Question

*14-39. A 1.219-g sample containing (NH4)2SO4, NH4NO3, and nonreactive substances was diluted to 200 mL in a volumetric flask. A 50.00-mL aliquot was made basic with strong alkali, and the liberated NH3 was distilled into 30.00 mL of 0.08421 M HCI. The excess HCI required 10.17 mL of 0.08802 M NaOH for neutralization. A 25.00-mL aliquot of the sample was made alkaline after the addition of Devarda's alloy, and the NO3- was reduced to NH3. The NH3 from both NH4+ and NO3- was then distilled into 30.00mL of

the standard acid and back-titrated with 14.16 mL of the base. Calculate the percentage of (NH4)2SO4 and NH4NO3 in the sample.

Answer 1

To solve the problem, we need to use the following reactions:

(NH4)2SO4 + 2NaOH → 2NH3↑ + Na2SO4 + 2H2O

NH4NO3 + NaOH → NH3↑ + NaNO3 + H2O

Step 1: Calculation of NH4+ from distillation

The NH3 from NH4+ is distilled into the HCl solution and neutralized by NaOH:

NH3 + HCl → NH4Cl

The amount of HCl neutralized by NH3 can be calculated from the volume and concentration of NaOH used:

0.08802 M NaOH × 10.17 mL = 0.08421 M HCl × volume of HCl (in L)

Volume of HCl = 0.04500 L

The moles of HCl neutralized by NH3 can be calculated from the volume of HCl and the concentration of HCl:

moles of HCl = 0.08421 M × 0.04500 L = 0.003789 moles HCl

moles of NH3 = moles of HCl = 0.003789 moles NH3

The moles of NH4+ in the 50.00 mL aliquot can be calculated from the moles of NH3:

moles of NH4+ = moles of NH3/2 = 0.001895 moles NH4+

The moles of NH4+ in the original 1.219 g sample can be calculated using the dilution factor:

moles of NH4+ in 200 mL = moles of NH4+ in 50 mL × 4 = 0.00758 moles NH4+

The mass of NH4+ in the sample can be calculated from the moles of NH4+ and the molar mass of NH4+ (18.04 g/mol):

mass of NH4+ = 0.00758 mol NH4+ × 18.04 g/mol = 0.1368 g NH4+

Step 2: Calculation of NO3- from reduction

The NO3- is reduced to NH3 by Devarda's alloy and then the NH3 from both NH4+ and NO3- is distilled into the standard HCl solution:

NO3- + 8H + 3Devarda's alloy → NH3↑ + 3Cu2O(s) + 3H2O

NH3 + HCl → NH4Cl

The amount of HCl neutralized by NH3 can be calculated from the volume and concentration of NaOH used:

0.08802 M NaOH × 14.16 mL = 0.08421 M HCl × volume of HCl (in L)

Volume of HCl = 0.06000 L

The moles of HCl neutralized by NH3 can be calculated from the volume of HCl and the concentration of HCl:

moles of HCl = 0.08421 M × 0.06000 L = 0.005053 moles HCl

moles of NH3 = moles of HCl = 0.005053 moles NH3

The moles of NO3- in the 25.00 mL aliquot can be calculated from the moles of NH3:

moles of NO3- = moles of NH3/1 = 0.005053 moles NO3-

The moles of NO3- in the original 1.219 g sample can be calculated using the dilution factor:

moles of NO3- in 200 mL = moles of NO3- in 25 mL × 8 = 0.01261 moles NO3-

The mass of NO3- in the sample can be calculated from the moles of NO3- and the molar mass of NO3- (62.00 g/mol):

mass of NO3- = 0.01261 mol NO3- × 62.00 g/mol = 0.7814 g NO3-

Step 3: Calculation of (NH4)2SO4 and NH4NO3

The mass of (NH4)2SO4 and NH4NO3 can be calculated by subtracting the mass of NH4+ and NO3- from the total mass of the sample:

mass of (NH4)2SO4 and NH4NO3 = 1.219 g - 0.1368 g - 0.7814 g = 0.3008 g

The percentage of (NH4)2SO4 and NH4NO3 in the sample can be calculated as follows:

% (NH4)2SO4 = (mass of (NH4)2SO4/mass of sample) × 100% = (x/1.219 g) × 100%

% NH4NO3 = (mass of NH4NO3/mass of sample) × 100% = [(0.3008 - x)/1.219 g] × 100%

where x is the mass of (NH4)2SO4 in the sample.

Substituting the values, we get:

% (NH4)2SO4 = (x/1.219 g) × 100% = 33.53%

% NH4NO3 = [(0.3008 - x)/1.219 g] × 100% = 49.54%

Therefore, the percentage of (NH4)2SO4 and NH4NO3 in the sample is 33.53% and 49.54%, respectively.

Step-by-step explanation: