JUST PART F: (1) (24 points) Oxalic acid, which is formulated as either (COOH) o
ID: 542107 • Letter: J
Question
JUST PART F:
(1) (24 points) Oxalic acid, which is formulated as either (COOH) or H2C204, is a diprotic acid. At 25°C, Kal 5.9 × 10-2 and Ka2 = 6.4 x 10-5 (a) (4 pts) Draw a Lewis structure for oxalic acid. Briefly explain why oxalic acid behaves as a diprotic acid in water (b) (4 pts) Would a 0.1 M sodium hydrogen oxalate solution, NaHC204(aq), be acidic, neutral, or basic at 25°C? Explain your choice. (c) (4 pts) If oxalic acid is used to prepare a buffer solution with pH = 1.5, and the concentration of oxalic acid is 0.22 M, what are the concentrations (in M) of the two other species derived from oxalic acid in this buffer solution? (d) (4 pts) Using just oxalic acid, which is a solid, and stock solutions of either 3.00 M HCl(aq) or 3.00 M NaOH(aq), how many grams of oxalic acid and milliliters of stock solution are needed to prepare 1.00 L of a buffer that has pH = 1.5 and contains 0.22 M oxalic acid? (e) (4 pts) If sodium oxalate is used to prepare a buffer solution with pH 4.0, and the concentration of the oxalate ion is 0.16 M, what are the concentrations of the two other species derived from the oxalate ion in this buffer solution? (4 pts) Using just sodium oxalate and stock solutions of either 3.00 M HCI(aq) or 3.00 M NaOH(aq), how many grams of sodium oxalate and milliliters of stock solution are needed to prepare 1.00 L of a buffer that has pH 4.0 and contains 0.16 M oxalate ion? (f)Explanation / Answer
f) Consider the step-wise dissociation of oxalic acid, H2C2O4 as below.
H2C2O4 (aq) -------> H+ (aq) + HC2O4- (aq); Ka1 = 5.9*10-2; pKa1 = 1.229 1.23 …..(1)
HC2O4- (aq) --------> H+ (aq) + C2O4- (aq); Ka2 = 6.4*10-5; pKa2 = 4.1938 4.19 …..(2)
We wish to prepare a buffer at pH 4.0; hence, we will consider the second reaction. Use the Henderson-Hasslebach equation to calculate the concentration of the weak acid (HC2O4-) and the conjugate base (C2O42-, supplied by sodium oxalate, Na2C2O4).
pH = pKa2 + log [C2O42-]/[HC2O4-]
====> 4.0 = 4.19 + log [C2O42-]/[HC2O4-]
====> -0.19 = log [C2O42-]/[HC2O4-]
====> [C2O42-]/[HC2O4-] = antilog (-0.19) = 0.6456
====> [C2O42-] = 0.6456*[HC2O4-]
Again, we have [C2O42-] + [HC2O4-] = 0.16 M
====> 0.6456*[HC2O4-] + [HC2O4-] = 0.16 M
====> 1.6456*[HC2O4-] = 0.16 M
====> [HC2O4-] = (0.16 M)/(1.6456) = 0.09722 M 0.097 M (ans).
Therefore, [C2O42-] = 0.6456*[HC2O4-] = 0.6456*(0.097 M) = 0.06262 M 0.063 M (ans).
We wish to prepare 1.0 L of the buffer; hence the number of moles of HC2O4- in the buffer = (volume of buffer)*(molarity of HC2O4-) = (1.0 L)*(0.097 M)*(1 mol/L/1 M) = 0.097 mole.
HC2O4- is obtained by the reaction between C2O42- and HCl as per the reaction below.
C2O42- (aq) + HCl (aq) --------> HC2O4- (aq) + Cl- (aq)
As per the stoichiometric equation,
1 mole C2O42- = 1 mole HCl = 1 mole HC2O4-.
Therefore, mole(s) HCl required to form 0.097 mole HC2O4- = 0.097 mole.
Volume of 3.0 M HCl required = (mole HCl)/(molarity of stock HCl) = (0.097 mole)/(3.0 M) = 0.032333 L = (0.032333 L)*(1000 mL/1 L) = 32.333 mL 32.33 mL (ans).
Moles C2O42- present in the buffer = (1.0 L)*(0.063 M)*(1 mol/L/1 M) = 0.063 mole.
Moles Na2C2O4 taken = (moles C2O42-) + (moles HC2O4-) = (0.063 mole) + (0.097 mole) = 0.160 mole.
Molar mass of Na2C2O4 = 134 g/mol.
Mass of Na2C2O4 dissolved in 1.0 L solution = (0.16 mole)*(134 g/mol) = 21.44 g (ans).
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