Question

A 2.137 g sample of a solid mixture containing only potassium carbonate ( MM=138.2058 g/mol ) and potassium bicarbonate ( MM=100.1154 g/mol ) is dissolved in distilled water. A volume of 31.70 mL of a 0.747 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:

Weight percent of potassium carbonate: 21,4 wt%

Weight percent of potassium bicarbonate: 78,6 wt%

Explanation:

Potassium carbonate (K₂CO₃) and potassium bicarbonate (KHCO₃) react with HCl thus:

K₂CO₃ + 2 HCl → 2 KCl + CO₂ + H₂O

KHCO₃ + HCl → KCl + CO₂ + H₂O

The moles of HCl that react with the mixture are:

0,03170L × $\frac{0,747mol}{L}$ = 0,0237 mol of HCl

If X are grams of K₂CO₃ and Y are grams of KHCO₃ it is possible to obtain these two equations:

0,0237 mol = 2X÷132,2058 + Y÷100,1154 (1)

2,137 g = X + Y (2)

Replacing (2) in (1):

Y = 1,679 g of KHCO₃

Thus,

X = 0,458 g of K₂CO₃

Weight percent of potassium carbonate:

$\frac{0,458g}{2,137g}$×100 = 21,4 wt%

Weight percent of potassium bicarbonate:

$\frac{1,679g}{2,137g}$×100 = 78,6 wt%

I hope it helps!