Consider how best to prepare one liter of a buffer solution with pH = 11.12 usin

Question

Consider how best to prepare one liter of a buffer solution with pH = 11.12 using one of the weak acid/conjugate base systems shown here.

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 potassium salt of the weak acid must be combined with how many grams of thepotassium 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 =

pKa

Explanation / Answer

HCO3- + H2O <---> CO3 2- + H3O+           pKa=10.32
to make a buffer solution of specific pH use acid/base pair of the reaction whose pKa is nearest to wanted pH here the nearest pKa is 10.32 of

HCO3-

CO32-

So we use this in following ratio:

pH=pKa + log[HCO3-]/[CO3 2-]

log[HCO3-]/[CO3 2-]=pH - pKa=11.12-10.32=0.80
[HCO3-]/[CO3 2-]=10^0.80=6.31

[HCO3-] = 6.31 [CO3 2-]

Acid = 6.31 Base

From the above ratio means that we can make this buffer using 6.31 moles of KHCO3 and 1 mol of K2CO3

to produce 1.00 L of a buffer that is 1.00 M

acid + base = 1.00 M

6.31 Base + base = 1.00

7.31 base = 1.00

Base = 0.136 mol

Acid = 1.00- base

Acid = 0.864 mol

Amount of acid or KHCO3

0.864 g* 100.115 g/mol

=86.5 g

Amount of base or K2CO3

0.136 mol * 138.205 g/mol

=18.80 g

HCO3-

CO32-

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