both will react

excess NaOH remaining = 1.25 mmol

Volume of Solution = 25 + 25 = 50 mL

[OH-] = 1.25 mmol/50 mL = 0.025 M

use:

pOH = -log [OH-]

= -log (2.5*10^-2)

= 1.6021

use:

PH = 14 - pOH

= 14 - 1.6021

= 12.3979

Answer: 12.40