Question

2.0 g of chalk was dissolved in 10 ml of 4N HCl and the solution was made up to 100 ml. 25 ml of this solution required 18.75 ml of 0.2 N NaOH solution for complete neutralization. Calculate the percentage of pure $CaC{O}_3$ in the sample of chalk?

Answer 1

In 10 ml 4N HCl solution, meqv of HCl = 40 meqv
18.75 ml of 0.2 N NaOH = 3.75 meqv of NaOH = 3.75 mev of HCl
So, 3.75*(100/25) = 15 meqv would be required to neutralise 100 ml of the solution.
So, meqv of HCl that reacts with $CaC{O}_3=40-15=25meqv$
So, meqv of $CaC{O}_3$in the sample = 25 meqv
So, mmol of $CaC{O}_3$in sample = 25/2 = 12.5 mmol (n-fac=2)
So, mass of $CaC{O}_3$ in sample = $12.5mmol\times 100g$ = 1.25 gm
So, % of pure $CaC{O}_3=\frac{1.25}{2}\times 100=62.5$