As part of a soil analysis on a plot of land, you want to determine the ammonium
ID: 1059698 • 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_4CI. A 4.985-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 NH_4^+ ions completely: B[C_6 H_5]_4^- + K^+ rightarrow KB [C_6H_5]_4(s) B[C_6 H_5]_4^- + NH_4^+ rightarrow NH_4 B [C_6H_5]_4 (s) The resulting precipitate amounted to 0.257 g. A new 300.0-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 0.185 g of precipitate. Find the mass percentages of NH_4CI and K_2CO_3 in the original solid.Explanation / Answer
Calculation of % of K2CO3
In the second experiment there will be only KB(C6H5)4 because NH4+ has already be removed by making it alkaline and the heating the solution.
Weight of KB(C6H5)4 obtained = 0.185 g
Formula mass of KB(C6H5)4 = 358.33; Formula weight of K2CO3 = 138.21
Since 500 mL solution contains soil sample: 4.98 g
Therefore, 300 mL will of solution will contain: 4.98 x 300/500 = 2.988 g
Since 358.33 g of KB(C6H5)4 can be formed from 138.21 g of K2CO3
So, 0.185 g would have come from: 0.185 x 138.21/358.33 = 0.071 g of K2CO3
Therefore, % K2CO3 will be: 100 x calculated weight of K2CO3/weight of sample = 100 x 0.071/2.988
= 2.38 %
Calculation of % of NH4Cl
Amount of soil sample dissolved in 150 mL aliquot = 150 x 4.98/500 = 1.494 g
Amount of precipitate (KB(C6H5)4 + NH4B(C6H5)4 = 0.257 g
Since 2.988 g o soil sample gives 0.185 g of KB(C6H5)4 (From above experiment)
Therefore, 1.494 g of soil sample will give: 1.494 x 0.185/2.988 = 0.0925 g of KB(C6H5)4
Therefore, amount of NH4B(C6H5)4 obtained = 0.257 – 0.0925 = 0.1645 g
Since, 337.27 g NH4B(C6H5)4 can be obtained from 53.492 g of NH4Cl
Therefore, 0.1645 g of NH4B(C6H5)4 would have been obtained from:
0.1645 x 53.492/337.27 = 0.026 g NH4Cl
So, % of NH4Cl = 100 x weight of NH4Cl/Weight of sample = 100 x 0.026/1.494 = 1.74%
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.