mcordovart
contestada

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
HELP

Respuesta :

Eduard22sly

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

Learn more: https://brainly.com/question/14600914

Otras preguntas