To determine the molarity of the given HCl solution by titrating it against a [tex]\(\left(\frac{M}{20}\right)\)[/tex] solution of Na[tex]\(_2\)[/tex]CO[tex]\(_3\)[/tex], we can use the principles of stoichiometry and the balanced chemical equation.
### Balanced Chemical Equation:
The balanced chemical equation for the reaction between HCl and Na[tex]\(_2\)[/tex]CO[tex]\(_3\)[/tex] is:
[tex]\[2 \text{HCl} + \text{Na}_2\text{CO}_3 \rightarrow 2 \text{NaCl} + \text{CO}_2 + \text{H}_2\text{O}\][/tex]
### Stoichiometry of the Reaction:
From the balanced equation, we see that 2 moles of HCl react with 1 mole of Na[tex]\(_2\)[/tex]CO[tex]\(_3\)[/tex].
### Given Data:
1. Volume of HCl solution = [tex]\(1.0 \, \text{L}\)[/tex] (assuming 1L if not provided)
2. Molarity of Na[tex]\(_2\)[/tex]CO[tex]\(_3\)[/tex] solution = [tex]\(\frac{1}{20} \, \text{M}\)[/tex]
### Determining the Molarity of HCl:
Since 2 moles of HCl react with 1 mole of Na[tex]\(_2\)[/tex]CO[tex]\(_3\)[/tex]:
[tex]\[
\text{Molarity of HCl} = 2 \times \text{Molarity of Na}_2\text{CO}_3
\][/tex]
Substitute the given molarity of Na[tex]\(_2\)[/tex]CO[tex]\(_3\)[/tex]:
[tex]\[
\text{Molarity of HCl} = 2 \times \left(\frac{1}{20}\right) \, \text{M}
\][/tex]
Simplify the expression:
[tex]\[
\text{Molarity of HCl} = 2 \times 0.05 \, \text{M}
\][/tex]
[tex]\[
\text{Molarity of HCl} = 0.1 \, \text{M}
\][/tex]
Therefore, the molarity of the given HCl solution is [tex]\(0.1 \, \text{M}\)[/tex].
