A 1.475-g sample containing NH4Cl (FM 53.492), K2CO3 (FM 138.21), and inert ingr

Question

A 1.475-g sample containing NH4Cl (FM 53.492), K2CO3 (FM 138.21), and inert ingredients was dissolved to give 0.100 L of solution. A 25.0-mL aliquot was acidified and treated with excess sodium tetraphenylborate, Na_B(C6H5)4 _, to precipitate K_ andNH4- ions completely: FM 358.33, FM 337.27 The resulting precipitate amounted to 0.617 g. A fresh 50.0-mL aliquot of the original solution was made alkaline and heated to drive off all the NH3: It was then
acidified and treated with sodium tetraphenylborate to give 0.554 g of precipitate. Find the weight percent of NH4Cl andK2CO3 in the original solid.

4. A 1.475-g sample containing NH,Cl (FW = 53.492). K-CO (FW = 138.21) and inert ingredients was dissolved to give 0.100 L of solution A 25.0 ml. aliquot was acidified and treated with excess sodium tetraphenylborate, Na·B(C6H). to precipitate K' and NH4 ions completely: (CaH54B' K+ (GH5ABK(s) (FW 358.33) (FW = 358.33) + (c,Hs),BNH4(s) {FW = 337.27) The resulting precipitate amounted to 0.617 g. A fresh 50.0-mL aliquot of the original tion was made alkaline and heated to drive off all the NH3: It was then acidified and treated with sodium tetraphenylborate to ive 0554 g of cipitate. Find the weight percent of NH,Cl and K CO in the original solid.

Explanation / Answer

weight of K+ tetraphenylborate precipitate in 50 ml aliquot = 0.554 g

moles of K+ = 0.554 g/358.33 g/mol = 0.00155 mols

So in 0.1 L of solution K+ = 2 x 0.00155 mol = 0.0031 mols

moles of K2CO3 = 0.0031/2 = 0.00155 mols

mass of K2CO3 = 0.00155 mol x 138.21 g/mol = 0.214 g

weight% of K2CO3 in sample = (0.214/1.475) x 100 = 14.51%

weight of K+ and NH4+ tetraphenylborate precipitated from 25 ml aliquot = 0.617 g

weight of NH4+ tetraphenyl borate in it = 0.617 - (0.554/2) = 0.34 g

mols of NH4+ = 0.00101 mols

So in 0.1 L NH4+ = 4 x 0.00101 = 0.00404 mols

mass of NH4Cl = 0.00404 mol x 53.492 g/mol = 0.216 g

weight% of NH4Cl in sample = (0.216/1.475) x 100 = 14.64%

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