Question

PLEASE SHOW STEPS CLEARLY

You need to prepare an acetate buffer of pH 5.82 from a 0.712 M acetic acid solution and a 2.78 M KOH solution. If you have 575 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.82? The pKa of acetic acid is 4.76.

Answer 1

Let volume of KOH be V mL

mol of KOH added = 2.78*V mmol

Before adding KOH

Before Reaction:

mol of CH3COONa = 0 mmol

mol of CH3COOH = 0.712 M *575.0 mL

mol of CH3COOH = 409.4 mmol

2.78*V KOH will react with 2.78*V of CH3COOH to form extra 2.78*V of base

After adding KOH

mol of CH3COOH = 409.4-2.78*V mmol

mol of CH3COONa = 0+2.78*V mmol

use:

pH = pKa + log {[conjugate base]/[acid]}

5.82 = 4.76+log {[CH3COONa]/[CH3COOH]}

log {[CH3COONa]/[CH3COOH]} = 1.06

[CH3COONa]/[CH3COOH] = 11.481536

So,

(2.78*V)/(409.4-2.78*V) = 11.481536

2.78*V = 4700.540926 - 31.918671*V

(2.78+31.918671)*V = 4700.540926-0

V = 135 mL