As part of a soil analysis on a plot of land, a scientist wants 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 5.025 g soil sample was dissolved to give 0.500 L of solution. A 100.0 mL aliquot was acidified and excess sodium tetraphenylborate was added to precipitate both K+ K + and NH+4 ions completely.
Compound Formula mass
B(C6H5)−4+K+⟶KB(C6H5)4(s) KB(C6H5)4 358.33 g/mol
B(C6H5)−4+NH+4⟶NH4B(C6H5)4(s) NH4B(C6H5)4 337.27 g/mol
K2CO3 138.21 g/mol
NH4Cl 53.492 g/mol
The resulting precipitate amounted to 0.277 g. A new 200.0 mL aliquot of the original solution was made alkaline and heated to remove all of the NH+4 as NH3. The resulting solution was then acidified, and excess sodium tetraphenylborate was added to give 0.105 g of precipitate.
Find the mass percentages of NH4Cl and K2CO3 in the original solid.
%NH4CL
%K2CO3

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mickymike92 mickymike92 · Answer 1 · 2020-10-15T00:11:54+02:00

Answer:

Mass percentage of NH₄Cl = 3.54%

Mass percentage of K₂CO₃ = 1.01%

Explanation:

If a 200.0 mL aliquot produced 0.105 g of KB(C₆H₅)₄, then a 100.0 mL aliquot would produce 1/2 * 0.105 g = 0.0525 g of KB(C₆H₅)₄.

Therefore, mass of NH₄B(C₆H₅)₄ in the 100.0 ml aliquot = (0.277 - 0.0525)g = 0.2245 g

Number of moles of NH₄B(C₆H₅)₄ in 0.2245 g = 0.2245 g/ 337.27 g/mol = 0.0006656 moles

In 500 ml solution, number of moles present = 0.0006656 * 500/100 = 0.003328 moles.

From equation of the reaction; mole ratio of NH₄⁺ and NH₄B(C₆H₅)₄ = 1:1

Similarly, mole ratio of NH₄⁺ and NH₄Cl = 1:1

Therefore, moles of NH₄Cl in 500 ml sample = 0.003328 moles

Mass of NH₄Cl = 0.003328 mol * 53.492 g/mol = 0.178 g

Mass percentage of NH₄Cl = (0.178/5.025) * 100% = 3.54%

Number of moles of KB(C₆H₅)₄ in 0.105 g (precipitated from 200.0 ml aliquot) = 0.105 g/ 358.33 g/mol = 0.000293 moles

In 500 ml solution, number of moles present = 0.000293 * 500/200 = 0.0007326 moles.

From equation of the reaction; mole ratio of K⁺ and KB(C₆H₅)₄ = 1:1

Similarly, mole ratio of K⁺ and K₂CO₃ = 2:1

Therefore, moles of K₂CO₃ in 500 ml sample = 0.0007326/2 moles = 0.0003663 moles

Mass of K₂CO₃ = 0.0003663 mol * 138.21 g/mol = 0.05063 g

Mass percentage of K₂CO₃ = (0.05063/5.025) * 100% = 1.01%