Question

If you mix 10mL of 0.25M sodium acetatewith 35 mL of 0.1M acetic acid, what will the pH of the resulting buffer be? The molarity?

Answer 1

i) pH

After mixing the solutions

[CH3COO^-] = 0.25M/(45ml/10ml) = 0.05556M

[CH3COOH] = 0.10M/( 45ml / 35ml ) = 0.07778M

Henderson - Hasselbalch equation which give pH of a buffer solution is

pH = pKa + log([A^-]/[HA])

where,

A^- = conjucate base , CH3COO^-

HA = weak acid , CH3COOH

pKa = pKa of weak acid , 4.75

substituting the values

pH = 4.75 + log(0.05556M/0.07778M)

pH = 4.75 - 0.15

pH = 4.60

ii) Molarity

Molarity of a buffer solution = molarity of weak acid + molarity of conjucate base

Molarity of the buffer = 0.07778M + 0.05556M

Molarity of the buffer = 0.1333M