Question

A student titrates a sample with HCl solution using methyl red as indicator and finds it contains 20% K2CO3. Later he realises his sample contained Na2CO3 instead of K2CO3. What is the % concentration of Na2CO3 in the sample?

Answer 1

According to the given conditions the sample can contain the same number of moles of K₂CO₃ or Na₂CO₃.
Let’s assume that he has 100 g of the sample. Then
m(K₂CO₃) = (20% × 100 g)/100% = 20 g
At the same time the amount K₂CO₃ is: ν(K₂CO₃) = m(K₂CO₃)/M(K₂CO₃) = 20 /138 mol = 0.1449 mol
When he realizes the he has got Na₂CO₃:
ν(K₂CO₃) = ν(Na₂CO₃) =m(Na₂CO₃)/M(Na₂CO₃) = 0.1449 mol
m(Na₂CO₃) = 0.1449 mol × 106 g/mol = 15.36 g
And, finally, the concentration of Na₂CO₃ is:
W(%) = m(Na₂CO₃)×100%/100 g = 15.36 %