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 5.045-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 0.281 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 0.111 g of precipitate. Find the mass percentages of NH4Cl and K2CO3 in the original solid.

B(C6H5)4^- + K+ → KB(C6H5) (s)
B(C6H5)4^- +NH4 → NH4B (C6H5)4(s)

The resulting precipitate amounted to 0.281 g. A new 300.0-mL aliquot of the original solution was made alkaline and heated to remove all of the NH'* as NH3.The resulting solution was then acidified and excess sodium tetraphenylborate was added to give 0.111 g of precipitate. Find the mass percentages of NH4Cl and K2C03 in the original solid.

Answer 1

Percentage of NH4Cl = 2.36%

Percentage of K2CO3 = 0.707%

Step-by-step explanation:

when 300mL aliquot gives 0.111 g precipitate, then a 150mL will produce precipitate = 0.111g/300 x 150 mL

=0.0555 g.

In the first aliquot, (0.281 g ) - (0.0555 g) = 0.2255 g NH4B(C6H5)4

So, % of NH4Cl = (0.2255 g NH4B(C6H5)4) / (337.27 g NH4B(C6H5)4/mol) x (53.492 g NH4Cl/mol) x (500 mL / 150 mL) / (5.045 g)

= 0.0236 = 2.36%

% of K2CO3 = (0.111 g KB(C6H5)4) / (358.33 g KB(C6H5)4/mol) x (1 mol K2CO3 / 2 mol K) x (138.21 g K2CO3/mol) x (500 mL / 300 mL) / (5.045 g) = 7.07*10^-3 =0.707%