# 1.5g chalk reactd wid 10ml of 4N-HCl.This solution was made to 100ml.25ml of this sol required 18.75 ml 0.2NaOH for reaction.Calc % of CaCO3 in chalk

unkyd | Certified Educator

The 18.75 ml of .2 N NaOH was to neutralize the unknown HCl solution.  For neutralization the equation:

Na Va = Nb Vb can be used to find the Normality of the Acid.

N = Normality   V = Volume   a = acid   b = base

so, Na = Normality of the acid

solving for Na

Na = Nb*Vb/Va = .2 N * 18.75 ml/25 ml

=  .15 N

Since the solution that was diluted to 100 ml had a Normality of .15 and it was diluted to one tenth its original concentration, the solution before dilution had a Normality of 1.5.

Since, Normality and Molarity are the same for HCl, you can calculate the number of moles of HCl that reacted with the chalk

Original Moles of HCL

Moles of HCL = M * V = 4 M * .01 liters = .04 Moles of HCl

After reaction

Moles of HCL = M * V = 1.5 M * .01 liters = .015 Moles of HCl

Means .025 moles of HCl reacted with CaCO3 in chalk

According to the reaction:

2 HCl + CaCO3 -> CaCl2  + H2CO3

so only .0125 moles of CaCO3 were in the chalk

.0125 moles CaCO3 * 100.1 g  CaCO3 / 1 mole CaCO3

= 1.25 g CaCO3

1.25 g CaCO3 / 1.5 g Chalk = .833

The chalk was 83.3 % CaCO3