Question

15.00 mL of 0.100 M NaOH are added to 50.00 mL of a buffer solution containing 1.0 mmol CH3COOH and 1.0 mmol CH3COO-. What is the new solution pH? (show work)

a) 11.89

b) 2.11

c) 7.87

d) can't tell/need more info.

Answer 1

mol of NaOH added = 0.1M *15.0 mL = 1.5 mmol

CH3COOH will react with OH- to form CH3COO-

Before Reaction:

mol of CH3COO- = 1 mmol

mol of CH3COOH = 1 mmol

Mol of base added is greater than mol of CH3COOH

So, all of CH3COOH will react

Remaining amount of base = 0.5 mmol

Total volume = 50.00 mL + 15.00 mL

= 65.00 mL

So,

[OH-] = Remaining amount of base / Total volume

= 0.5 mmol / 65.00 mL

= 7.69*10^-3 M

Given:

[OH-] = 7.69*10^-3 M

use:

pOH = -log [OH-]

= -log (7.69*10^-3)

= 2.11

use:

PH = 14 - pOH

= 14 - 2.11

= 11.89

Answer: 11.89