Question

How many ml of 4.00 M HCl would you need to add to 500.0 ml of a 150. mM sodium acetate solution to make a buffer with a pH of 4.5? The pKa for acetic acid is 4.756.

Answer 1

Volume of HCl required = 12.1 mL

Explanation

molarity of acetate = 150. mM = 0.150 M

volume of acetate solution = 500.0 mL

initial moles acetate = (molarity of acetate) * (volume of acetate solution)

initial moles acetate = (0.150 M) * (500 mL)

initial moles acetate = 75.0 mmol

Let moles HCl added = x

moles acetic acid formed = moles HCl added

moles acetic acid formed = x

moles acetic acid remaining = (initial moles acetate) - (moles acetic acid formed)

moles acetic acid remaining = 75.0 mmol - x

According to Henderson - Hasselbalch equation,

pH = pK_a + log([conjugate base] / [weak acid])

pH = pK_a + log(moles acetic acid remaining / moles acetic acid formed)

4.5 = 4.756 + log(75.0 mmol - x / x)

log(75.0 mmol - x / x) = 4.5 - 4.756

log(75.0 mmol - x / x) = -0.256

(75.0 mmol - x) / x = 10^-0.256

(75.0 mmol - x) / x = 0.5546

(75.0 mmol / x) - 1 = 0.5546

(75.0 mmol / x) = 0.5546 + 1

(75.0 mmol / x) = 1.5546

x = 75.0 mmol / 1.5546

x = 48.243 mmol

moles HCl added = x = 48.243 mmol

volume HCl added = (moles HCl added) / (molarity HCl)

volume HCl added = (48.243 mmol) / (4.00 M)

volume HCl added = 12.06 mL