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 150.0-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^- rightarrow NH_4B(C_6H_5)_4(s) The resulting precipitate amounted to 0.285 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 0.167 g of precipitate. Find the mass percentages of NH_4Cl and K_2CO_3 in the original solid.

Explanation / Answer

m = 5.075 g of sample in V = 0.5 L

then, V = 150 mL is taken.. i.e. 150/500 = 0.3, for 0.3*5.075 = 1.5225 g of soil in 150 mL

excess sodium is precipitated later...

m = 0.285 g of KB(C6H5)4

mol = mass/MW = 0.285/358.33= 0.0007953

so... ratio is 1:1 so we have 0.0007953 mol of B(C6H5)4-

this is in 150 mL..

so x = 150/500 = 0.3

C = 0.0007953 /0.3 = 0.002651 mol in 500 mL... or 5.075 g

so

mol of K --> 0.002651

mol of K2CO3 = 0.002651/2 = 0.0013255

mass of K2CO3 = 138.21*0.0013255 = 0.18319 g

% K2CO3 = 0.18319 / 5.075 *100 = 3.60965%

Q2)

V = 300 mL of sample so 300/500 = 0.6 of sample... 0.6*5.075 = 3.045 g of soil...

mass of precip = 0.167

mol = 0.167/337.27 = 0.000495 mol of NH4+

so..

0.000495 mol are in 300 mL...

in 500 mL = 500/300*0.000495 = 0.000825 mol of NH4+

then

ratio is 1:1 so NH4Cl = 0.000825 as well

so

mass = mol*MW = 0.000825*53.492 = 0.0441309 g of NH4Cl

so

% mass = 0.0441309/5.075*100 = 0.8695743 %

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