You split in half a solution of 2.0 grams sodium acetate and 10 mL 3.0 M acetic acid that was later added with 50 mL water (total 60 mL). Find the pH

Question

You split in half a solution of 2.0 grams sodium acetate and 10 mL 3.0 M acetic acid that was later added with 50 mL water (total 60 mL). One solution will be added with 1 mL of 6.0 M HCl and the other with 1 mL 6.0 M NaOH. Find the pH.

Answer 1

Molar mass sodium acetate (CH3COONa) = 82.0 g / mole

Acid = acetic acid (CH3COOH)

Conjugate base = sodium acetate

moles acetic acid = 10 ml x 1 L / 1000 ml x 3.0 mol / L = 0.03 moles

moles sodium acetate = 2.0 g x 1 mol / 82.0 g = 0.0244 moles

Final volume = 10 ml + 50 ml = 60 ml = 0.060 L

[acetic acid] = 0.03 moles / 0.060 L = 0.50 M acetic acid

[sodium acetate] = 0.0244 moles / 0.060 L = 0.407 M sodium acetate

If you add 1 ml 6 M HCl, it will react with the CH3COO- to produce CH3COOH

[HCl added] = 1 ml x 1 L/1000 ml x 6 mol/L = 0.006 mols HCl/0.030 L = 0.2 M

[CH3COO-] = 0.407 - 0.2 = 0.207 M

[CH3COOH] = 0.50 + 0.2 = 0.70 M

pH = 4.75 + log (0.207/0.70) = 4.75 + log 0.296

pH = 4.75 - 0.529

pH = 4.22

If you add 1 ml 6 M NaOH, it will react with the CH3COOH to produce CH3COO-

[NaOH] added = 1 ml x 1 L / 1000 ml x 6 mol / L = 0.006 mol NaOH / 0.030 L = 0.2 M

[CH3COO-] = 0.407 + 0.2 = 0.607 M

[CH3COOH] = 0.50 - 0.2 = 0.30 M

pH = 4.75 + log (0.607 / 0.30) = 4.75 + log 2.02

pH = 4.75 + 0.306

pH = 5.06