1(a) (15 points) A 0.3600-g mixture of oxalic acid (H2C204, diprotic) and benzoi

Question

1(a) (15 points) A 0.3600-g mixture of oxalic acid (H2C204, diprotic) and benzoic acid (CeHsCOOH, mono- protic) is dissolved in water and then treated with an excess of 0.1000 M NaOH(aq). The resulting basic solution is then titrated with 0.1000 M HCl(aq). If 0.100 L of the NaOH solution was first added and 0.040 L of theHCLsolution was subsequently required to neutralize the solution (ie, bring it to pH = 7), find themass percentage of benzoic acid in the original 0.3600-g mixture. (b) (5 ponts) In answering part (a) one can neglect the re-protonation of the oxalate (C20) and benzoate (C6????-) anions by the HCI in the neutralization process, why? Explain your answer fully. (Hint: Consider the ratios [C2O2-)/[HC201 and [CcH,coo-??6?????? at pH-7.) (c) (15 points) 200 mL of an aqueous solution 0.200 M in ammonium nitrate (NH NO3) is titrated with 0.1000 M NaOH(aq). Determine the pH of the solution (i) before the titration begins and (ii) at the point.

Explanation / Answer

(a) and (b) The no. of moles of NaOH added = 0.1 L * 0.1 mol/L = 0.01 mol

The no. of moles of Na2C2O4 + C6H5COONa = 0.004 mol

The no. of moles of HCl added = 0.04 L * 0.1 mol/L = 0.004 mol

(i) pH = pKa2 + Log([C2O42-]/[HC2O4-])

i.e. 7 = 4.19 + Log([C2O42-]/[HC2O4-])

i.e. Log([C2O42-]/[HC2O4-]) = 7-4.19 = 2.81

i.e. [C2O42-]/[HC2O4-] = 102.81 = 645.65

(ii) pH = pKa + Log([C6H5COO-]/[C6H5COOH])

i.e. 7 = 4.20 + Log([C6H5COO-]/[C6H5COOH])

i.e. Log([C6H5COO-]/[C6H5COOH]) = 7-4.2 = 2.8

i.e. [C6H5COO-]/[C6H5COOH] = 102.8 = 630.96

(c) The no. of moles of ammonium nitrate = 200*10-3 L * 0.2 mol/L = 0.04 mol

(i) Before titration begins:

Formula: pH = 7 - 1/2(pKb + Log[NH4NO3])

i.e. pH = 7 - 1/2 (4.74 + Log 0.04)

i.e. pH = 3.93

(ii) At the equivalence point:

The no. of moles of NH4NO3 = no. of moles of NaOH = 0.04 mol

i.e. The volume of NaOH solution = 0.04 mol/(0.1 mol/L) = 0.4 L

i.e. The no. of moles of NH4OH produced = 0.04 mol, [NH4OH] = 0.04 mol/(0.4+0.2) L = 0.0667 M

Formula: pOH = 1/2 (pKb - Log[NH4OH])

i.e. pOH = 1/2 (4.74 - Log 0.0667)

i.e. pOH = 2.96

pH + pOH = 14

i.e. pH + 2.96 = 14

i.e. pH = 14-2.96 = 11.04

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