Question

QUESTION 2.6

How many mL of 0.1 M HCl are required to react completely with 1 g mixture of and containing equimolar amounts of both?

Answer 1

Step 1:

Let the mixture contains x g of sodium carbonate and 1−x g of sodium bicarbonate.

The molar masses of sodium carbonate and sodium bicarbonate are 106 g/mol and 84 g/mol respectively.

The number of moles of sodium carbonate and sodium bicarbonates are $\frac{x}{106}and\frac{1-x}{84}$ respectively.

Step 2:

Since, it is an equimolar mixture.

$\frac{x}{106}=\frac{1-x}{84}$

$84x=106-106x$

$x=0.5579$

Step 3:

Number of moles of sodium carbonate = $\frac{0.5579}{106}=0.005263$

Number of moles of sodium hydrogen carbonate = $\frac{1-0.5579}{84}=0.005263$

One mole of sodium carbonate will react with 2 moles of HCl and 1 mole of sodium bicarbonate will react with 1 mole of HCl.

Step 4:

Total number of moles of HCl that will completely neutralize the mixture

= $2\times 0.005263+0.005263=0.01578moles$

Volume of 0.1 M HCl required = $\frac{0.01578}{0.1}=0.158L=158mL$