Problem 67 - Chemistry and Chemical Reactivity

Question

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

Step-by-Step Solution

Answer

44.64 mL of 0.812 M HCl is required.

Step 1: Determine the moles of NaOH

First, we need to calculate the number of moles of NaOH present in 1.45 grams. The molar mass of NaOH is calculated as follows: \( \text{Molar mass of } \text{NaOH} = \text{Molar mass of } \text{Na} + \text{Molar mass of } \text{O} + \text{Molar mass of } \text{H} = 22.99 + 16.00 + 1.01 = 40.00 \, \text{g/mol} \). Therefore, the number of moles of NaOH is \( \frac{1.45 \, \text{g}}{40.00 \, \text{g/mol}} = 0.03625 \, \text{mol} \).

Step 2: Use stoichiometry to find moles of HCl

From the balanced chemical equation, the reaction between NaOH and HCl is 1:1, meaning 1 mole of HCl reacts with 1 mole of NaOH. Thus, the moles of HCl needed are equal to the moles of NaOH, which is \( 0.03625 \, \text{mol} \).

Step 3: Calculate volume of HCl solution

Now, use the molarity formula \( M = \frac{n}{V} \) to find the volume of the HCl solution required. Rearrange this to find \( V = \frac{n}{M} \), where \( n = 0.03625 \, \text{mol} \) and \( M = 0.812 \, \text{M} \). Thus, \( V = \frac{0.03625}{0.812} \approx 0.04464 \, \text{L} \). Convert this to milliliters: \( 0.04464 \, \text{L} \times 1000 \, \frac{\text{mL}}{\text{L}} = 44.64 \, \text{mL} \).

Key Concepts

Stoichiometry

Stoichiometry helps us understand the quantitative relationships within a chemical reaction. It's based on the law of conservation of mass and the use of balanced chemical equations to determine the proportions of reactants and products. For example, in our titration exercise, the reaction between sodium hydroxide \( \text{NaOH} \)and hydrochloric acid \( \text{HCl} \)is given by: \( \text{NaOH}_{(aq)} + \text{HCl}_{(aq)} \rightarrow \text{H}_{2}\text{O}_{(\ell)} + \text{NaCl}_{(aq)} \).

• This equation tells us that one mole of \( \text{NaOH} \)reacts with one mole of \( \text{HCl} \)to form one mole of \( \text{NaCl} \)and one mole of \( \text{H}_2\text{O} \).
• So, if you know the amount of one substance (like \( \text{NaOH} \), which is given by its mass), you can find the required moles of the other reactant (\( \text{HCl} \) in this example) using their molar ratios.

Here, stoichiometry is crucial for determining the exact amount of \( \text{HCl} \)needed to reach the equivalence point of the titration.

Molarity

Molarity is a way to express the concentration of a solution. It's defined as the number of moles of solute per liter of solution. In the context of our titration problem, knowing the molarity of \( \text{HCl} \) is essential.

• The given problem tells us the molarity \( (0.812 \text{ M}) \), which means 0.812 moles of \( \text{HCl} \) are present in one liter of the solution.
• This value is used to calculate the volume of \( \text{HCl} \)needed for the reaction with \( \text{NaOH} \).

The formula to determine how much \( \text{HCl} \)you need is given by \( M = \frac{n}{V} \), where \( n \) is the number of moles and \( V \) is the volume in liters. Rearranging gives \( V = \frac{n}{M} \). This expression helped us find that the required volume was 44.64 mL, connecting the stoichiometric calculations with molarity.

Equivalence Point

The equivalence point in a titration is the moment when the amount of titrant added is stoichiometrically equivalent to the quantity of analyte present in the sample. Basically, it's the point at which the moles of \( \text{HCl} \)added equal the moles of \( \text{NaOH} \)we started with, resulting in a complete reaction with no excess reactants.

• In the balanced chemical equation, \( \text{NaOH}_{(aq)} + \text{HCl}_{(aq)} \rightarrow \text{H}_{2}\text{O}_{(\ell)} + \text{NaCl}_{(aq)} \), we see that they react in a 1:1 ratio.
• This means when 0.03625 moles of \( \text{NaOH} \)are present, it's neutralized by the same amount of \( \text{HCl} \).

The equivalence point is crucial because it signifies the completion of the reaction, determining how much titrant is necessary. This understanding is fundamental in titrations and forms the basis of determining unknown concentrations accurately.