Dilute hydrochloric acid was titrated with sodium carbonate solution. ( I am struggling with this kind of questions T_T)

Question

• 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.

ii . Calculate how many moles of hydrochloric acid were used. ............................ mol

iii. Use your answer to (ii) and the equation for the reaction to calculate the number of moles of sodium carbonate that reacted. ............................ mol

iv Use your answer to (iii) to calculate the concentration of the sodium carbonate solution in mol/dm3 .

Answer 1

The first thing you should always do is to write the correctly balanced equation for the reaction taking place:

2HCl + Na2CO3 ==> 2NaCl + H2CO3 ==> H2O + CO2

From this you can see that it takes TWO moles HCl for each ONE mole Na2CO3.

ii. moles HCl used = 10.0 ml x 1 L/1000 ml x 0.100 mol/L = 0.00100 moles HCl used

iii. 0.00100 moles HCl x 1 mol Na2CO3 / 2 moles HCl = 0.000500 moles Na2CO3 present

iv. [Na2CO3] = moles/liter = 0.000500 moles / 0.0162 L = 0.0309 M