Question

As part of a soil analysis on a plot of land, you want to determine the ammonium content using gravimetric analysis with sodium tetraphenylborate, Na B(C6H5)4–. Unfortunately, the amount of potassium, which also precipitates with sodium tetraphenylborate, is non-negligible, and must be accounted for in the analysis. Assume that all potassium in the soil is present as K2CO3, and all ammonium is present as NH4Cl. A 4.935-g soil sample was dissolved to give 0.500 L of solution. A 150.0-mL aliquot was acidified and excess sodium tetraphenylborate was added to precipitate both K and NH4 ions completely:The resulting precipitate amounted to 0.231 g. A new 300.0-mL aliquot of the original solution was made alkaline and heated to remove all of the NH4 as NH3.The resulting solutionwas then acidified and excess sodium tetraphenylborate was added to give 0.157 g of precipitate. Find the mass percentages of NH4Cl and K2CO3 in the original solid.

Answer 1

Write down the ionic reactions taking place.

NH₄⁺ (aq) + B(C₆H₅)₄^- (aq) ---------> NH₄B(C₆H₅)₄ (s) ……(1)

K⁺ (aq) + B(C₆H₅)₄^- (aq) -------> KB(C₆H₅)₄ (s) ……(2)

Again, K₂CO₃ and NH₄Cl are the sources of K⁺ and NH₄⁺. Write down the dissociation reactions.

K₂CO₃ (aq) ------> 2 K⁺ (aq) + CO₃^2- (aq) ……(3)

NH₄Cl (aq) -------> NH₄⁺ (aq) + Cl^- (aq) ……(4)

Excess NaB(C₆H₅)₄ was added to precipitate both NH₄⁺ and K⁺ as their tetraphenylborate salts. Therefore, the combined mass of the precipitates = 0.231 g.

In the next case, the solution was boiled to remove NH₄⁺ as NH₃. NaB(C₆H₅)₄ was added to precipitate K⁺ as KB(C₆H₅)₄; the mass of the precipitate = 0.157 g. The mass of the KB(C₆H₅)₄ formed = 0.157 g.

Molar mass KB(C₆H₅)₄ = 358.3249 g/mol.

Mole(s) of KB(C₆H₅)₄ formed = (0.157 g)/(358.3249 g/mol) = 4.3815*10^-4 mole.

As per the stoichiometry of the reactions (2) and (3) shown above,

1 mole KB(C₆H₅)₄ = 1 mole K⁺.

2 mole K⁺ = 1 mole K₂CO₃

Therefore, 4.3815*10^-4 mole KB(C₆H₅)₄ = 4.3815*10^-4 mole K⁺ = (4.3815*10^-4 mole K⁺)*(1 mole K₂CO₃/2 mole K⁺) = 2.19075*10^-4 mole K₂CO₃.

Molar mass of K₂CO₃ = 138.205 g/mol; therefore, mass of K₂CO₃ in the 300 mL of the sample = (2.19075*10^-4 mole)*(138.205 g/mol) = 0.030278 g.

Mass of K₂CO₃ in 500 mL of the original sample = (0.030278 g)*(500 mL/300 mL) = 0.05046 g.

Mass percentage of K₂CO₃ in the original sample = (0.05046 g)/(4.935 g)*100% = 1.022% (ans).

Mass of NH₄B(C₆H₅)₄ in the sample = (0.231 -1/2*0.157) g (the ½ is added because we take 150 mL aliquot and 0.157 g K salt was obtained from 300 mL aliquot) = 0.1525 g.

Molar mass of NH₄B(C₆H₅)₄ = 337.2651 g/mol.

Moles of NH₄B(C₆H₅)₄ formed = (0.1525 g)/(337.2651 g/mol) = 4.5217*10^-4 mole.

As per stoichiometric equations (1) and (4),

1 mole NH₄B(C₆H₅)₄ = 1 mole NH₄⁺ = 1 mole NH₄Cl.

Therefore, 4.5217*10^-4 mole NH₄B(C₆H₅)₄ = 4.5217*10^-4 mole NH₄Cl.

Molar mass of NH₄Cl = 53.491 g/mol; therefore, mass of NH₄Cl in 150 mL aliquot = (4.5217*10^-4 mole)*(53.491 g/mol) = 0.02419 g.

Mass of NH₄Cl in 500 mL sample = (0.02419 g)*(500 mL/150 mL) = 0.08063 g.

Mass percentage NH₄⁺ in the sample = (0.08063 g)/(4.935 g)*100 = 1.634% (ans).