Question

A 0.573 g portion of an ammonium salt containing a single mole of ammonium ion for each mole of the salt was reacted with 50.00 mL of 0.2000 M NaOH. After the reaction the solution was boiled to remove ammonia created during the reaction. The resulting solution was back titrated to the end point with 26.80 mL of 0.1000 M H_2C_2O_4. 1. Calculate the number of moles of hydroxide ion in the 50.00 mL of 0.2000 M sodium hydroxide solution added to the ammonium salt. 2. From the volume and concentration of oxalic acid used during the back titration, calculate the number of moles of oxalic acid used. 3. By using the balanced chemical equation for the back titration [Reaction (14-2)1 and the number of moles of oxalic acid from step 2, calculate the number of moles of excess hydroxide. 4. Using the number of moles of sodium hydroxide from step 1 (the moles of hydroxide initially added to the sample) and the number of moles of excess hydroxide from step 3, calculate the number of moles of hydroxide that reacted with the ammonium salt. 5. Using the number of moles of sodium hydroxide that reacted with the ammonium salt (the result of the calculation in step 4), and the balanced chemical equation for the reaction between hydroxide and the ammonium [Reaction (14-1)], what is the number of moles of ammonium ion in the 0.573 g sample? 6. Since each mole of salt contains one mole of ammonium ion, the answer obtained in step 5 is also the number of moles of ammonium salt. Use the mass of the sample (g) and the number of moles of salt in the sample to calculate the molar mass of the ammonium salt (g/mol).

Answer 1

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⁺ (1)

1. 50.00mL of 0.2000M NaOH contains:

0.05000L × (0.2000 mol / 1L) = 0.01000 moles of NaOH

2. Oxalic acid (H₂C₂O₄) reacts with NaOH, moles of oxalic acid used during titration are:

0.02680L × (0.1000 mol / 1L) = 0.002680 moles of H₂C₂O₄

3. The reaction of oxalic acid with NaOH is:

H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + H₂O (2)

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) = 0.005360 moles of NaOH

4. Thus, moles of NaOH that reacted with ammonium salt are:

0.01000 moles of NaOH - 0.005360 moles of NaOH = 0.004640 moles of NaOH

5. Based in reaction (1), moles of NaOH ≡ moles of NH₄⁺. That means moles of ammonium salt in 0.573g of sample are 0.004640 moles of ammonium salt

6. As moles of ammonium ion are the same than ammonium salt. Molar mass of the ammonium salt is:

0.573g / 0.004640 moles = 123.5 g/mol