Problem 69 - Chemistry and Chemical Reactivity

Question

What volume of \(0.812 \mathrm{M} \mathrm{HCl}\), in milliliters, is required to titrate \(1.45 \mathrm{g}\) of \(\mathrm{NaOH}\) to the equivalence point? \(\mathrm{NaOH}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{NaCl}(\mathrm{aq})\)

Step-by-Step Solution

Answer

44.654 mL of 0.812 M HCl is needed to titrate 1.45 g of NaOH.

Step 1: Calculate the moles of NaOH

First, we need to determine how many moles of NaOH we have from the given mass. The molar mass of NaOH is approximately 40.00 g/mol. Use the formula: \( \text{moles of NaOH} = \frac{\text{mass of NaOH (g)}}{\text{molar mass of NaOH (g/mol)}} \). Substituting the values, \( \text{moles of NaOH} = \frac{1.45 \text{ g}}{40.00 \text{ g/mol}} = 0.03625 \text{ moles} \).

Step 2: Use the reaction stoichiometry

According to the balanced chemical equation, \( \text{NaOH} + \text{HCl} \rightarrow \text{H}_2\text{O} + \text{NaCl} \), there is a 1:1 molar ratio between NaOH and HCl. Therefore, the moles of HCl required will also be 0.03625 moles, the same as the moles of NaOH.

Step 3: Calculate the volume of HCl solution required

To find the volume of the HCl solution needed, use the formula: \( \text{Volume (L)} = \frac{\text{moles of HCl}}{\text{concentration (M)}} \). Substitute the known values: \( \text{Volume (L)} = \frac{0.03625 \text{ moles}}{0.812 \text{ M}} = 0.044654 L \). Convert this volume from liters to milliliters by multiplying by 1000: \( 0.044654 \times 1000 = 44.654 \text{ mL} \).

Key Concepts

Mole Calculations

When solving chemistry problems, knowing how to calculate moles is key. The mole is a fundamental unit in chemistry used to quantify amounts of a chemical substance. To figure out moles, you use the mass of a substance along with its molar mass.
In this exercise, we started by calculating the moles of NaOH present. The formula you need is simple:

• \( \text{moles} = \frac{\text{mass of substance (g)}}{\text{molar mass (g/mol)}} \)

For NaOH, the mass given was 1.45 g, and the molar mass is roughly 40.00 g/mol. Plug these numbers into the formula:

• \( \text{moles of NaOH} = \frac{1.45 \text{ g}}{40.00 \text{ g/mol}} = 0.03625 \text{ moles} \)

Understanding mole calculations helps us bridge between the macroscopic quantities we can measure and the microscopic world of atoms and molecules. This foundational skill aids in determining the amounts needed in reaction stoichiometry.

Reaction Stoichiometry

Stoichiometry involves using balanced chemical equations to determine the relationships between reactants and products in a reaction. This is essential in predicting how much of each reactant is needed to form a given amount of product or vice versa.
In the balanced chemical equation for the neutralization reaction given:

• \( \text{NaOH} + \text{HCl} \rightarrow \text{H}_2\text{O} + \text{NaCl} \)

There is a 1:1 molar ratio between NaOH and HCl. This means that one mole of NaOH reacts with one mole of HCl. In our calculation:

• The moles of HCl required is the same as the moles of NaOH, which is 0.03625 moles.

This stoichiometric relationship ensures precise calculations in chemical reactions, preserving matter in accordance with the law of conservation of mass. Knowing stoichiometry is crucial for lab work and industrial processes where precise reactions are necessary.

Acid-Base Reaction

An acid-base reaction is a chemical process where an acid reacts with a base to produce water and a salt. This type of reaction is fundamental in chemistry, especially in the process of titration.
Titration is a technique used to determine the amount of a reactant in a solution. For our problem, we're titrating NaOH, a strong base, with HCl, a strong acid. During titration, the acid reacts with the base and reaches an equivalence point where they neutralize each other.
In this case, the reaction is:

• \( \text{NaOH} + \text{HCl} \rightarrow \text{H}_2\text{O} + \text{NaCl} \)

At the equivalence point, the amount of acid equals the amount of base, which forms the basis of the calculation to find the volume of HCl needed.
Understanding acid-base reactions, especially in terms of titrations, is essential for lab chemistry. It helps in determining concentrations and understanding the behavior of acids and bases in various chemical processes.