Question

What would be the approximate pH of the final solution if you added 1.0 mL of 1.0 M acetic acid (pKa = 4.76, Ka = 1.74 x 10–5 ) to one liter of 1 M NaOH?

Answer 1

Given:
M(CH3COOH) = 1 M
V(CH3COOH) = 1 mL
M(NaOH) = 1 M
V(NaOH) = 1 L = 1000 mL


mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)
mol(CH3COOH) = 1 M * 1 mL = 1 mmol

mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 1 M * 1000 mL = 1000 mmol


We have:
mol(CH3COOH) = 1 mmol
mol(NaOH) = 1000 mmol

1 mmol of both will react

excess NaOH remaining = 999 mmol
Volume of Solution = 1 + 1000 = 1001 mL
[OH-] = 999 mmol/1001 mL = 0.998002 M

use:
pOH = -log [OH-]
= -log (0.998002)
= 0.000869


use:
PH = 14 - pOH
= 14 - 0.000869
= 13.9991
Answer: 13.999