As part of a soil analysis on a plot of land, you want to determine the ammonium

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 4.925-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 and NH4 ions completely:

B(C6H5)4- + K+ -> KB(C6H5)4(s)

B(C6H5)4- + NH4+ -> NH4B(C6H5)4(s)

The resulting precipitate amounted to 0.211 g. A new 200.0-mL aliquot of the original solution was made alkaline and heated to remove all of the NH4 as NH3.The resulting solution was then acidified and excess sodium tetraphenylborate was added to give 0.193 g of precipitate. Find the mass percentages of NH4Cl and K2CO3 in the original solid.

Explanation / Answer

If a 200.0 mL aliquot produced 0.193 g of KB(C6H5)4, then a 100.0 mL aliquot would produce half that much, namely 0.0675 g of KB(C6H5)4.

So in the first aliquot:
(0.211 g total) - (0.0675 g KB(C6H5)4) = 0.1435 g NH4B(C6H5)4

(0.1435 g NH4B(C6H5)4) / (337.2679 g NH4B(C6H5)4/mol) x (1 mol NH4{+} / 1 mol NH4B(C6H5)4) x
(1 mol NH4Cl / 1 mol NH4{+}) x (53.4917 g NH4Cl/mol) x (500 mL / 125.0 mL) / (5.015 g) = 0.0177 =
1.77% NH4Cl

(0.193 g KB(C6H5)4) / (358.3277 g KB(C6H5)4/mol) x (1 mol K / 1 mol KB(C6H5)4) x
(1 mol K2CO3 / 2 mol K) x (138.20569 g K2CO3/mol) x (500 mL / 250.0 mL) / 4.925 g) = 0.1246 =
1.24% K2CO3

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