Question

Suppose that 10.1 \(\mathrm{mL}\) of \(\mathrm{HNO}_{3}\) is neutralized by 71.4 \(\mathrm{mL}\) of a \(4.2 \times 10^{-3} \mathrm{M}\) solution of KOH in a titration. Calculate the concentration of the HNO \(_{3}\) solution.

Short answer

The concentration of HNO₃ is 0.0297 M.

Step-by-step solution

- Write the neutralization reaction

The balanced chemical equation for the neutralization reaction is: \[ \text{HNO}_3 + \text{KOH} \rightarrow \text{KNO}_3 + \text{H}_2\text{O} \]This reaction indicates that one mole of \text{HNO}_3 reacts with one mole of \text{KOH}.

- Identify the known values

Given values:- Volume of \(\text{HNO}_3\): 10.1 mL- Volume of \(\text{KOH}\): 71.4 mL- Concentration of \(\text{KOH}\): \(4.2 \times 10^{-3}\) M

- Calculate the moles of KOH used

To find the moles of KOH, use the formula:\[ \text{Moles of KOH} = \text{Concentration} \times \text{Volume} \]Convert the volume of KOH from mL to L (71.4 mL = 0.0714 L):\[ \text{Moles of KOH} = 4.2 \times 10^{-3} \text{M} \times 0.0714 \text{L} \]\[ \text{Moles of KOH} = 3.0 \times 10^{-4} \text{ moles} \]

- Determine the moles of HNO3 neutralized

From the balanced chemical equation, the reaction ratio is 1:1. Therefore, the moles of \(\text{HNO}_3\) neutralized are the same as the moles of \(\text{KOH}\) used.\[ \text{Moles of HNO}_3 = 3.0 \times 10^{-4} \text{ moles} \]

- Calculate the concentration of HNO3

To find the concentration of \(\text{HNO}_3\), use the formula:\[ \text{Concentration} = \frac{\text{Moles}}{\text{Volume}} \]Convert the volume of \(\text{HNO}_3\) from mL to L (10.1 mL = 0.0101 L):\[ \text{Concentration of HNO}_3 = \frac{3.0 \times 10^{-4} \text{ moles}}{0.0101 \text{ L}} \]\[ \text{Concentration of HNO}_3 = 2.97 \times 10^{-2} \text{ M} \]

Key concepts

Neutralization Reaction

In chemistry, a neutralization reaction occurs when an acid reacts with a base to form water and a salt. Typically, this is represented by the equation: \(\text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water}\). For the given exercise, the balanced chemical equation is: \[ \text{HNO}_3 + \text{KOH} \rightarrow \text{KNO}_3 + \text{H}_2\text{O} \] This equation tells us that one mole of nitric acid (\text{HNO}_3) reacts with one mole of potassium hydroxide (\text{KOH}) to produce one mole of potassium nitrate (\text{KNO}_3) and water (\text{H}_2\text{O}). The 1:1 molar ratio is crucial for understanding the stoichiometry of the reaction. When working with titration problems, recognizing this ratio helps in determining how much of one reactant is needed to completely react with a given amount of another reactant.

Concentration Calculation

Concentration is an essential concept in chemistry, often expressed in terms of molarity (M), which is the number of moles of solute per liter of solution. In the exercise, we need to calculate the concentration of \text{HNO}_3. Using the general formula:

\(\text{Concentration (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}}\).

First, we need to determine the moles of \text{KOH} and then use the reaction stoichiometry to find the moles of \text{HNO}_3. Once we have that, the given volume of \text{HNO}_3 can be converted from mL to L, and then we apply the concentration formula. This approach helps in understanding how the amount of a substance in a solution influences the overall concentration.

Mole Concept

The mole is a fundamental unit in chemistry that quantifies the amount of a substance. One mole corresponds to approximately \(6.022 \times 10^{23}\) particles, which is Avogadro's number. In the given problem, we calculate the moles of \text{KOH} using its concentration and volume. The formula:

\[ \text{Moles of solute} = \text{Concentration (M)} \times \text{Volume (L)} \] allows us to determine the moles of \text{KOH}. Given: Concentration of KOH = \(4.2 \times 10^{-3} \text{ M}\) and Volume of KOH = 71.4 mL (converted to 0.0714 L), the moles of \text{KOH} can be calculated as: \(4.2 \times 10^{-3} \text{ M} \times 0.0714 \text{ L} = 3.0 \times 10^{-4} \text{ moles}\). By understanding the mole concept, you can properly relate the quantities of substances involved in chemical reactions.

Volume Conversion

Volume conversion is often necessary when dealing with different units in chemistry problems. Usually, volumes are given in milliliters (mL), but standard calculations for concentration use liters (L). The conversion factor to remember is:

\[ 1 \text{ L} = 1000 \text{ mL} \] For the exercise, the volumes of both \text{HNO}_3 and \text{KOH} need to be converted from mL to L.
The given volume of \text{HNO}_3 is 10.1 mL, which converts to: \( \frac{10.1 \text{ mL}}{1000} = 0.0101 \text{ L} \). Similarly, the volume of \text{KOH} is 71.4 mL, converting to: \( \frac{71.4 \text{ mL}}{1000} = 0.0714 \text{ L} \). Accurate volume conversion ensures precise calculations and is a vital step in many chemistry problems.