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A hydrochloric acid solution is standardized by titrating 0.2329 g of primary standard sodium carbonate to a methyl red end point by boiling the carbonate solution near the end point to remove carbon dioxide. If 42.87 mL acid is required for the titration, what is its molarity
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The molarity of the acid, HCl solution is 0.103 M

We'll begin by calculating the number of mole in 0.2329 g of Na₂CO₃

Mass of Na₂CO₃ = 0.2329 g

Molar mass of Na₂CO₃ = (23×2) + 12 + (16×3) = 106 g/mol

Mole of Na₂CO₃ =?

Mole = mass / molar mass

Mole of Na₂CO₃ = 0.2329 / 106

Mole of Na₂CO₃ = 0.0022 mole

  • Next, we shall determine the number of mole of HCl needed to react with 0.0022 mole of Na₂CO₃

Na₂CO₃ + 2HCl —> 2NaCl + H₂O + CO₂

From the balanced equation above,

1 mole of Na₂CO₃ reacted with 2 moles of HCl.

Therefore,

0.0022 mole of Na₂CO₃ will react with = 0.0022 × 2 = 0.0044 mole of HCl

  • Finally, we shall determine the molarity of the HCl.

Mole of HCl = 0.0044 mole

Volume = 42.87 mL = 42.87 / 1000 = 0.04287 L

Molarity of HCl =?

Molarity = mole / Volume

Molarity of HCl = 0.0044 / 0.04287

Molarity of HCl = 0.103 M

Thus, the molarity of the acid, HCl solution is 0.103 M

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