Question

0.5 g of a mixture of K2C03 and Li2C0 3 required 30
mL of 0.25 N HCl solution for neutralisation. What is
the % composition of the mixture ?​

Answer 1

Step 1:

Weight of K2CO3=ag

Weight of Li2CO3=bg

a+b=0.5------(1)

For reaction

Meq.of K2CO3+Meq of Li2CO3=Meq of HCl

a×1000138/2+b×100074/2=30×0.25

74a+138b=38.295------(2)

Step 2:

By equation (1) & (2)

a=0.48g

b=0.02g

% of K2CO3=0.480.5×100

⇒96%

% of Li2CO3=0.020.5×100

⇒4%