A 2.118 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 30.20 mL of a 0.763 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.

Question

A  2.118  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  30.20  mL of a  0.763  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.

DrBob222 · Accepted Answer

The HCl reacts completely with both as follows:
K2CO3 + 2HCl ==> 2KCl + H2O + CO2
KHCO3 + HCl ==> KCl + H2O + CO2
You set up two equations and solve them simultaneously.
Let x = grams K2CO3
and y = grams KHCO3.
eqn 1 is x + y = 2.118
--------------------------------------
eqn 2 comes from
mols HCl to titrate K2CO3 + mols HCl to titrate KHCO3 = mols HCl used.
(2x/MM K2CO3) + (y/MM KHCO3) = mols HCl x L HCl

Solve these two equations simultaneously. When you get x and y then
% K2CO3 = (g K2CO3/wt sample)*100 = ?
% KHCO3 = (g KHCO3/wt sample)*100 = ?
Post your work if you get stuck.

mike · Answer

@DrBob222 I am stuck, x=1.590 g of k2CO3=> x+y=2.118 1.590+y=2.118 => y=0.528 g of KHCO3 then 0.528 g to moles 2x+0.00527 mole=0.0230 mol => x=1.225 g of KCO3