In: Chemistry
You need to prepare an acetate buffer of pH 6.24 from a 0.885 M acetic acid solution and a 2.38 M KOH solution. If you have 475 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 6.24? The pKa of acetic acid is 4.76.
Answer: _____ mL
Let volume of KOH be V mL
mol of KOH added = 2.38*V mmol
Before adding KOH
Before Reaction:
mol of CH3COONa = 0 mmol
mol of CH3COOH = 0.885 M *475.0 mL
mol of CH3COOH = 420.375 mmol
2.38*V KOH will react with 2.38*V of CH3COOH to form extra 2.38*V of base
After adding KOH
mol of CH3COOH = 420.375-2.38*V mmol
mol of CH3COONa = 0+2.38*V mmol
use:
pH = pKa + log {[conjugate base]/[acid]}
6.24 = 4.76+log {[CH3COONa]/[CH3COOH]}
log {[CH3COONa]/[CH3COOH]} = 1.48
[CH3COONa]/[CH3COOH] = 30.1995
So,
(0+2.38*V)/(420.375-2.38*V) = 30.1995
0+2.38*V = 12695.122 - 71.8749*V
(2.38+71.8749)*V = 12695.122-0
V = 171 mL
Answer: 171 mL