Question

2.65 g of sodium carbonate was dissolved in water and made up to 500 mL in a standard flask. 20 mL portions of this solution required 18.50 mL of a solution of hydrochloric acid when titrated using methyl orange indicator.Calculate:(ii) The concentration of the hydrochloric acid in mol/L.

Answer 1

Step 1 - Writing the chemical equation

The reaction that is taking place in this experiment is:

$Na_2CO_3+2HCl\rightarrow H_2O+CO_2+2NaCl$

Note the proportion between the reactants is:

one mole of Na2CO3 reacts with 2 moles of HCl

Step 2 - Calculating the concentration of Na2CO3 in mol/L

The molar concentration of Na2CO3 can be calculated by the following formula:

$[Na_2CO_3\rbrack=\frac{m}{MM\times V}$

In this equation, m is the mass, MM the molar mass and V the total volume of solution.

We know that:

$\begin{gathered} m=2.65g \\ MM=106g/mol \\ V=0.5L \end{gathered}$

So, setting the values in the formula:

$[Na_2CO_3\rbrack=\frac{2.65}{106\times0.5}=0.05\text{ mol/L}$

Step 3 - Finding how many moles of Na2CO3 reacted

Now let's find how many moles there were in 20ml of this solution:

$\begin{gathered} 1\text{ L contains ---- 0.05 moles} \\ 0.02\text{ L contains ---- x moles} \\ \\ x=0.001\text{ moles of Na2CO3} \\ \end{gathered}$

Step 4 - Finding how many moles of HCl reacted

Let's remember that one mole of Na2CO3 reacts with 2 moles of HCl. Therefore, if 0.001 moles of Na2CO3 reacted, 0.002 moles of HCl also reacted.

Step 5 - Calculating the concentration of HCl in mol/L

The 0.002 moles that reacted were present in 18.5ml of HCl solution. Therefore:

$[HCl\rbrack=\frac{n}{V}=\frac{0.002\text{ moles}}{0.0185\text{ L}}=0.108\text{ mol/L}$

Answer: 0.108 mol/L