You need to prepare an acetate buffer of pH 5.54 from a 0.677 M acetic acid solu

Question

You need to prepare an acetate buffer of pH 5.54 from a 0.677 M acetic acid solution and a 2.04 M KOH solution. If you have 825 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.54? The pKa of acetic acid is 4.76.

You need to prepare an acetate buffer of pH 5.54 from a 0.677 M acetic acid solution and a 2.04 M KOH solution. If you have 825 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 5.54? The pKa of acetic acid is 4.76 Number mL

Explanation / Answer

Let volume of KOH be V mL

mol of KOH added = 2.04*V mmol

Before adding KOH

Before Reaction:

mol of CH3COO- = 0 mmol

mol of CH3COOH = 0.677 M *825.0 mL

mol of CH3COOH = 558.525 mmol

2.04*V KOH will react with 2.04*V of CH3COOH to form extra 2.04*V of base

After adding KOH

mol of CH3COOH = 558.525-2.04*V mmol

mol of CH3COO- = 0+2.04*V mmol

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

5.54 = 4.76+log {[CH3COO-]/[CH3COOH]}

log {[CH3COO-]/[CH3COOH]} = 0.78

[CH3COO-]/[CH3COOH] = 6.026

So,

(0+2.04*V)/(558.525-2.04*V) = 6.0256

0+2.04*V = 3365.4459 - 12.2922*V

(2.04+12.2922)*V = 3365.4459-0

V = 235 mL

Answer: 235 mL

Feel free to comment below if you have any doubts or if this answer do not work

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