Question

# A chemist chooses to standardize an approximately 0.015 M HCl solution to the methyl orange endpoint...

A chemist chooses to standardize an approximately 0.015 M HCl solution to the methyl orange endpoint using Na2CO3 (s) as a primary standard. If 22.7 mL of the 0.015 M HCl neutralizes 0.0144 grams of Na2CO3 (molar mass= 105.99 g/mol), what is the standardized molar concentration of the HCl solution?

2H3O+(aq) + CO32-(aq) --> H2CO3(aq) + 2H2O(l) --> CO2(aq) + 3H2O(l)

2H3O+(aq) + CO32-(aq) --> H2CO3(aq) + 2H2O(l) --> CO2(aq) + 3H2O(l)

from this balanced equation it is clear that one mole of carbonate required 2 moles of H+

first calculate the no of moles of Na2CO3 = weight of Na2CO3 / molar mass of Na2CO3

= 0.0144 g / 105.99 g/mol

= 0.0001358 mol

no of moles of HCl required is 2 x 0.0001358 mol = 0.0002717 mol

the volume of the HCl used = 22.7 mL = 0.0227 L

now Molarity of HCl = moles of HCl / volume of HCl in liters

= 0.0002717 mol / 0.0227 L

= 0.012 M

so actual concentration is 0.012M

#### Earn Coins

Coins can be redeemed for fabulous gifts.