Question

2.122 g of a solid mixture containing only potassium carbonate (FW = 138.2058 g/mol) and potassium bicarbonate (FW = 100.1154 g/mol) is dissolved in distilled water. 34.16 mL of a 0.762 M HCl standard solution is required to titrate the mixture to a bromocresol green end point.Calculate the weight percent of potassium carbonate and potassium bicarbonate in the mixture.

Answer 1

Answer:

The weight percent of potassium carbonate is 50,8 wt% and of potassium bicarbonate 49,2 wt%

Explanation:

The reactions of potassium carbonate (K₂CO₃) and potassium bicarbonate (KHCO₃) with HCl produce:

K₂CO₃ + 2HCl → 2KCl + CO₂ + H₂O

KHCO₃ + HCl → KCl + CO₂ + H₂O

That means that you need 2 moles of HCl to titrate potassium carbonate and 1 mol to titrate potassium bicarbonate.

The moles of HCl to titrate the mixture are:

0,03416L×$\frac{0,762mol}{1L}$ = 0,02603 mol of HCl

If X is mass of K₂CO₃ and Y is mass of KHCO₃ in the mixture, the moles of HCl to titrate the mixture are equals to:

0,02603 mol = 2X×$\frac{138,2058 g}{1mol}$ + Y×$\frac{100,1154 g}{1mol}$ (1)

As the mass of the mixture is 2,122g:

2,122g = X + Y (2)

Replacing (2) in (1):

0,02603 mol = 0,01447 (2,122-Y) + 9,988x10⁻³Y

0,02603 mol = 0,0307 - 0,01447Y + 9,988x10⁻³Y

-4,6778x10⁻³ = -4,4827x10⁻³Y

1,044g = Y -mass of potassium bicarbonate-

Thus:

X = 1,078g -mass of potassium carbonate-

The weight percent of potassium carbonate is:

$\frac{1,078g}{2,122g}$×100 = 50,8 wt%

The weight percent of potassium bicarbonate is:

$\frac{1,044g}{2,122g}$×100 = 49,2 wt%

I hope it helps!