Question

Prepare a buffer by direct addition.

Consider how best to prepare one liter of a buffer solution with pH = 3.62 using one of the weak acid/conjugate base systems shown here.

Weak Acid	Conjugate Base	Ka	pKa
HC2O4-	C2O42-	6.4 x 10-5	4.19
H2PO4-	HPO42-	6.2 x 10-8	7.21
HCO3-	CO32-	4.8 x 10-11	10.32

How many grams of the sodium salt of the weak acid must be combined with how many grams of the sodium salt of its conjugate base, to produce 1.00 L of a buffer that is 1.00 M in the weak base?

grams sodium salt of weak acid =

grams sodium salt of conjugate base =

Answer 1

Sol :-

Given

pH = 3.62

we know that

pH of the buffer should be in the range

pK_a -1 < pH < pK_a + 1

pK_a -1 < 3.62

pK_a < 4.62

So,

The buffer combination should be HC₂O₄^- / C₂O₄^2-

Weak acid --> NaHC₂O₄

conjugate base --> Na₂C₂O₄

Now

Given 1 M in weak base

So, 1 M of Na₂C₂O₄ in 1 L buffer

We know that,

No. of moles = concentration x volume (L)

No. of moles of Na₂C₂O₄ = 1 x 1 = 1

Also

Mass = moles x molar mass

so,

Mass of Na₂C₂O₄ = 1 x 134 = 134 g

For buffers

pH = pK_a + log [ base / acid ]

3.62 = 4.19 + log [ 1 / NaHC₂O₄]

[NaHC₂O₄] = 3.72 M

No. of moles = molarity x volume

No. of moles of NaHC₂O₄ = 3.72 M x 1.0 L = 3.72 mol

mass = moles x molar mass

so,

mass of NaHC₂O₄ = 3.72 x 112 = 416.64 g

So,

Grams sodium salt of weak acid ( NaHC₂O₄) = 416.64 g

Grams sodium salt of conjugate base ( Na₂C₂O₄) = 134 g