Problem 68 - Chemistry and Chemical Reactivity

Question

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

Step-by-Step Solution

Answer

42.5 mL of 0.955 M HCl is needed.

Step 1: Determine Moles of Na2CO3

First, find the number of moles of \( \text{Na}_2\text{CO}_3 \). Use the formula: \( \text{Moles} = \frac{\text{mass}}{\text{molar mass}} \). The molar mass of \( \text{Na}_2\text{CO}_3 \) is (2*23)+(12)+(3*16) = 106 g/mol. Thus, moles of \( \text{Na}_2\text{CO}_3 = \frac{2.152 \text{ g}}{106 \text{ g/mol}} = 0.0203 \text{ mol} \).

Step 2: Use Stoichiometry to Find Moles of HCl

According to the balanced chemical equation, \(\text{Na}_2\text{CO}_3\) reacts with \(2\) moles of HCl. This means for every mole of \(\text{Na}_2\text{CO}_3\), there are \(2\) moles of \( \text{HCl} \) needed. Therefore, moles of \( \text{HCl} = 2 \times 0.0203 = 0.0406 \text{ mol} \).

Step 3: Calculate Volume of HCl Solution Required

Using the molarity formula, \( \text{Molarity} = \frac{\text{moles}}{\text{volume in L}} \), rearrange to find the volume: \( \text{Volume in L} = \frac{\text{moles}}{\text{Molarity}} \). \( \text{Volume in L} = \frac{0.0406}{0.955} = 0.0425 \text{ L} \). Convert this to milliliters: \( 0.0425 \times 1000 = 42.5 \text{ mL} \).

Key Concepts

Stoichiometry

Stoichiometry is all about understanding the quantities and relationships in chemical reactions. In our exercise, stoichiometry helps determine how much hydrochloric acid (HCl) is needed to fully react with a given amount of sodium carbonate (\( \text{Na}_2\text{CO}_3 \)).
This concept involves using

• Balanced chemical equations
• Molar ratios
• Mole calculations

The balanced equation provided is \[ \text{Na}_2\text{CO}_3(aq) + 2 \ \text{HCl}(aq) \rightarrow \text{H}_2\text{O}(l) + \text{CO}_2(g) + 2 \ \text{NaCl}(aq) \]. It tells us that one molecule of \( \text{Na}_2\text{CO}_3 \) reacts with two molecules of HCl.
This 1:2 ratio means for every mole of \( \text{Na}_2\text{CO}_3 \) there are two moles of HCl required. Once you know the moles of one reactant, you can calculate the moles of all others involved using these ratios.
By understanding stoichiometry, you can predict how much of each substance you need to fully complete a reaction, ensuring efficiency and preventing waste.

Molarity

Molarity is a measure of the concentration of a solute in a solution. It tells us how many moles of a substance are present in one liter of solution.Molarity (\(M\)) is calculated using the formula
\[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
In the problem, we need to find out how much 0.955 M HCl is necessary. Molarity is useful for finding out exactly how much of a chemical is contained within a specific volume of solution.
Knowing the molarity helps in performing accurate chemical calculations and understanding how different solutions will react with each other.

Chemical Reactions

Chemical reactions represent the process where reactants are transformed into products. The equation provided — \( \text{Na}_2\text{CO}_3(aq) + 2 \ \text{HCl}(aq) \rightarrow \text{H}_2\text{O}(l) + \text{CO}_2(g) + 2 \ \text{NaCl}(aq) \) — is a balanced equation representing the reaction in this exercise.
The

• Reactants are \( \text{Na}_2\text{CO}_3 \) and HCl.
• Products formed are water \( \text{H}_2\text{O} \), carbon dioxide \( \text{CO}_2 \), and sodium chloride \( \text{NaCl} \).

In a chemical reaction, the total mass and number of atoms stay the same due to the law of conservation of mass.
Balancing a chemical equation is crucial as it ensures that the number of atoms for each element is the same on both sides of the equation.
This balance allows us to use stoichiometry, as previously discussed, to determine the correct proportions of reactants and products.

Equivalence Point

The equivalence point in titration is when the amount of titrant added is perfectly sufficient to react with the analyte in the solution.
In simpler terms, it's when the exact amount of one solution is added to neutralize or completely react with a specific amount of another.
In the reaction we have:

• \( \text{Na}_2\text{CO}_3 \) as the analyte;
• HCl as the titrant.

The equivalence point is crucial because it signals that the reaction is complete.
The measurement occurs when all the analyte has reacted, not less, not more, ensuring accuracy in chemical analysis and solution preparation.
Understanding this concept helps ensure that you have applied the right amounts to achieve a neutral or complete reaction at the endpoint of a titration.