Problem 52 - General Chemistry: Principles and Modern Applications

Question

How many milliliters of \(0.0844 \mathrm{MBa}(\mathrm{OH})_{2}\) are required to titrate \(50.00 \mathrm{mL}\) of \(0.0526 \mathrm{M} \mathrm{HNO}_{3} ?\)

Step-by-Step Solution

Answer

You would require approximately 15.6 mL of \(0.0844 \mathrm{MBa}(\mathrm{OH})_{2}\) to titrate 50.00 mL of \(0.0526 \mathrm{M} \mathrm{HNO}_{3}\).

Step 1: Analyze the chemical reaction

In the reaction between nitric acid and barium hydroxide, 1 mole of barium hydroxide reacts with 2 moles of nitric acid. You can write it down as follows: \(Ba(OH)_2 + 2HNO_3 \rightarrow Ba(NO_3)_2 + 2H_2O\)

Step 2: Calculate moles of nitric acid

First, calculate the moles of nitric acid (\(HNO_3\)) in the metioned volume. The moles of \(HNO_3\) can be calculated using the given concentration and volume. You can use the formula: moles = concentration (in M) * Volume(in Liters). The volume needs to be converted from mL to Liters, where 1mL = 0.001L. Hence, moles of \(HNO_3\)= \(0.0526 \mathrm{M} * 50.00 \mathrm{mL}*0.001 \mathrm{L/m}L\) = \(0.00263 \mathrm{mol}\)

Step 3: Calculate volume of \(Ba(OH)_2\)

Since 2 moles of nitric acid react with 1 mole of barium hydroxide, the number of moles of \(Ba(OH)_2\) required to completely react with \(0.00263 \mathrm{mol}\) of \(HNO3\) will be half of that, i.e., \(0.001315 \mathrm{mol}\). Now, let V be the volume in liters of \(Ba(OH)_2\). As concentration \(C = \frac{moles}{volume}\), so rearranging for volume, we get \(V = \frac{moles}{C}\). Substituting for moles and C, we get: \(V = \frac{0.001315 \mathrm{mol}}{0.0844 \mathrm{M}} = 0.0156 \mathrm{L}\). Converting this to mL (as 1L = 1000mL), we get \(V = 0.0156 * 1000 = 15.6 \mathrm{mL}\).

Key Concepts

Barium Hydroxide

Barium hydroxide is a chemical compound with the formula \(Ba(OH)_2\). It is a strong base commonly used in titration, a technique often employed to determine the concentration of an unknown solution.

Barium hydroxide is composed of barium ions \(Ba^{2+}\) and hydroxide ions \(OH^-\). It fully dissociates in water to release these ions. This dissociation is crucial in titrations, as the hydroxide ions react with hydrogen ions \(H^+\) from acids, neutralizing them to form water.

• The high solubility of barium hydroxide in water makes it useful in titrations requiring a strong base.
• In the presence of acids, it combines readily to form water and a salt, in this case, barium nitrate \(Ba(NO_3)_2\).

Understanding these characteristics of barium hydroxide helps liven up the core aspect of its role in titration processes.

Nitric Acid

Nitric acid, with the chemical formula \(HNO_3\), is a strong and corrosive acid. It is highly reactive and plays an essential role in various chemical reactions including titration with bases like barium hydroxide.

When nitric acid dissolves in water, it fully ionizes to produce hydrogen ions \(H^+\) and nitrate ions \(NO_3^-\). This property is integral for its use in chemical reactions:

• It acts as a strong acid component in neutralization reactions.
• The dissociation of \(H^+\) ions is the key factor that allows it to readily interact with the hydroxide ions from bases.

Upon reaction with barium hydroxide, a salt, barium nitrate \(Ba(NO_3)_2\), and water are formed. This formation is the result of hydrogen ions from nitric acid reacting with hydroxide ions from the base. The neutrality of the net reaction indicates a complete neutralization process.

Mole Calculation

The concept of mole calculation is fundamental when performing titrations, as it allows for the quantification of reactants and products in a chemical reaction.

Moles express the amount of substance and can be calculated using the formula:
\[ \text{moles} = \text{concentration (in M)} \times \text{Volume (in Liters)} \]
For example, to calculate the moles of nitric acid in a solution, you need to know both the concentration and the volume of the solution.

• Convert the volume from milliliters to liters by multiplying with \(0.001\), since \(1 \text{ mL} = 0.001 \text{ L}\).
• The resulting multiplication gives the number of moles of \(HNO_3\): \(0.0526 \text{ M} \times 50.00 \text{ mL} \times 0.001 = 0.00263 \text{ mol}\).

This concept enables further calculations, such as determining the required amount of another reactant using stoichiometry from the balanced reaction equation.

Chemical Reactions

Chemical reactions are processes in which substances interact to form new products. In the context of titration between barium hydroxide and nitric acid, the chemical reaction can be represented by the equation:

\[ Ba(OH)_2 + 2HNO_3 \rightarrow Ba(NO_3)_2 + 2H_2O \]
This balanced equation shows that one mole of barium hydroxide reacts with two moles of nitric acid to produce one mole of barium nitrate and two moles of water.

• The coefficients of the reactants and products are crucial for mole calculation, dictating stoichiometric ratios.
• A balanced equation ensures that the law of conservation of mass is obeyed, indicating what quantities of reactants are required or what quantities of products are created.

In a titration, this knowledge is used to determine unknown concentrations by measuring the volume and using stoichiometry to relate the reactants and products. It allows chemists to predict how much barium hydroxide is necessary to completely neutralize a given amount of nitric acid, as seen in the exercise.