Question

You need to prepare an acetate buffer of pH 5.79 from a 0.645 M acetic acid...

You need to prepare an acetate buffer of pH 5.79 from a 0.645 M acetic acid solution and a 2.53 M KOH solution. If you have 625 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a bufer of pH 5.79? The pKa of acetic acid is 4.76.

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Answer #1

the buffer equation

pH= pKa + log(A-/HA)

5.79 = 4.76 + log(ratio)

ratio = 10^(5.79-4.76) = 10.715

[A-] = 10.715*[HA]

assume:

initially:

[HA] = 0.645*625 = 403.125 mmol

[A-] = 0

after addition of:

KOH = MV = 2.53*Vbase

[HA] = 403.125 - 2.53*Vbase

[A-] = 2.53*Vbase

then

we know that

[A-] = 10.715*[HA]

2.53*Vbase = 10.715*( 403.125 - 2.53*Vbase)

solve for Vbase

2.53*Vbase = 10.715*( 403.125 - 2.53*Vbase)

2.53*Vbase =4319.484 - 27.1089*Vbase

Vbase ( 2.53 + 27.1089) = 4319.484

Vbase = (4319.484 )/( ( 2.53 + 27.1089) ) = 145.736

Vbase = 145.736mL of KOH is required

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