As part of a soil analysis on a plot of land, a scientist wants 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 5.025 g soil sample was dissolved to give 0.500 L of solution. A 100.0 mL aliquot was acidified and excess sodium tetraphenylborate was added to precipitate both K+ K + and NH+4 ions completely.
Compound Formula mass
B(C6H5)−4+K+⟶KB(C6H5)4(s) KB(C6H5)4 358.33 g/mol
B(C6H5)−4+NH+4⟶NH4B(C6H5)4(s) NH4B(C6H5)4 337.27 g/mol
K2CO3 138.21 g/mol
NH4Cl 53.492 g/mol
The resulting precipitate amounted to 0.277 g. A new 200.0 mL aliquot of the original solution was made alkaline and heated to remove all of the NH+4 as NH3. The resulting solution was then acidified, and excess sodium tetraphenylborate was added to give 0.105 g of precipitate.
Find the mass percentages of NH4Cl and K2CO3 in the original solid.
%NH4CL
%K2CO3

Mass percentage of NH₄Cl = 3.54%

Mass percentage of K₂CO₃ = 1.01%

Explanation:

If a 200.0 mL aliquot produced  0.105 g of KB(C₆H₅)₄, then a 100.0 mL aliquot would produce 1/2 * 0.105 g = 0.0525 g of KB(C₆H₅)₄.

Therefore, mass of NH₄B(C₆H₅)₄ in the 100.0 ml aliquot = (0.277 - 0.0525)g = 0.2245 g

Number of moles of NH₄B(C₆H₅)₄ in 0.2245 g = 0.2245 g/ 337.27 g/mol = 0.0006656 moles

In 500 ml solution, number of moles present = 0.0006656 * 500/100 = 0.003328 moles.

From equation of the reaction; mole ratio of  NH₄⁺ and NH₄B(C₆H₅)₄ = 1:1

Similarly, mole ratio of  NH₄⁺ and NH₄Cl = 1:1

Therefore, moles of NH₄Cl in 500 ml sample = 0.003328 moles

Mass of NH₄Cl  = 0.003328 mol * 53.492 g/mol = 0.178 g

Mass percentage of NH₄Cl = (0.178/5.025) * 100% = 3.54%

Number of moles of KB(C₆H₅)₄ in 0.105 g (precipitated from 200.0 ml aliquot) = 0.105 g/ 358.33 g/mol = 0.000293 moles

In 500 ml solution, number of moles present = 0.000293 * 500/200 = 0.0007326 moles.

From equation of the reaction; mole ratio of  K⁺ and KB(C₆H₅)₄ = 1:1

Similarly, mole ratio of  K⁺ and K₂CO₃ = 2:1

Therefore, moles of K₂CO₃ in 500 ml sample = 0.0007326/2 moles =  0.0003663 moles

Mass of  K₂CO₃ = 0.0003663 mol * 138.21 g/mol = 0.05063 g

Mass percentage of K₂CO₃ = (0.05063/5.025) * 100% = 1.01%

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.