Question

2.176 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. 30.06 mL of a 0.765 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

K2CO3 + 2HCl ---> H2CO3 + 2KCl

1 mol K2CO3 = 2 MOL HCl

KHCO3 + HCl ---> H2CO3 + KCl

1 mol KHCO3 = 1 mol HCl

so that,

total no of mol of HCl reacted = 30.06*0.765/1000 = 0.023 mol

no of mol K2CO3 reacted = x/2

no of mol of KHCO3 = 0.023 - x

2.176 = (x/2*138.2)+((0.023-x)*100.11)

x = 0.00408

mass of K2CO3 = x/2*138.2

   = 0.00408/2*138.2

   = 0.282 g

mass of KHCO3 = (0.023-x)*100.11

              = (0.023-0.00408)*100.11

              = 1.894 g

percentage of K2CO3 = 0.282/2.176*100 = 12.96%

percentage of KHCO3 = 1.894/2.176*100 = 87%