Design a buffer that has a pH of 6.65 using one of the weak acid/conjugate base

Question

Design a buffer that has a pH of 6.65 using one of the weak acid/conjugate base systems shown below.

HC2O4-

C2O42-

6.4×10-5

4.19

H2PO4-

HPO42-

6.2×10-8

7.21

HCO3-

CO32-

4.8×10-11

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 = g

grams sodium salt of conjugate base = g

Weak Acid Conjugate Base Ka pKa

HC2O4-

C2O42-

6.4×10-5

4.19

H2PO4-

HPO42-

6.2×10-8

7.21

HCO3-

CO32-

4.8×10-11

10.32

Explanation / Answer

we know that

for a buffer

pKa should be in the range

pH -1 < pKa < pH + 1

given

pH = 6.65

so

5.65 < pKa < 7.65

so

from the given table

the buffer combination is

H2P04- / HP042-

here

weak acid is HP042-

weak base is H2P04-

now

for buffers

pH = pKa + log [ conjugate base / acid ]

so

pH = pKa + log [HP042- / H2P04-]

6.65 = 7.21 + log [ 1 / H2P04-]

[H2P04-] = 3.6308 M

now

given 1 L of buffer

we know that

moles = concentration x volume (L)

so

moles of Na2HP04 = 1 x 1 = 1

moles of NaH2P04 = 3.6308 x 1 = 3.6308

now

mass = moles x molar mass

so

mass of Na2HP04 = 1 x 142 = 142 g

mass of NaH2P04 = 3.6308 x 120 = 435.7 g

so

grams sodium salt of weak acid = 435.7 g

grams sodium salt of conjugate base = 142 g

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