Prepare a buffer by direct addition Consider how best to prepare one liter of a
ID: 705574 • Letter: P
Question
Prepare a buffer by direct addition Consider how best to prepare one liter of a buffer solution with pH- 4.68 using one of the weak acid/conjugate base systems shown here Weak Acid ??,04 H2PO4 HCO3 Conjugate Base C20,2 ? ??,2 ??? Ka 6.4 x 10-5 6.2 x 10-8 4.8 x 10-11 pKa 4.19 7.21 10.32 How many grams of the potassium salt of the weak acid must be combined with how many grams of the potassium salt of its conjugate base, to produce 1.00 L of a buffer that is 1.00 M in the weak base? grams potassium salt of weak acid- grams potassium salt of conjugate base-Explanation / Answer
Answer
Suitable weak acid and conjucate base combination is HC2O4-/C2O42-
mass of KHC2O4 required = 41.460g
mass of K2C2O4 required = 166.21g
Explanation
HC2O4-/C2O42- is the suitable buffer for the target pH 4.68
because pKa is near to the target pH
Henderson-Hasselbalch equation is
pH = pKa + log([A-] /[HA])
where,
[A-] =[ C2O42-] = 1.00M
[HA]= [HC2O4-]
pKa = 4.19
4.68 = 4.19 + log([C2O42-] /[HC2O4-])
log([C2O42-] /[HC2O4-]) = 0.49
[C2O42-] /[HC2O4-] = 3.09
given concentration of C2O42- = 1.00M
HC2O4- = 1.00/3.09 = 0.3236
Volume of buffer = 1.00L
Therefore
moles of KHC2O4 required = 0.3236
moles of K2C2O4 required = 1.00
molar mass of KHC2O4 = 128.12g/mol
mass of KHC2O4 required = 128.12g/mol × 0.3236mol = 41.460g
molar mass of K2C2O4 = 166.21g/mol
mass of K2C2O4 required = 166.21g/mol × 1.00mol = 166.21g