Question

1.18 g of a mixture of Na2CO3 and K2SO4 is present in 250 ml aqueous solution. 25 ml of this solution is completely neutralised by 20 ml of. 0.05 M H2SO4. Calculate the percentage composition of the mixture.

Answer 1

Answer:

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×V1=N2×V2

0.1×40=N2×20

N2=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