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(C_6H_5)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 K_2CO_3, and all ammonium is present as NH_4CI. A 5.075-g soil sample was dissolved to give 0.500 L of solution. A 125.0-mL aliquot was acidified and excess sodium tetraphenylborate was added to precipitate both K^+ and NF_4^+ ions completely: The resulting precipitate amounted to 0.287 g. A new 250.0-mL aliquot of the original solution was made alkaline and heated to remove all of the NH_4^+ as NH_3.The resulting solution was then acidified and excess sodium tetraphenylborate was added to give 0.107 g of precipitate. Find the mass percentages of NH_4CI and K_2CO_3 in the original solid.

Explanation / Answer

If a 250.0 mL aliquot produced 0.107 g of KB(C6H5)4, then a 125.0 mL aliquot would produce half that much, namely 0.0535 g of KB(C6H5)4.

So in the first aliquot:
(0.287 g total) - (0.0535 g KB(C6H5)4) = 0.2335 g NH4B(C6H5)4

(0.2335 g NH4B(C6H5)4) / (337.27 g NH4B(C6H5)4/mol) x (1 mol NH4{+} / 1 mol NH4B(C6H5)4) x
(1 mol NH4Cl / 1 mol NH4{+}) x (53.492 g NH4Cl/mol) x (500 mL / 125.0 mL) / (5.075 g) = 0.0291 =
2.91% NH4Cl

(0.107 g KB(C6H5)4) / (358.33 g KB(C6H5)4/mol) x (1 mol K / 1 mol KB(C6H5)4) x
(1 mol K2CO3 / 2 mol K) x (138.21 g K2CO3/mol) x (500 mL / 250.0 mL) / (5.075 g) = 8.13*10^0-3 =
0.81% K2CO3

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