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(C6H5)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 K2CO3, and all ammonium is present as NH4Cl. 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 NH4 ions completely:The resulting precipitate amounted to 0.209 g. A new 300.0-mL aliquot of the original solution was made alkaline and heated to remove all of the NH4 as NH3.The resulting solution was then acidified and excess sodium tetraphenylborate was added to give 0.143 g of precipitate. Find the mass percentages of NH4Cl and K2CO3 in the original solid.

Explanation / Answer

In the second aliquot (300.0 mL), when the solution was made alkaline and heated to remove all of the NH4 as NH3, only K+(aq) will be remained in the solution.

When sodium tetraphenylborate reacts with K+ ion it forms potassium tetraphenylborate.

NaB(C6H5)4 + K+(aq) ---- > KB(C6H5)4 (s)(ppt)+ Na+ (aq)

1 mol 1 mol 1 mol 1 mol

When NH4+(aq)reacts with sodium tetraphenylborate it forms ammonium tetraphenylborate.

NaB(C6H5)4 + NH4+(aq) ---- > NH4B(C6H5)4 (s)(ppt)+ Na+ (aq)

1 mol 1 mol 1 mol 1 mol

Since there is only  K+(aq) in the second aliquot (300.0mL), the 0.143 g of precipitate formed will be only due to  KB(C6H5)4 (s).

Hence mass of  KB(C6H5)4 (s) formed in the second aliquot = 0.143 g

Molecular mass of  KB(C6H5)4 (s) = 358 gmol-1

Atomic mass of K = 39.10 g

Hence mass of potassium present in  0.143 g of KB(C6H5)4 (s)

=  (0.143 g of KB(C6H5)4 ) x (39.10 g K / 358 g  KB(C6H5)4 ) = 0.0156 g K

Moles of K present  in  0.143 g of KB(C6H5)4 (s) = 0.0156 g K / 39.10 gmol-1  = 3.99x10-4 mol

2 moles of K is present in 1 mol of K2CO3.

Hence moles of K2CO3 present in 300 mL of aliquot = 3.99x10-4 mol / 2 = 2.00x10-4 mol K2CO3.

Molecular mass of K2CO3 = 138.2 gmol-1

Hence mass of K2CO3 in 300 mL of aliquot = 2.00x10-4 mol K2CO3 x138.2 gmol-1 = 0.0276 g K2CO3

Mass of soil present in 300.0 mL of aliquot = 4.985 g soil x (300.0 mL / 500.0 mL) = 2.991 g soil

Hence mass percentage of K2CO3 in the soil = (0.0276 g K2CO3 / 2.991 g soil) x 100

= 0.923 % K2CO3 (answer)

300.0 mL of the aliquot gives 0.143 g precipitate of  KB(C6H5)4 (s)

Hence the mass of KB(C6H5)4 (s) that would be precipitated by 150.0 mL of aliquot taken initially

= ( 0.143 g KB(C6H5)4 (s) ) x (150.0 mL / 300.0 mL = 0.0715 g KB(C6H5)4 (s)

0.209g of the ppt formed by 150.0 mL of aliquot contains both KB(C6H5)4 (s) and NH4B(C6H5)4 (s)

Hence the mass of NH4B(C6H5)4 (s) present in 0.209 g of ppt = 0.209g - 0.0715 g = 0.137 g NH4B(C6H5)4 (s)

Molecular mass of  NH4B(C6H5)4 (s) = 337.26 g/mol

mass of NH4+ present in  0.137 g NH4B(C6H5)4 (s)

= 0.137g  NH4B(C6H5)4 (s) x (18 g NH4+ / 337.26 g NH4B(C6H5)4 (s) ) = 0.00731 g NH4+

Moles of NH4+ =  0.00731 g NH4+ / 18 gmol-1  = 4.06x10-4 mol NH4+

Hence moles of NH4Cl = 4.06x10-4 mol NH4Cl

Molecular mass of NH4Cl = 53.5 g/mol

Hence mass of NH4Cl present in 150.0 mL aliquot = 4.06x10-4 mol NH4Cl x 53.5 g/mol = 0.0217 g NH4Cl

Mass of soil present in 150.0 mL aliquot = 4.985 gx (150.0mL / 500.0 mL) = 1.50 g soil

Hence mass percentage of NH4Cl = (0.0217 g NH4Cl / 1.50 g soil) x 100 = 1.45 % NH4Cl (answer)

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