Question

Dilute hydrochloric acid was titrated with sodium carbonate solution. • 10.0cm3 of 0.100mol/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. • 16.2cm3 of sodium carbonate solution were required to react completely with the acid. Calculate how many moles of hydrochloric acid were used

Answer 1

Given:

10 ml of 0.1 molar Dilute HCL is titrated  with 16.2 ml of sodium carbonate solution .

To Find :

How many moles of HCL will be used.

Solution:

Na2CO3 (aq) + 2 HCl (aq) → 2 NaCl (aq) + CO2 (g) + H2O (l)

We Know that ,

Molarity = [n/v]*100

For HCL solution

0.1 = [n/10]*100

n = 0.01 moles

For 1 mole of sodium carbonate  2 moles of  HCL is required

=> For 1 mole of  HCL , 1/2 mole of sodium carbonate  is required

=>  For 0.01 mole of  HCL , 0.005 mole of sodium carbonate  is required

Hence,  0.01 mole of  HCL & 0.005 mole of sodium carbonate  are used

Answer 2

the other solution is incorrect