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 4.975-g soil sample was dissolved to give 0.500 L of solution. A 125.0-mL aliquot was acidified and excess sodium tetraphenylborate was added to precipitate both K^+ and NH_4^+ ions completely: The resulting precipitate amounted to 0.249 g. A new 250.0-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 0.131 g of precipitate. Find the mass percentages of NH_4CI and K_2CO_3 in the original solid.

Explanation / Answer

0.131 g precipitate is nothing but KB(C6H5)4 solid

0.131g / (358.33 g/mole ) = 0.0003655 moles of KB(C6H5)4  

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

So moles of K2CO3 = 0.0003655 / 2

= 0.0001827 moles of K+ ions

0.0001827 moles * (138.21 g/mole) = 0.025 g K2CO3

in 250 ml aliquote, 0.131 g KB(C6H5)4 is present

So in 125 ml aliquote, 0.0655 g KB(C6H5)4 would be present

0.249 g of precipitate is obtained for both ions, out of which 0.0655 g of KB(C6H5)4  is present. So the remaining [0.249 - 0.0655 g=] 0.1835 g of NH4B(C6H5)4

0.1835 g / (337.27 g/mole) = 0.000544 moles of  NH4B(C6H5)4 also equivalent to moles of NH4Cl since

1 mole of NH4Cl -----> 1 mole of NH4B(C6H5)4

0.000544 moles NH4Cl * (53.492 g/mole) = 0.0291 g

mass % of K2CO3 = 0.025 g K2CO3 * 100

4.975 g soil

= 0.5 %

mass % of NH4Cl = 0.0291 NH4Cl * 100

4.975 g soil

= 0.58%

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.