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.
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
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