0.265 g of pure anhydrous $Na X_{2} CO X_{3}$ (106 g/mol) is dissolved in water and titrated with a HCl solution to the methyl orange end point, requiring 22.0 mL. Find the molarity of the HCl (three decimal places).

Numerical answer type, answer in M, tolerance 0.0006.

Answer: 0.227 M

Moles of

Na X_{2} CO X_{3}

= 0.265/106 = 2.50 mmol. At the methyl orange end point each carbonate needs 2

H X^{+}

, so HCl = 5.00 mmol. Molarity = 5.00/22.0 = 0.227 M.