A buffer prepared by dissolving oxalic acid dihydrate (H_2C_2O_4 middot 2H_2O) a
ID: 491165 • Letter: A
Question
A buffer prepared by dissolving oxalic acid dihydrate (H_2C_2O_4 middot 2H_2O) and disodium oxalate (Na_2C_2O_4) in 1.00 L of water has a pH of 5.334. How many grams of oxalic add 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 0.288 M? Oxalic acid has pK_a values of 1.250 (pK_a1) and 4.266 (pK_a2). Mass of H_2C_2O_4 middot 2H_2O = Incorrect. When H_2C_2O_4 and C_2O_4^2- react, 2 moles of HC_2O_4^- are formed, not one. there are x moles of H_2C_2O_4 in solution and y moles of C_2O_4^2- in solution. We know the final solution contains a mixture of C_2O_4^2- and HC_2O_4^- since the pH (5.334) is larger than pK_a2 for oxalic acid. This means that all the H_2C_2O_4 is converted to HC_2O_4^- and the amount of C_2O_4^2- is reduced by x moles. Substitute these expressions for the final moles of C_2O_4^2- and HC_2O_4^- into the Henderson - Hasselbalch equation.Explanation / Answer
According to Hinderson -Hasselbalch equation
pH = pKa + log [Na2C2O4/h2C2O4]
As the pH of the solution is greater than the second ionization of oxalic acid, so we have to consider pka2
5.334 = 4.266 + log [Na2C2O4/h2C2O4]
or, [Na2C2O4/h2C2O4] = 11.69
Calculate the decimal fraction
[Na2C2O4] = 11.69/1+11.69 = 0.92
[H2C2O4] = 1/1+11.69 = 0.079
Molarity of Na2C2O4 = 0.92 *0.288 M = 0.265 M
moles of Na2C2O4 = 0.265 M * 1L = 0.265 mol
mass of Na2C2O4 = 0.265 mol * 133.99g/mol = 35.5 gm
[H2C2O4] = 0.079 * 0.288 = 0.022 M
moles of H2C2O4 = 0.22 mol
mass of H2C2O4 =126.07 g/mol * 0.22 mol = 2.83 gm
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