How would you prepare 200 mL of 0.1 M potassium acetate buffer at pH 4.5 (please...
How would you prepare 200 mL of 0.1 M potassium acetate buffer at pH 4.5 (please give with clear detailed steps)
Homework Answers
Concentration of buffer = 0.1 M
[acetic acid] + [acetate] = Concentration of buffer
[acetic acid] + [acetate] = 0.1 M ...(1)
pKa of acetic acid = 4.76
According to Henderson - Hasselbalch equation,
pH = pKa + log([conjugate base] / [weak acid])
pH = pKa + log([acetate] / [acetic acid])
4.5 = 4.76 + log([acetate] / [acetic acid])
log([acetate] / [acetic acid]) = 4.5 - 4.76
log([acetate] / [acetic acid]) = -0.26
[acetate] / [acetic acid] = 10-0.26
[acetate] / [acetic acid] = 0.55 ...(2)
Solving equations (1) and (2) simultaneously,
[acetic acid] = 0.0645 M
[acetate] = 0.0355 M
moles acetic acid = (molarity acetic acid) * (volume of buffer in Liter)
moles acetic acid = (0.0645 M) * (0.400 L)
moles acetic acid = 0.0258 mol
mass acetic acid = (moles acetic acid) * (molar mass acetic acid)
mass acetic acid = (0.0258 mol) * (60.05 g/mol)
mass acetic acid = 1.55 g
moles acetate = (molarity acetate) * (volume of buffer in Liter)
moles acetate = (0.0355 M) * (0.400 L)
moles acetate = 0.0142 mol
mass potassium acetate = (moles acetate) * (molar mass potassium acetate)
mass potassium acetate = (0.0142 mol) * (98.14 g/mol)
mass potassium acetate = 1.39 g
To prepare the buffer solution, mix 1.55 grams of acetic acid and 1.39 grams of potassium acetate in 400 mL water.
moles acetate =
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