Question

2.164 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. 32.22 mL of a 0.766 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

Moles of HCl used = M x V   = 0.766 x 0.0322 = 0.02468 mol

all HCl moles is used by NaHCO3 and Na2CO3

Let K2CO3 moles = m , then HCl moles ued to titrate Na2CO3 = 2m ( since 2HCl reacts with1 K2CO3)

Let KHCO3 moles = n , then HCl moles = n (1KHCO3 is neutralised by 1HCl)

Total HCl moles = 2m+n = 0.02468 ..............(1)

Mass of K2CO3 = moles x molar mass of K2CO3

                 = 138.2 m

mass of KHCO3 = moles x molar mass of KHCO3 = 100 n

total mass of K2CO3 + KHCO3 = 138.2m + 100 n = 2.164 ...........(2)

solving eq (1) and eq (2 ) we get

m = 0.004919 , n = 0.01484

K2CO3 mass = 138.2 ( 0.004919) = 0.68 g

KHCO3 mass = 2.164 - 0.68 = 1.484 g

% of K2CO3 = ( mass of K2CO3 / sample mass) x 100 = ( 0.68/2.164) x 100 = 31.4 %

% of KHCO3 = 100 - 31.4 = 68.6 %