Question

# Prepare a buffer by direct addition. Consider how best to prepare one liter of a buffer...

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 =

Sol :-

Given

pH = 3.62

we know that

pH of the buffer should be in the range

pKa -1 < pH < pKa + 1

pKa -1 < 3.62

pKa < 4.62

So,

The buffer combination should be HC2O4- / C2O42-

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 = pKa + 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

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