Answer:
1. 0.01000 moles of NaOH
2. 0.002680 moles of H₂C₂O₄
3. 0.005360 moles of NaOH
4. 0.004640 moles of NaOH
5. 0.004640 moles of ammonium salt
6. 123.5 g/mol
Explanation:
Ammonium ion (NH₄⁺) reacts with NaOH, thus:
NH₄⁺ + NaOH → NH₃ + H₂O + Na⁺ <em>(1)</em>
1. 50.00mL of 0.2000M NaOH contains:
0.05000L × (0.2000 mol / 1L) = <em>0.01000 moles of NaOH</em>
2. Oxalic acid (H₂C₂O₄) reacts with NaOH, moles of oxalic acid used during titration are:
0.02680L × (0.1000 mol / 1L) = <em>0.002680 moles of H₂C₂O₄</em>
3. The reaction of oxalic acid with NaOH is:
H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + H₂O <em>(2)</em>
That means one mole of acid reacts with 2 moles of NaOH. Thus, moles of NaOH that remains after reaction with ammonium salt are:
0.002680 moles of H₂C₂O₄ × (2 mol NaOH / 1 mol H₂C₂O₄) = <em>0.005360 moles of NaOH</em>
4. Thus, moles of NaOH that reacted with ammonium salt are:
0.01000 moles of NaOH - 0.005360 moles of NaOH =<em> 0.004640 moles of NaOH</em>
5. Based in reaction (1), moles of NaOH ≡ moles of NH₄⁺. That means moles of ammonium salt in 0.573g of sample are <em>0.004640 moles of ammonium salt</em>
6. As moles of ammonium ion are the same than ammonium salt. Molar mass of the ammonium salt is:
0.573g / 0.004640 moles = <em>123.5 g/mol</em>