: Preparing a 0.050 M buffered solution using a conjugate weak acid/weak base pair of sodium...
- : Preparing a 0.050 M buffered solution using a conjugate weak acid/weak base pair of sodium citrate tribasic and sodium citrate dibasic. You will prepare 100 mL of buffered solution with pH = 6.00. The formal concentration of this buffer should be 0.050 F (i.e., the total concentration of the weak acid and its conjugate base is 0.050 M). Find the total amount (g or mL) of each species you will have to add to 100 mL of water to get your assigned pH.
Sodium CItrate Tribasic ( MW: 294.1 grams)
Sodium CItrate Dibasic ( MW: 263.11 grams)
pKa3 of citric acid: 6.396.
Homework Answers
Sodium Citrate Dibasic is the weak acid and Sodium Citrate Tribasic is the conjugate base
Total concentration of buffer = 0.050 M
[Sodium Citrate Dibasic] + [Sodium Citrate Tribasic] = 0.050 M ...(1)
According to Henderson-Hasselbalch equation,
pH = pKa + log([conjugate base] / [weak acid])
pH = pKa3 + log([Sodium Citrate Tribasic] / [Sodium Citrate Dibasic])
6.00 = 6.396 + log([Sodium Citrate Tribasic] / [Sodium Citrate Dibasic])
log([Sodium Citrate Tribasic] / [Sodium Citrate Dibasic]) = 6.00 - 6.396
log([Sodium Citrate Tribasic] / [Sodium Citrate Dibasic]) = -0.396
[Sodium Citrate Tribasic] / [Sodium Citrate Dibasic] = 10-0.396
[Sodium Citrate Tribasic] / [Sodium Citrate Dibasic] = 0.4 ... (2)
Solving equations (1) and (2) simultaneously, we get
[Sodium Citrate Dibasic] = 0.03567 M
[Sodium Citrate Tribasic] = 0.01433 M
moles Sodium Citrate Dibasic = (molarity Sodium Citrate Dibasic) * (volume of buffer in Liter)
moles Sodium Citrate Dibasic = (0.03567 M) * (0.1 L)
moles Sodium Citrate Dibasic = 0.003567 mol
mass Sodium Citrate Dibasic = (moles Sodium Citrate Dibasic) * (molar mass Sodium Citrate Dibasic)
mass Sodium Citrate Dibasic = (0.003567 mol) * (263.11 g/mol)
mass Sodium Citrate Dibasic = 0.938 g
moles Sodium Citrate Tribasic = (molarity Sodium Citrate Tribasic) * (volume of buffer in Liter)
moles Sodium Citrate Tribasic = (0.01433 M) * (0.1 L)
moles Sodium Citrate Tribasic = 0.001433 mol
mass Sodium Citrate Tribasic = (moles Sodium Citrate Tribasic) * (molar mass Sodium Citrate Tribasic)
mass Sodium Citrate Tribasic = (0.001433 mol) * (294.1 g/mol)
mass Sodium Citrate Tribasic = 0.421 g
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