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?
Solution and Explanation
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.
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Concepts Used:
Concentration of Solutions
It is the amount of solute present in one liter of solution.
Concentration in Parts Per Million - The parts of a component per million parts (106) of the solution.
Mass Percentage - When the concentration is expressed as the percent of one component in the solution by mass it is called mass percentage (w/w).
Volume Percentage - Sometimes we express the concentration as a percent of one component in the solution by volume, it is then called as volume percentage
Mass by Volume Percentage - It is defined as the mass of a solute dissolved per 100mL of the solution.
Molarity - One of the most commonly used methods for expressing the concentrations is molarity. It is the number of moles of solute dissolved in one litre of a solution.
Molality - Molality represents the concentration regarding moles of solute and the mass of solvent.
Normality - It is the number of gram equivalents of solute present in one liter of the solution and it is denoted by N.
Formality - It is the number of gram formula present in one litre of solution.