As part of a soil analysis on a plot of land, you want to determine the ammonium
ID: 530460 • 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 so is present as K_2CO_3, and all ammonium is present as NH_4Cl. A 5.025-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_4B(C_6H_5)_4(s) The resulting precipitate amounted to 0.233 g. A new 300.0-mL aliquot of the original solution was made alkaline and heated to remove all of the NH_46+ as NH_3. The resulting solution was then acidified and excess sodium tetraphenylborate was added to give 0.193 g of precipitate Find the mass percentages of NH_4Cl and K_2CO_3 in the original solid Number 6.26 % NH_4Cl Number 24.74 % K_2CO_3Explanation / Answer
Ans. Part 1: All NH4+ is removed as NH3
Mass of KB(C6H5)4 precipitated = 0.193 g
Moles of KB(C6H5)4 precipitated =Mass / Molar mass
= 0.193 g / (358.33 g/ mol)
= 5.3861 x 10-4 mol
K2CO3 ----------> 2K+ + CO32-
1 mol K2CO3 yields 2 mol K+. Or, 1 mol K+ is equivalent to 0.5 mol K2CO3.
See the stoichiometry, 1 mol K+ (= 0.5 mol K2CO3) forms 1 mol precipitate of KB(C6H5)4. Thus, 300.0 mL of original aliquot consists of 5.3861 x 10-4 mol K+ or 2.6931 x 10-4 mol K2CO3(= 0.5 x 5.3861 x 10-4 mol).
So,
Moles of K+ in 150.0 mL aliquot = (1/2) x moles of K+ in 300.0 mL aliquot
= (1/2) x 5.3861 x 10-4 mol
= 2.6931 x 10-4 mol
And, moles of K2CO3 in 150.0 mL aliquot = (1/2) x mol of K2CO3 in 300.0 mL aliquot
= (1/2) x 2.6931 x 10-4 mol K2CO3
= 1.34655 x 10-4 mol
Similarly,
Moles of KB(C6H5)4 precipitated in 150.0 mL= (1/2) x moles of Ppt. in 300.0 mL aliquot
= (1/2) x 5.3861 x 10-4 mol = 2.6931 x 10-4 mol
Therefore, 150.0 mL aliquot would form 2.6931 x 10-4 mol ppt. of KB(C6H5)4.
Part B:
Total mass of ppt. in 150.0 mL aliquot = mass of KB(C6H5)4 + Mass of NH4B(C6H5)4
Or, 0.233 g = (Moles x Molar mass) of KB(C6H5)4 + Mass of NH4B(C6H5)4
Or, 0.233 g = [2.6931 x 10-4 mol x (358.33 g/ mol)]+ Mass of NH4B(C6H5)4
Or, 0.233 g - 0.0965 g = Mass of NH4B(C6H5)4
Or, Mass of NH4B(C6H5)4 = 0.1365 g
Therefore, mass of NH4B(C6H5)4 in precipitate = 0.1365 g
Moles of NH4B(C6H5)4 = Mass / Molar mass
= 0.1365 g / (337.27 g/mol)
= 4.0472 x 10-4 mol
Now,
1 mol NH4Cl yields 1 mol NH4+.
1 mol NH4+ forms 1 mol NH4B(C6H5)4 precipitate.
So, moles of NH4+ in 150.0 mL aliquot = Moles of NH4B(C6H5)4 = 4.0472 x 10-4 mol
Part C: In 150.0 mL original aliquot –
Moles of K2CO3 in 150.0 mL aliquot = 1.34655 x 10-4 mol - from Part A
Moles of NH4+ in 150.0 mL aliquot = 4.0472 x 10-4 mol - from Part B
Mass of K2CO3 in 150 mL = 1.34655 x 10-4 mol x (138.21 g/ mol)
= 0.01861 g
Mass of K2CO3 in 500 mL = (500 mL/ 150 mL) x 0.01861 g
= 0.0620 g
Since, original soil sample is diluted to 500.0 mL, the amount of K2CO3 in original soil sample is equal to its amount in 500.0 mL aliquot.
Therefore, mass of Mass of K2CO3 in soil = 0.0620 g
% K2CO3 in soil = (Mass of K2CO3 / Mass of soil sample) x 100
= (0.0620 g / 5.025 g) x 100
= 1.234 %
And,
Mass of NH4Cl in 150 mL = 4.0472 x 10-4 mol x (53.492 g/ mol)
= 0.0216 g
Mass of NH4Cl in 500 mL = (500 mL/ 150 mL) x 0.0216 g
= 0.0720 g
% NH4Cl in soil = (Mass of NH4Cl / Mass of soil sample) x 100
= (0.0720 g / 5.025 g) x 100
= 1.433 %
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