Question

Aluminum solutions have to be kept at fairly low pH to prevent formation of hydroxide complex ions. A buffer was prepared using 0.25 MHC_2H_3O_2 (Ka - 1 76 times 10^-5) and mixing in some KC_2H_3O_2. How many grams of KC_2H_3O_2 must be added to 50.0 mL of 0.10 M HC_2H_3O_2, solution to make a pH = 4.0 buffer?

Answer 1

Let us assume first preparation of 1 L of given AcOH/AcOK buffer.
Henderson-Hasselbalch equation gives pH of this buffer solution by the formuala,
pH = pKa + log([Salt]/[Acid])
For AcOH/AcOK buffer.
pH = pKa + log([AcOK]/[AcOH])
We want pH = 4.00
Ka for AcOH = 1.76 x 10-5
Hence pKa = -log(1.76 x 10-5)
pKa = 4.75
Using this let us calculate the mole ratio ([AcOK]/[AcOH]),
4.00 = 4.75 + log([AcOK]/[AcOH])
log ([AcOK]/[AcOH]) = 4.00 - 4.75
log ([AcOK]/[AcOH]) = -0.75
([AcOK]/[AcOH]) = 10-0.75
([AcOK]/[AcOH]) = 0.1778
[AcOK] = 0.1778 x [AcOH]
But given that,
[AcOH] = 0.10 M/L
Hence,
[AcOK] = 0.1778 x 0.10
[AcOK] = 0.01778 M/L
It means ,
0.01778 moles of AcOK in 1000 mL
Hence for 50 mL (i.e. 1000/50 = 20) solution,
No. of moles of AcOK = 0.01778/20 = 8.89 x 10-4 moles.
Now,
Molar mass of AcOK = 98.14 g/mol
Mass of 2.22 x 10-3 moles AcOK = 8.89 x 10-4 x 98.14 = 0.087 g.
0.0872 g of Potassium acetate AcOK needed to be dissolved for 50.0 mL of 0.1 M acetic acid.
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