Question

A buffer prepared by dissolving oxalic acid dihydrate (H2C2O4·2H2O) and disodium oxalate (Na2C2O4) in 1.00 L of water has a pH of 5.043. How many grams of oxalic acid dihydrate (MW = 126.07 g/mol) and disodium oxalate (MW = 133.99 g/mol) were required to prepare this buffer if the total oxalate concentration is 0.195 M? Oxalic acid has pKa values of 1.250 (pKa1) and 4.266 (pKa2).

Answer 1

Using henderson-hasslebalch equation-

pH = pKa + log[salt]/[acid]

5.043 = 4.266 + log[Na2C2O4]/[H2C2O4.2H2O]

10^0.777 = [Na2C2O4]/[H2C2O4.2H2O]

5.984 = [Na2C2O4]/[H2C2O4.2H2O]

Also given,

[Na2C2O4] + [H2C2O4.2H2O] = 0.195M

Solving these two equations we get,

[H2C2O4.2H2O] = 0.0279M

[Na2C2O4] = 0.1671M

Moles = Molarity *Volume

so

Moles of Na2C2O4 = 0.1671M*1L = 0.1671moles

Moles of H2C2O4.2H2O = 0.0279M*1L = 0.0279moles

Mass in grams = moles*molar mass

Mass of oxalic acid dihydrate(H2C2O4.2H2O) = 0.0279moles*126.07g/mol = 3.517g

Mass of disodium oxalate(Na2C2O4) = 0.1671moles*133.99g/mol = 22.39g