Question

What is the molarity of a \(\mathrm{HCl}\) solution if \(5.00 \mathrm{~mL}\) is neutralized by \(28.6 \mathrm{~mL}\) of a \(0.145 \mathrm{M} \mathrm{NaOH}\) solution? $$ \mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$

Short answer

The molarity of the \(\mathrm{HCl}\) solution is \(0.8294 \; \mathrm{M}\).

Step-by-step solution

- Write the balanced chemical equation

The balanced chemical equation for the reaction is: \[ \mathrm{HCl}(aq) + \mathrm{NaOH}(aq) \to \mathrm{NaCl}(aq) + \mathrm{H_2O}(l) \]

- Identify the relevant information

Given data:- Volume of \(\mathrm{HCl}\): \(5.00 \; \mathrm{mL}\)- Volume of \(\mathrm{NaOH}\): \(28.6 \; \mathrm{mL}\)- Molarity of \(\mathrm{NaOH}\): \(0.145 \; \mathrm{M}\)

- Convert volume of \(\mathrm{NaOH}\) used to liters

The volume needs to be in liters to use in molarity calculations. \[ 28.6 \; \mathrm{mL} = 28.6 \times 10^{-3} \; \mathrm{L} = 0.0286 \; \mathrm{L} \]

- Calculate moles of \(\mathrm{NaOH}\)

Using the molarity equation: \( M = \frac{n}{V} \), where \(n\) is the number of moles and \(V\) is the volume in liters. \[ n_{\mathrm{NaOH}} = M \times V = 0.145 \; \mathrm{M} \times 0.0286 \; \mathrm{L} = 0.004147 \; \mathrm{mol} \]

- Find moles of \(\mathrm{HCl}\)

From the balanced equation, the mole ratio of \(\mathrm{HCl}\) to \(\mathrm{NaOH}\) is 1:1. Thus, \(n_{\mathrm{HCl}} = n_{\mathrm{NaOH}}\). \[n_{\mathrm{HCl}} = 0.004147 \; \mathrm{mol} \]

- Convert volume of \(\mathrm{HCl}\) solution to liters

The volume needs to be in liters to use in molarity calculations. \[5.00 \; \mathrm{mL} = 5.00 \times 10^{-3} \; \mathrm{L} = 0.00500 \; \mathrm{L} \]

- Calculate the molarity of the \(\mathrm{HCl}\) solution

Again using the molarity equation: \( M = \frac{n}{V} \). \[ M_{\mathrm{HCl}} = \frac{0.004147 \; \mathrm{mol}}{0.00500 \; \mathrm{L}} = 0.8294 \; \mathrm{M} \]

Key concepts

neutralization reaction

In chemistry, a neutralization reaction is where an acid and a base react to form water and a salt. These reactions are important because they help us understand how substances interact in solution to produce neutral compounds. In the exercise, the balanced chemical equation is provided: \( \text{HCl}(aq) + \text{NaOH}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l) \). Here, hydrochloric acid (\(\text{HCl}\)) reacts with sodium hydroxide (\(\text{NaOH}\)) to form sodium chloride (table salt, \(\text{NaCl}\)) and water (\(\text{H}_2\text{O}\)). This type of reaction is typical for a strong acid and a strong base. It is essential to balance the chemical equation to make sure the number of each type of atom is the same on both sides of the reaction.
The neutralization reaction also involves the release of heat as energy, known as an exothermic reaction.

mole ratio

The mole ratio is a crucial concept in stoichiometry that helps us understand the proportions of reactants and products involved in a chemical reaction. In the balanced equation \( \text{HCl}(aq) + \text{NaOH}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l) \), the mole ratio of \(\text{HCl}\) to \(\text{NaOH}\) is 1:1. This means that one mole of \(\text{HCl}\) reacts with one mole of \(\text{NaOH}\) to produce one mole of \(\text{NaCl}\) and one mole of \(\text{H}_2\text{O}\). In our exercise, we used this 1:1 mole ratio to find that the moles of \(\text{HCl}\) are equal to the moles of \(\text{NaOH}\). This makes it straightforward to calculate the amount of the unknown solution when you know the amount of the other reactant.
It is always important to first ensure the chemical equation is balanced properly before using mole ratios for calculations.

molarity equation

Molarity (\(M\)) is a way to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution. The molarity equation is: \( M = \frac{n}{V} \), where \( n \) is the number of moles of the solute and \( V \) is the volume of the solution in liters.
In the given exercise, we use this equation to find the molarity of the \(\text{HCl}\) solution. First, we calculate the moles of \(\text{NaOH}\) using its volume and molarity: \( n_{\text{NaOH}} = M_{\text{NaOH}} \times V_{\text{NaOH}} \). Then, because of the 1:1 mole ratio, \( n_{\text{HCl}} = n_{\text{NaOH}} \). Finally, we use the volume of the \(\text{HCl}\) solution to find its molarity: \( M_{\text{HCl}} = \frac{n_{\text{HCl}}}{V_{\text{HCl}}} \). Understanding and using this equation properly is crucial in various fields, from laboratory work to industrial applications.

volume conversion

Volume conversion is essential when dealing with different units in chemical calculations. In the provided exercise, volumes are initially given in milliliters (mL). However, the molarity equation requires volumes to be in liters (L). Therefore, we need to convert milliliters to liters by dividing by 1000:
\( 28.6 \text{ mL} = 28.6 \times 10^{-3} \text{ L} = 0.0286 \text{ L} \)
\( 5.00 \text{ mL} = 5.00 \times 10^{-3} \text{ L} = 0.00500 \text{ L} \)
This step is critical because using the wrong units can lead to incorrect calculations and results. Always make sure to use consistent units throughout your calculations to get accurate results.