Question

Citric acid (192.13 g/mol) is a triprotic acid with the following chemical formula: H3C6O7H5.

The acid dissociation constants for citric acid are:

Ka1 = 7.1 x 10^-4 M; Ka2 = 1.7 x 10^-5 M; Ka3 = 6.4 x 10^-6 M

Describe how you would prepare at least 100.00 mL (you may make more) of a citrate buffer with a pH of about 4.40. You have available distilled water, volumetric flasks and any sodium salts you desire. You must indicate the amount in grams of material needed (except for water, you may use volume). Your designed buffer must have a reasonable buffer capacity and no solution you make can have a concentration greater than 0.500 M.

Answer 1

Ka1 = 7.1 x 10-4

we know that

pKa= -logKa

so

pKa1 = -log 7.1 x 10-4 = 3.1487

pKa2 = -log 1.7 x 10-5 = 4.76955

pKa3 = -log 6.4 x 10-6 = 5.1938

given

pH of buffer = 4.4

when choosing a buffer its pKa should close to the pH

from the above we get

pKa2 is close to pH

so the required equation is

H2C6H5O7- + H2O ----> HC6H5O72- + H3O+

according to hasselbach henderson

for a acid buffer

pH = pKa + log [ conjugate base / acid ]

4.4 = 4.76955 + log[ HC6H5072- / H2C6H507-]

[ HC6H5072- / H2C6H507-] = 0.427

so

the ratio of conc of [ HC6H5072- / H2C6H507-] = 0.427

we know that

moles = conc x volume

as the final volume is same for both of them

ratio of moles = ratio of conc

so

moles of HC6H5072- / H2C6H507- = 0.427

aslo given

conc should not be greater than 0.5

we got

[ HC6H5072- / H2C6H507-] = 0.427

so [H2C6H507-] > [ HC6H5072-]

so let [H2C6H507-] = 0.5 M

final volume = 100 ml = 0.1 L

we know that

moles = conc x volume of solution (L)

moles of [H2C6H507-] = 0.5 x 0.1 = 0.05

mass of [H2C6H507-] = moles x molar mass

= 0.05 x 191.13

mass of [H2C6H507-] = 9.5565 g

from the above

moles of HC6H5072- / H2C6H507- = 0.427

so

moles of HC6H5072- = 0.427 x 0.05 = 0.02135

mass of HC6H5072- = moles x molar mass

= 0.02135 x 190.13

mass of HC6H5072- = 4.059 g

so finally to get the required buffer

take 50 ml of water then add 4.059 g of Na2(HC6H507) and then

add 9.5565 g of NaH2C6H507 . mix

now add extra water to make the final volume to 100 ml