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_6 H_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_2 CO_3, and all ammonium is present as NH_4 CI. 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_6 H_5)_4(s) B(C_6 H_5)^-_4 + NH^+_4 rightarrow NH_4 B(C_6 H_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_4^+ 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_4 CI and K_2 CO_3 in the original solid.

Explanation / Answer

Let the sample contain x g K2CO3 and y g NH4Cl.

Moles of K2CO3 in the sample = (x g)/(138.21 g/mol) = (x/138.21) mole.

Moles of NH4Cl in the sample = (y g)/(53.492 g/mol) = (y/53.492) mole.

Consider the dissociation of K2CO3 and NH4Cl as below:

K2CO3 -------> 2 K+ + CO32-

1 mole K2CO3 = 2 moles K+

Therefore, (x/138.21) mole K2CO3 = (2x/138.21) mole K+

NH4Cl ------> NH4+ + Cl-

1 mole NH4Cl = 1 mole NH4+

Therefore, (y/53.492) mole NH4Cl = (y/53.492) mole NH4+

Note down the reactions with tetraphenylborate as below.

B(C6H5)4- + K+ ------> KB(C6H5)4

1 mole K+ = 1 mole KB(C6H5)4

Therefore, (2x/138.21) mole K+ = (2x/138.21) mole KB(C6H5)4

Mass of KB(C6H5)4 produced = (2x/138.21)*(358.33) g

B(C6H5)4 + NH4+ ------> NH4B(C6H5)4

1 mole NH4+ = 1 mole NH4B(C6H5)4

Therefore, (y/53.492) mole K+ = (y/53.492) mole KB(C6H5)4

Mass of KB(C6H5)4 produced = (y/53.492)*(337.27) g

As per the problem,

(2x/138.21)*(358.33) + (y/53.492)*(337.27) = 0.233 ……(1)

Again, we have,

(2x/138.21)*(358.33) = 0.193

===> x = 0.193*138.21/(2*358.33) = 0.0372

Put the value of x in (1) and obtain

(2*0.0372/138.21)*(358.33) + (y/53.492)*(337.27) = 0.233

===> 0.1928 + (y/53.492)*(337.27) = 0.233

===> (y/53.492)*(337.27) = 0.0402

===> y = 0.0402*53.492/337.27 = 0.00637

The sample contains 0.0372 g K2CO3 and 0.00637 g NH4Cl.

% K2CO3 = (0.0372 g)/(5.025 g)*100 = 0.740 (ans)

% NH4Cl = (0.00637 g)/(5.025 g)*100 = 0.1267 0.127 (ans).

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