Answer:
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%
