Question

Dilute hydrochloric acid was titrated with sodium carbonate solution. • 10.0 cm3 of 0.100 mol / dm3 hydrochloric acid were placed in a conical flask. • A few drops of methyl orange indicator were added to the dilute hydrochloric acid. • The mixture was titrated with sodium carbonate solution. • 16.2 cm3 of sodium carbonate solution were required to react completely with the acid. What colour would the methyl orange indicator be in the hydrochloric acid? Calculate how many moles of hydrochloric acid were used. Use your answer to (b)(ii) and the equation for the reaction to calculate the number of moles of sodium carbonate that reacted.

Answer 1

Solution :

It is given that dilute hydrochloric acid is titrated with sodium carbonate solution.

$10 \ cm^3$ of a 0.100 $mol/dm^3$ of hydrocholric acid is used.

Thus, 10 mL of the 0.1 molar dilute hydrochloric acid is titrated with 16.2 mL of Sodium carbonate solution.

The equation would be :

$Na_2CO_3 (aq) + 2HCl(aq) \rightarrow 2NaCl(aq) + CO_2(g) +H_2O (l)$

Now we know that that the

Molarity = $\frac{n}{v} \times 100$

For hydrochloric acid solution,

$0.1=\frac{n}{10} \times 100$

∴ n = 0.01 moles.

Now for one mole of $Na_2CO_3$, two moles of HCl acid is used.

For one mole of HCl, one-half mole of $Na_2CO_3$ is required.

Therefore, for 0.01 mole of HCl, we require 0.005 mole of $Na_2CO_3$.

Hence, 0.01 mole of hydrochloric acid and 0.005 mole of $Na_2CO_3$ solution is used.

The color turns red when methyl orange indicator is used in HCl acid.