A pH 4.01 buffer is to be prepared by combining 0.125 M acetic acid with 0.150...
A pH 4.01 buffer is to be prepared by combining 0.125 M acetic acid with 0.150 M sodium acetate. How many mL of each substance is needed to prepare 250.0 mL of this buffer?
Homework Answers
Let volume of acetic acid needed = x mL
then, volume of sodium acetate needed = (total volume) - (volume of acetic acid)
volume of sodium acetate needed = 250 mL - x mL
volume of sodium acetate needed = (250 - x) mL
moles acetic acid needed = (concentration acetic acid) * (volume of acetic acid needed)
moles acetic acid needed = (0.125 M) * (x mL)
moles acetic acid needed = 0.125x mmol
moles sodium acetate needed = (concentration sodium acetate) * (volume of sodium acetate needed)
moles sodium acetate needed = (0.150 M) * (250 - x) mL
moles sodium acetate needed = (37.5 - 0.150x) mmol
Ka acetic acid = 1.8 x 10-5
pKa = -log(Ka)
pKa = -log(1.8 x 10-5)
pKa = 4.74
According to Henderson - Hasselbalch equation,
pH = pKa + log([conjugate base] / [weak acid])
pH = pKa + log(moles sodium acetate needed / moles acetic acid needed)
Substituting the values,
4.01 = 4.74 + log((37.5 - 0.150x) / 0.125x)
log((37.5 - 0.150x) / 0.125x) = 4.01 - 4.74
log((37.5 - 0.150x) / 0.125x) = -0.73
(37.5 - 0.150x) / 0.125x = 10-0.73
(37.5 - 0.150x) / 0.125x = 0.184
37.5 - 0.150x = (0.184) * (0.125x)
37.5 - 0.150x = 0.023x
37.5 = 0.150x + 0.023x
37.5 = 0.173x
x = (37.5) / (0.173)
x = 216.7
volume of acetic acid needed = x mL
volume of acetic acid needed = 216.7 mL
volume of sodium acetate needed = (250 - x) mL
volume of sodium acetate needed = (250 - 216.7) mL
volume of sodium acetate needed = 33.3 mL
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