Question

What volume of \(0.101 M \mathrm{HNO}_{3}\) is required to neutralize each of the following solutions? a. \(12.7 \mathrm{mL}\) of \(0.501 \mathrm{M} \mathrm{NaOH}\) b. \(24.9 \mathrm{mL}\) of \(0.00491 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) c. \(49.1 \mathrm{mL}\) of \(0.103 \mathrm{M} \mathrm{NH}_{3}\) d. 1.21 L of \(0.102 M\) KOH

Short answer

The required volumes of \(0.101 M HNO_3\) to neutralize the given base solutions are: a. \(63.0 \, mL\) for \(12.7\, mL\) of \(0.501M\, NaOH\) b. \(2.42 \, mL\) for \(24.9 \, mL\) of \(0.00491M\, Ba(OH)_2\) c. \(50.0 \, mL\) for \(49.1\, mL\) of \(0.103M\, NH_3\) d. \(1.222 \, L\) for \(1.21\, L\) of \(0.102M\, KOH\)

Step-by-step solution

Write down the balanced chemical equations

First, we need to write down the balanced equations for each neutralization reaction between the given base and \(HNO_3\). Remember that in a neutralization reaction, an acid reacts with a base to form a salt and water. a. \(NaOH + HNO_3 \rightarrow NaNO_3 + H_2O\) b. \(Ba(OH)_2 + 2HNO_3 \rightarrow Ba(NO_3)_2 + 2H_2O\) c. \(NH_3 + HNO_3 \rightarrow NH_4NO_3\) d. \(KOH + HNO_3 \rightarrow KNO_3 + H_2O\)

Calculate moles of the given base

In this step, we will calculate the moles of the given base solution using the formula: moles = volume(L) × concentration(moles/L) a. Moles of \(NaOH\): \(12.7 \times 10^{-3} L \times 0.501 M = 6.3627 \times 10^{-3} \, moles\) b. Moles of \(Ba(OH)_2\): \(24.9 \times 10^{-3} L \times 0.00491 M = 1.22359 \times 10^{-4} \, moles\) c. Moles of \(NH_3\): \(49.1 \times 10^{-3} L \times 0.103 M = 5.0573 \times 10^{-3} \, moles\) d. Moles of \(KOH\): \(1.21 L \times 0.102 M = 0.12342 \, moles\)

Determine moles of HNO3 needed for neutralization

Now, we will find the moles of \(HNO_3\) required according to balanced chemical equations: a. Moles of \(HNO_3\) = \(6.3627 \times 10^{-3} \, moles\) b. Moles of \(HNO_3\) = \(2 \times 1.22359 \times 10^{-4}\, moles = 2.44718 \times 10^{-4} \, moles\) c. Moles of \(HNO_3\) = \(5.0573 \times 10^{-3} \, moles\) d. Moles of \(HNO_3\) = \(0.12342 \, moles\)

Calculate the volume of HNO3 required

To find the volume of \(0.101 M\, HNO_3\) needed for neutralization, we use the formula: volume(L) = moles / concentration(moles/L) a. Volume of \(HNO_3\): \(\frac{6.3627 \times 10^{-3} \, moles}{0.101 M} = 6.3 \times 10^{-2} L = 63.0 \, mL\) b. Volume of \(HNO_3\): \(\frac{2.44718 \times 10^{-4} \, moles}{0.101 M} = 2.42 \times 10^{-3} L = 2.42 \, mL\) c. Volume of \(HNO_3\): \(\frac{5.0573 \times 10^{-3} \, moles}{0.101 M} = 5.0 \times 10^{-2} L = 50.0 \, mL\) d. Volume of \(HNO_3\): \(\frac{0.12342 \, moles}{0.101 M} = 1.222 L\) This means: a. \(63.0 \, mL\) of \(0.101M\, HNO_3\) is required to neutralize \(12.7\, mL\) of \(0.501M\, NaOH\) b. \(2.42 \, mL\) of \(0.101M\, HNO_3\) is required to neutralize \(24.9 \, mL\) of \(0.00491M\, Ba(OH)_2\) c. \(50.0 \, mL\) of \(0.101M\, HNO_3\) is required to neutralize \(49.1\, mL\) of \(0.103M\, NH_3\) d. \(1.222 \, L\) of \(0.101M\, HNO_3\) is required to neutralize \(1.21\, L\) of \(0.102M\, KOH\)

Key concepts

Acid-Base Titration

Acid-base titrations are a fundamental procedure used in chemistry to determine the concentration of an unknown acid or base solution. By carefully adding a reactant with a known concentration, called a titrant, to the unknown solution until the reaction is complete, which is often indicated by a color change from an indicator, one can calculate the unknown concentration using stoichiometry.

For example, if we're titrating a base with an acid, as the problem describes, we would measure how much acid (titrant) is needed to completely neutralize the base. The point at which neutralization occurs is known as the equivalence point. This concept is crucial when performing calculations to determine the necessary volume of acid to neutralize a base, as illustrated in the provided exercise.

Molarity Calculations

Molarity, denoted by the symbol 'M', is a measure of concentration in chemistry that indicates the number of moles of a solute per liter of solution. It is given by the formula: \( M = \frac{{moles\text{{ of solute}}}}{{volume\text{{ of solution in liters}}}} \).

Knowing how to calculate molarity is essential when working with solutions in chemistry, such as determining how much of an acid is needed to neutralize a base, as seen in our exercise. This calculation also becomes the cornerstone for determining the volumes or amounts of reactants in various chemical reactions.

Stoichiometry

Stoichiometry is the part of chemistry that involves using balanced chemical equations to calculate mass, volume, and number of moles of reactants and products. Using the coefficients from the balanced equations, stoichiometry allows the conversion between moles of different substances involved in a reaction.

In the context of the exercise, stoichiometry is used to determine the moles of acid needed to neutralize a given amount of a base. For instance, with the reaction involving barium hydroxide, the stoichiometry indicates that two moles of nitric acid are needed for every mole of barium hydroxide. This is a crucial step to ensure the correct volume of acid is used for neutralization.

Balanced Chemical Equations

A balanced chemical equation is a representation of a chemical reaction where the number of atoms for each element is the same on both the reactant and product sides. Balancing chemical equations is critical to understanding the quantitative relationships in a chemical reaction and thus is essential to stoichiometry.

The exercise provided incorporates four balanced chemical equations for neutralization reactions. These equations are necessary to accurately calculate the stoichiometry of the reaction, which in turn is used to solve for the volumes of acid required to neutralize the bases. A thorough understanding of how to balance equations is vital for performing accurate chemical calculations in such neutralization reactions.