You need to prepare an acetate buffer of pH 6.24 from a 0.885 M acetic acid...

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

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Expert Solution

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

queen_honey_blossom answered 1 year ago

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