Question

Dilute hydrochloric acid reacts with sodium carbonate solution.

2HCl (aq) + Na 2 CO 3 (aq) → 2NaCl (aq) + H 2 O(l) + CO 2 (g)
(b) Dilute hydrochloric acid was titrated with sodium carbonate solution.

• 10.0 cm 3 of 0.100 mol / dm 3 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 cm 3 of sodium carbonate solution were required to react completely with the
acid.
a) Calculate how many moles of hydrochloric acid were used.

............................ mol [1]
b) 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

Answer:-

2HCl(aq)+Na2CO3(aq)â2NaCl(aq)+H2O(l)+CO2(g)

Part A:

We can simply proceed by equating the moles of reactants on the two sides of the equation.

Let volume of HCl needed be x L.

Then,

Moles of Na2Co3=Moles of HCl

0.500M * 0.750 L = 2 * 2.50M * x L

On solving, x=0.075 L

Part B.

In this part, like in part A, we equate the moles on the two sides of the equation,

Let the concentration of HCl be xM.

Moles of HCl= Moles of Na2Co3

12.1g / 44 g/Mol= 2 * 0.651 L * x M

On solving, x= 0.037M