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Question

0.50 g of a mixture of K2CO3 and Li2CO3 requires 30 mL of a 0.25 N HCl solution for neutralisation. What is the percentage composition of the mixture?

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Solution

K2CO3138Eq.mass 69+2HCl2×36.536.52KCl+H2O+CO2
Li2CO374Eq.mass 37+2HCl2×36.536.52LiCl+H2O+CO2
Let x g of K2CO3 be present in the mixture.
Mass of Li2CO3=(0.50x)
No. of g equivalents of K2CO3=x69
No. of g equivalents of Li2CO3=(0.50x)37
No. of g equivalents in 30 mL of 0.25 N HCl
=Normality×Volume1000=0.25×301000=3400
At equivalence point,
x69+(0.50x)37=3400
So, x=0.48
K2CO3=0.48 g; or 96%
Li2CO3=0.02 g; or 4%.

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