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? $$\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{NaCl}(\mathrm{aq})$$

Step-by-Step Solution

Answer

42.5 mL of \(0.955 \text{ M HCl}\) is needed.

Step 1: Determine the Molar Mass of Na2CO3

Calculate the molar mass of \( \text{Na}_2\text{CO}_3 \). It consists of 2 sodium (Na) atoms, 1 carbon (C) atom, and 3 oxygen (O) atoms. Using the atomic masses: \( \text{Na} = 22.99 \text{ g/mol}, \text{C} = 12.01 \text{ g/mol}, \text{O} = 16.00 \text{ g/mol} \),\[ \text{Molar mass of } \text{Na}_2\text{CO}_3 = 2(22.99) + 12.01 + 3(16.00) = 105.99 \text{ g/mol} \].

Step 2: Calculate Moles of Na2CO3

Using the mass of \( \text{Na}_2\text{CO}_3 \) given, calculate the number of moles. \[ \text{Moles of } \text{Na}_2\text{CO}_3 = \frac{2.152 \text{ g}}{105.99 \text{ g/mol}} = 0.0203 \text{ moles} \].

Step 3: Use Stoichiometry to Find Moles of HCl Required

From the chemical equation, \(1\) mole of \( \text{Na}_2\text{CO}_3 \) reacts with \(2\) moles of \( \text{HCl} \). Therefore, double the moles of \( \text{Na}_2\text{CO}_3 \) to find the moles of \( \text{HCl} \):\[ \text{Moles of } \text{HCl} = 2 \times 0.0203 = 0.0406 \text{ moles} \].

Step 4: Calculate Volume of HCl Solution Required

Use the molarity formula to find the volume of \(0.955 \text{ M HCl} \):\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]Rearrange to find volume:\[ \text{Volume in liters} = \frac{0.0406 \text{ moles}}{0.955 \text{ M}} = 0.0425 \text{ liters} \].

Step 5: Convert Volume from Liters to Milliliters

Since 1 liter is equal to 1000 milliliters, convert the volume:\[ \text{Volume in milliliters} = 0.0425 \text{ liters} \times 1000 \text{ mL/L} = 42.5 \text{ mL} \].

Key Concepts

Molarity

Molarity is a key concept in chemistry, which defines the concentration of a solute in a solution. It is expressed as moles of solute per liter of solution. Molarity is denoted by the unit M (Molar). For example, in this exercise, we use a 0.955 M HCl solution.
Molarity comes in handy during titration, as it allows us to calculate how much of a reagent is needed to react with a given amount of a substance. The formula for molarity is:

• \( \text{Molarity} (M) = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \)

Having the molarity of a solution helps determine the exact volume required for a reaction to reach its equivalence point, ensuring the precise amount of reactants are present.

Stoichiometry

Stoichiometry is like the math behind chemistry. It involves the calculation of reactants and products in chemical reactions. In our exercise, we focus on the stoichiometric relationship between \( \text{Na}_2\text{CO}_3 \) and HCl.From the equation:

• \( 1 \text{ mole of Na}_2\text{CO}_3 \) reacts with \( 2 \text{ moles of HCl} \)

Using this relationship, we can predict the amount of HCl needed by simply multiplying the moles of \( \text{Na}_2\text{CO}_3 \) by 2, as shown in the original solution. Understanding stoichiometry is crucial for balancing chemical equations and performing accurate chemical calculations.

Molar Mass

The molar mass of a substance is its mass per mole of particles. It's expressed in grams per mole (g/mol). To compute it, sum up the atomic masses of all atoms in a molecule. In this exercise, \( \text{Na}_2\text{CO}_3 \) is broken down into its constituent atoms:

• 2 Sodium (Na) atoms: \( 22.99 \text{ g/mol} \)
• 1 Carbon (C) atom: \( 12.01 \text{ g/mol} \)
• 3 Oxygen (O) atoms: \( 16.00 \text{ g/mol} \)

Adding these together gives the molar mass as \( 105.99 \text{ g/mol} \). Knowing molar mass allows us to convert a substance's mass to moles, which is necessary for stoichiometry and determining reactivity.

Conversion of Units

Converting units is a vital skill in chemistry to ensure calculations are in the correct form. In our exercise, we're dealing with conversion from liters to milliliters. Understanding this conversion is straightforward:

• 1 liter = 1000 milliliters

So, to convert a volume measured in liters to milliliters, multiply by 1000. Therefore, if the calculated volume of HCl is \( 0.0425 \text{ liters} \), converting it into milliliters involves:

• \( 0.0425 \text{ liters} \times 1000 \text{ mL/L} = 42.5 \text{ mL} \)

These conversions ensure the final answer is expressed in a practical unit suitable for experiments.