Consider how best to prepare one liter of a buffer solution with pH = 10.92 usin

Question

Consider how best to prepare one liter of a buffer solution with pH = 10.92 using one of the weak acid/conjugate base systems shown here.

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 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 =

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

Explanation / Answer

To determine the best buffer combination, calculate first the ratio (conjugate base/acid) which is calculated as:

ratio = 10pH-pKa

The smallest number is the best buffer to use.

Now, let's do this with every system:

ratio 1 = 10(10.92-4.19) = 5.37x106
ratio 2 = 10(10.92-7.21) = 5128.61
ratio 3 = 10(10.92-1032) = 3.98

So the system CO32- / HCO32- will be the best system to prepare the desired buffer.

Now, as we know that the weak base has to be 1 M, then the acid would have to be:
ratio = [CO32-] / [HCO3-]
[HCO3-] = 1 / 3.98 = 0.25 M

Finally, the mass for these compounds would have to be:
mHCO3- = 0.25 mol/L * 1 L * (39+1+12+48) = 25 g
mCO32- = 1 * 1 * (2*39+12+48) = 132 g

Hope this helps

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