A buffer prepared by dissolving oxalic acid dehydrate (H2_C_2O_4 2H_2O) and diso

Question

A buffer prepared by dissolving oxalic acid dehydrate (H2_C_2O_4 2H_2O) and disodium oxalate (Na_2C_2O_4) in 1.00 L of water has pH of 5.212. How many grams of oxalic acid dehydrate (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 0.260 M? Oxalic acid has pK_a values of 1.250 (pK_a1) and 4,266 (pK_a2). H_2C_2O_4 + C_2O_4^2^- rightarrow 2 HC_2O_4^- We know the solution contains a mixture of C_2O_4^2- and HC_2O_4^- since the pH of the solution is larger than pK_a2 for oxalic acid. The total oxalate concentration is the sum of all oxalate containing species in the solution. Use the Henderson-Hasselbalch equation for the acid-base equilibrium between HC_2O_4^- and C_2O_4^2- and the equation for the total oxalate concentration to solve for the moles of H_2O_4^2- required to prepare the solution.

Explanation / Answer

pH= pKa + log([salt]/[acid])......Henderson–Hasselbalch equation

since oxallic acid is dibasic we are having two pKa, means 1st dissociation of oxallicx acid to HC2O4-,

and second dissociation of HC2O4- to C2O42-

so we have two equation

5.212= 1.25 + log([HC2O4-]/[H2C2O4])

and

5.212= 4.266 + log([C2O42-]/[HC2O4-])

hence

[HC2O4-]/[H2C2O4] = 9.61 x 103

and

[C2O42-]/[HC2O4-] = 8.83

also we know that

[C2O42-] + [HC2O4-] + [H2C2O4] =0.26

on sloving these equations i got

[C2O42-]= 0.233mol ==> 0.234 x 134 = 31.2 g

[HC2O4-]= 0.02645 mol

[H2C2O4]= 2.62x 10-6 ===> 0.33 mg

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