Question

The K_a of acetic acid, CH₃COOH, is 1.8 × 10^-5. A buffer solution was made using an unspecified amount of acetic acid and 0.30 moles of NaOOCCH₃ in enough water to make 1.50 L of solution. The pH of the solution is 4.55. How many moles of CH₃COOH were used?

Could you please showing working on how to get the answer of 0.47.

Answer 1

Ka of CH3COOH = 1.8 x 10^-5

pKa = - log Ka
       = - log (1.8 x 10^-5)
       = 4.74

Since it is a buffer, its pH can be calculated using Henderson-Hasselbalch equation. Mathematically henderson hasselbalch equation is expressed as;

pH = pKa + log { [salt] / [acid] }

or pH + pKa = log { [CH₃COONa] / [CH₃COOH] }

Now, Moles of CH3COONa = 0.30 moles
Volume = 1.50 L
So, [CH3COONa] = (0.30 moles) / 1.50 L = 0.20 M

pH = 4.55

Substituting the values, we get;

pH = pKa + log { [CH₃COONa] / [CH₃COOH] }

or, 4.55 = 4.74 + log { 0.2 / [CH₃COOH] }

or, log { 0.2 / [CH₃COOH] } = 4.55 – 4.74

or, log { 0.2 / [CH₃COOH] } = -0.19

or, { 0.2 / [CH₃COOH] } = 10^-0.19

or, { 0.2 / [CH₃COOH] } = 0.64565

or, [CH₃COOH] = 0.2 / 0.64565

or, [CH₃COOH] = 0.31 M

Again, volume = 1.5 L

So, moles of CH₃COOH = 0.31 M x 1.5 L
  = 0.465 moles
  = 0.47 moles