Question

$1.20g$ sample of ${Na}_2{CO}_3$ and $K_2{CO}_3$ was dissolved in water to form $100ml$ of a solution. $20ml$ of this solution required $40ml$ of $0.1N$ $HCl$ for complete neutralization. Calculate the weight of ${Na}_2{CO}_3$ in the mixture.

Answer 1

Total weight of the mixture(Na2CO3+K2CO3)=1.20g

Volume of the solution=100ml

Volume of solution taken for neutralisation V2= 20 ml

Volume of HCl solution V1=40 ml

Normality of HCl, N1=0.1 N

To calculate the mass of Na2CO3

$N1\times V1=N2\times V20.1\times 40=N2\times 20N2=0.2N$

1 l of solution contain no. of g equivalent=0.2

0.1 l of solution contain no. of g equivalent=0.1*0.2=0.02 g equivalent

No. of g equivalent=sum of (g equivalent of Na2CO3+g equivalent of K2CO3)

No. of g equivalent=mass/equivalent mass

equivalent mass=molar mass/valence

Molar mass of Na2Co3=106 g

Molar mass of K2CO3=138 g

Equivalent mass of Na2CO3=106/2=53 g

Equivalent mass of K2CO3=138/2=69g

Consider mass of Na2CO3=x g

Mass of K2CO3=(1.2-x)g

So we can write

0.02=x/53+1.2-x/69

x=0.59 g=0.6g