**Problem:**

A student is preparing for the acid base titration, he was standardized a hydrochloric acid solution by titrating it with 0.3023 g of primary standard sodium carbonate with methyl red as indicator, and boiling the carbonate solution near the end point to remove carbon dioxide. If he needed 37.5 mL of acid required for the titration, so what is the hydrochloric acid molarity?

**Solution 1**

1. First we calculate the mol of sodium carbonate, but we need to calculate of its formula weight

Formula weight of Na2CO3 = ( 2 x Ar Na) + (1 x Ar C) + ( 3 x Ar O)

= ( 2 x 22.99) + ( 1 x 12.01) + ( 3 x 16.00)

= 105.99 g/mol

Mol of Na2CO3 = g/formula weight

= 0.3023 / 105.99

= 2.8522. 10

^{-3}mol

= 2.8522 mmol

2. Write the chemical reaction

2HCl + Na2CO3 2NaCl + H2O + CO2

From the reaction we know that mol HCl : mol Na2CO3 = 2 : 1 so the mol of HCl can be calculated

Mol HCl = 2 x mol Na2CO3

= 2 x 2.8522 mmol

= 5.7044 mmol

3. because mol and volume of HCl is known, we can calculate its molarity

Molarity of HCl = mol/volume

= 5.7044 / 37.5

= 0.1521 M

**Solution 2**

We can solve the problem by calculate the mole equivalent, remember that one equivalent substance A will always react with one equivalent substance B in this case acid and base.

Mol equivalent Na2CO3 = 2 x 2.8522 = 5.7044 meq

Since, meq acid = meq base

NxV acid = NxV base

N acid = ( NxV base ) / V acid

= 5.7044 / 37.5

= 0.1521 N

= 0.1521 M

(Remember 1 N of HCl is the same with 1 M of HCl, because 1 mol HCl = 1 meq of HCl)