A 0.573 g portion of an ammonium salt containing a single mole of ammonium ion f

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. 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. From the volume and concentration of oxalic acid used during the back titration, calculate the number of moles of oxalic acid used. 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. 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. 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? 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).

Explanation / Answer

1. Molarity of NaOH = 0.2000 M

Volume of NaOH solution = 50.00 ml = 0.050 L

Number of moles of NaOH = molarity*volume (L) = 0.2000*0.050 = 0.01000 mol

2. Volume of oxalic acid = 26.80 ml = 0.0268 L

Molarity of oxalic acid = 0.1000 M

Number of moles of oxalic acid = molarity*volume (L) = 0.1000*0.0268 = 0.00268 mol

3. Balanced chemical equation,

H2C2O4 + 2NaOH ----------> C2O4Na2 + 2 H2O

Neutralization of 1 mol of oxalic acid needs 2 mol of NaOH.

So, 0.00268 mol of oxalic acid will require (2*0.00268) = 0.00536 mol NaOH

Number of moles of excess hydroxide = 0.00536 mol

4. Number of moles of hydroxide reacted with ammonium

= total number of moles of NaOH - moles of excess hydroxide

= 0.01000- 0.00536 = 0.00464 mol

5. Balanced chemical equation of NaOH with NH4+

OH- + NH4+ ------> NH4OH

1 mol of OH- is used in neutralization of 1 mol of NH4+

So, 0.00464 mol of OH- is used in neutralization of 0.00464 mol of NH4+.

So, number of moles of NH4+ ion in 0.573 g sample is 0.00464 mol.

6. Number of moles of NH4+ = number of moles of ammonium salt = 0.00464 mol

0.00464 mol of salt = 0.573 g

So, 1 mol of salt = 0.573 g/0.00464 mol = 123.49 g/mol

Molar mass of salt = 123.49 g

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