Question

How many mL of 0.1 M HCL are required to react completely with 1g mixture of $\text{Na}_2\text{CO}_3$ and $\text{NaHCO}_3$ containing equimolar amounts of both?

Answer 1

Let the amount of $\text{Na}_2\text{CO}_3$ in the mixture be $x \,g$.

Then, the amount of $\text{NaHCO}_3$ in the mixture is $(1 - x) g.$

Molar mass of $\text{Na}_2\text{CO}_3= 2\times23+1\times 12+3\times16$

$= 106\, \text{g mol}^{-1}$

∴Number of moles $\text{Na}_2\text{CO}_3 = \frac{x}{106} \text{mol}$

Molar mass of $\text{NaHCO}_3 = 1 \times23+ 1\times1\times12+3\times16$
$= 84\, \text{g mol}^{-1}$

∴Number of moles of $\text{NaHCO}_3= \frac{(1-x)}{84} \text{mol}$
According to the question,

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

$⇒84x=106-106x$
$⇒190x = 106$
$⇒x=0.5579$
Therefore, number of moles of $\text{Na}_2\text{CO}_3= \frac{0.5579}{106} \text{mol}$
$=0.0053\text{mol}$

And, number of moles of $\text{NaHCO}_3= \frac{1-0.5579}{84}$
$= 0.0053 \text{mol}$

$\text{HCI}$ reacts with $\text{Na}_2\text{CO}_3$ and $\text{NaHCO}_3$ according to the following equation.
$2\text{HCI} + \text{Na}_2\text{CO}_3 → 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$
$2 \text{mol}$ $1 \text{mol}$
$\text{HCI} + \text{NaHCO}_3→ \text{NaCI} +\text{H}_2\text{O} + \text{CO}_2$
$1mol$ $1 mol$

1 mol of $\text{Na}_2\text{CO}_3$ reacts with $2 mol$ of $\text{HCI}.$

Therefore, $0.0053 mol$ of $\text{Na}_2\text{CO}_3$ reacts with $2 \times 0.0053\, mol = 0.0106\, mol.$

Similarly, $1 mol$ of $\text{NaHCO}_3$ reacts with $1 mol$ of $\text{HCI}.$

Therefore, $0.0053 mol$ of $\text{NaHCO}_3$, reacts with $0.0053 mol$ of $\text{HCL}$

Total moles of $\text{HCI}$ required $= (0.0106 +0.0053) mol$
$=0.0159 mol$

In 0.1 M of $\text{HCI}$,
$0.1 mol$ of $\text{HCI}$ is preset in 1000 mL of the solution.

Therefore, 0.0159 mol of $HCI$ is present in $\frac{(1000\times 0.0159)}{0.1}mol$
= 159 mL of the solution
Hence, = 159 mL of 0.1 M of $\text{HCI}$ is required to react completely with 1 g mixture of $\text{Na}_2\text{CO}_3$ and $\text{NaHCO}_3$ containing equimolar amounts of both.