As part of a soil analysis on a plot of land, you want to determine the ammonium
ID: 520332 • Letter: A
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_4Cl. A 5.055-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 NH_4^+ ions completely: B(C_6H_5)_4^- + K^+ rightarrow KB(C_6H_5)_4 (s) B(C_6H_5)_4^- + NH_4^+ rightarrow NH_4B(C_6H_5)_4 (s) The resulting precipitate amounted to 0.265 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 0.113 g of precipitate. Find the mass percentages of NH_4Cl and K_2CO_3 in the original solid.Explanation / Answer
We have,
when NH4 was removed by NH3 formation,
mass of KB(C6H5)4 formed = 0.113 g in 0.250 L aliquot
moles K+ present = 0.113g/358.33 g/mol = 3.15 x 10^-4 mol
So in 0.50 L mols of K+ present = 3.15 x 10^-4 mol x 0.5 L/0.25 L = 6.3 x 10^-4 mol
mass of K2CO3 present = 6.3 x 10^-4 mol x 138.21 g/mol/2 = 0.0435g
mass% K2CO3 in soil sample = 0.0435 g x 100/5.055 g = 0.86%
Total mass of precipitate in 0.125 L aliquot = 0.265 g
So in 0.250 L would be = 0.265 g x 0.250 L/0.125 L = 0.530 g
mass of NH4B(C6H5)4 present = 0.530 - 0.113 = 0.417 g
moles NH4+ present = 0.417 g/337.37 g/mol = 1.24 x 10^-3 mol
moles of NH4+ present in soil sample = 1.24 x 10^-3 mol x 0.5 L/0.25 L = 2.48 x 10^-3 mol
mass of NH4Cl present = 2,48 x 10^-3 mol x 53.492 g/mol = 0.133 g
mass% NH4Cl in soil sample = 0.133 g x 100/5.055 g = 2.63%
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