Problem 70 - Chemistry and Chemical Reactivity

Question

What volume of \(0.955 \mathrm{M} \mathrm{HCl},\) in milliliters, is required to titrate \(2.152 \mathrm{g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) to the equivalence point? $$\begin{aligned} \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow & \\\ \mathrm{H}_{2} \mathrm{O}(\ell) &+\mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{NaCl}(\mathrm{aq}) \end{aligned}$$

Step-by-Step Solution

Answer

42.5 mL of 0.955 M HCl is needed.

Step 1: Write the Balanced Chemical Equation

The balanced chemical equation for the reaction of sodium carbonate with hydrochloric acid is: \[ \text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow \text{H}_2\text{O} + \text{CO}_2 + 2\text{NaCl} \] This shows that 1 mole of \(\text{Na}_2\text{CO}_3\) reacts with 2 moles of \(\text{HCl}\).

Step 2: Calculate the Moles of Na2CO3

First, find the molar mass of \(\text{Na}_2\text{CO}_3\): \[\text{Molar mass of Na}_2\text{CO}_3 = (2 \times 23) + 12 + (3 \times 16) = 106 \, \text{g/mol} \] Then, calculate the moles of \(\text{Na}_2\text{CO}_3\): \[\text{Moles} = \frac{2.152 \, \text{g}}{106 \, \text{g/mol}} = 0.0203 \, \text{mol} \]

Step 3: Calculate the Moles of HCl Needed

From the balanced chemical equation, 1 mole of \(\text{Na}_2\text{CO}_3\) reacts with 2 moles of \(\text{HCl}\). So, the moles of \(\text{HCl}\) required are: \[ \text{Moles of HCl} = 0.0203 \, \text{mol} \times 2 = 0.0406 \, \text{mol} \]

Step 4: Use the Molarity Equation to Find Volume of HCl

The molarity equation \( C = \frac{n}{V} \) can be rearranged to find the volume: \[ V = \frac{n}{C} \] where \( n = 0.0406 \, \text{mol} \) and \( C = 0.955 \, \text{M} \). Calculate the volume: \[ V = \frac{0.0406 \, \text{mol}}{0.955 \, \text{mol/L}} \approx 0.0425 \, \text{L} \] Convert this to milliliters: \[ V = 0.0425 \, \text{L} \times 1000 \, \text{mL/L} = 42.5 \, \text{mL} \]

Key Concepts

Stoichiometry

Stoichiometry is a central concept in chemistry that involves the calculation of reactants and products in chemical reactions. It provides a way to quantify the relationships between the substances that participate in chemical processes.
In stoichiometry, we use balanced chemical equations to determine the ratios of moles between different reactants and products. This ratio is vital as it allows us to predict the amount of one reactant needed to completely react with another, or the amount of product that will be formed.

• The balanced equation for sodium carbonate reacting with hydrochloric acid is \( ext{Na}_2 ext{CO}_3 + 2 ext{HCl} \rightarrow ext{CO}_2 + ext{H}_2 ext{O} + 2 ext{NaCl}\).
• This indicates that 1 mole of \( ext{Na}_2 ext{CO}_3\) reacts with 2 moles of \( ext{HCl}\).

This stoichiometric ratio is key to determining how much of each reactant is needed or the amount of product formed in reactions, ensuring precise measurements in chemical experiments.

Molarity

Molarity is a measure of the concentration of a solute in a solution, and is expressed as moles of solute per liter of solution (mol/L). It's an essential concept when it comes to preparing solutions and performing reactions with accurate chemical proportions.
To calculate molarity, we use the formula \(C = \frac{n}{V} \), where \(n\) is the number of moles of solute and \(V\) is the volume of solution in liters.

• For instance, in this exercise, we're dealing with a solution of \( ext{HCl}\) with a molarity of 0.955 M.
• We can manipulate this formula to find other values, like the volume of the solution if we know the amount in moles, i.e., \(V = \frac{n}{C} \).

Understanding molarity is fundamental when working with solutions as it affects how reactants interact and ensures that reactions proceed as expected.

Chemical Equation Balancing

Balancing chemical equations is a critical skill in chemistry that ensures the law of conservation of mass is upheld. Every chemical reaction must have the same number of each type of atom on both sides of the equation.

• Take the equation for this exercise: \( ext{Na}_2 ext{CO}_3 + 2 ext{HCl} \rightarrow ext{CO}_2 + ext{H}_2 ext{O} + 2 ext{NaCl}\).
• Initially, one might write the reaction without coefficients, but balancing requires adjusting the coefficients so that each element has an equal number of atoms on both sides.

Properly balanced equations like this allow us to perform further stoichiometric calculations accurately.
Balancing is a fundamental step and is necessary to reliably predict the quantities of reactants and products involved in a chemical reaction.

Acid-Base Titration

Acid-base titration is an analytical method for determining the concentration of an acid or base in a solution. By adding a reactant of known concentration and volume, we can calculate the unknown concentration using stoichiometry and the equation of the reaction.

• In this setup, \( ext{HCl}\) is titrated against \( ext{Na}_2 ext{CO}_3\).
• The point at which the reaction is complete is known as the equivalence point. At this stage, the moles of acid equal the moles of base as dictated by their stoichiometric ratio in the balanced equation.

Using the balanced equation and molarity concept, the volume of \( ext{HCl}\) needed to reach this point can be calculated, ensuring the reaction is complete. This shows the practical application of stoichiometry and molarity in real-world chemical analysis.