Problem 132 - Introductory Chemistry Atoms First

Question

A solution is prepared by dissolving \(0.250\) mole of \(\mathrm{Ba}(\mathrm{OH})_{2}\) in enough water to get \(4.00 \mathrm{I}\). of solution. What are the \(\mathrm{OH}^{-}\) and \(\mathrm{H}_{3} \mathrm{O}^{+}\) molar concentrations?

Step-by-Step Solution

Answer

The concentration of Ba(OH)₂ in the solution is: \(Concentration = \frac{0.250 moles}{4.00 L}\). Knowing that one molecule of Ba(OH)₂ will produce two molecules of OH⁻ ions, the concentration of OH⁻ ions can be calculated as: \( Concentration \, of \, OH^- = 2 \times \frac{0.250 \, moles}{4.00 \, L}\). Using the ion product of water (Kw) and the relationship between OH⁻ and H3O⁺ concentrations, the concentration of H₃O⁺ ions can be found: \([H_3 O^+] = \frac{1 \times 10^{-14}}{2 \times \frac{0.250 \, moles}{4.00 \, L}} \). Calculate the values to obtain the molar concentrations of OH⁻ and H₃O⁺ ions.

Step 1: Calculate the concentration of Ba(OH)₂

To find the concentration of Ba(OH)₂, we can use the formula: Concentration = (moles of solute) / (volume of solution in Liters) We are given the moles of solute (0.250 moles) and the volume of the solution (4.00 L). Therefore, the concentration of Ba(OH)₂ can be calculated as: \[Concentration = \frac{0.250 \, moles}{4.00 \, L}\]

Step 2: Calculate the concentration of OH⁻ ions produced

We know that one molecule of Ba(OH)₂ will produce two molecules of OH⁻ ions when dissolved in water: Ba(OH)₂ → Ba²⁺ + 2OH⁻ So, the concentration of OH⁻ ions can be determined by multiplying the concentration of Ba(OH)₂ by 2: \[Concentration \, of \, OH^- = 2 \times Concentration \, of \, Ba(OH)_2\] Substitute the value of Ba(OH)₂ concentration from step 1: \[Concentration \, of \, OH^- = 2 \times \frac{0.250 \, moles}{4.00 \, L}\]

Step 3: Calculate the concentration of H3O⁺ ions

We can use the relationship between the concentrations of OH⁻ and H3O⁺ and the ion product of water (Kw) to find the concentration of H3O⁺. The relationship is: Kw = [OH⁻] * [H3O⁺] The ion product of water, Kw, is 1 × 10⁻¹⁴ at 25℃. Rearrange the equation to find the concentration of H3O⁺: [H3O⁺] = Kw / [OH⁻] Now, substitute the value of OH⁻ concentration from step 2 and the value of Kw: \[ [H_3 O^+] = \frac{1 \times 10^{-14}}{2 \times \frac{0.250 \, moles}{4.00 \, L}} \] Calculate the values in each step, and you will get the molar concentrations of OH⁻ and H₃O⁺ ions.

Key Concepts

Chemical Solution Preparation

Understanding how to prepare a chemical solution is a fundamental skill in chemistry. It involves dissolving a substance, known as the solute, into a liquid, typically water, which is called the solvent. The process starts by determining the desired molar concentration of the solute in the solution, which is expressed in moles per liter (M or mol/L).

To create a solution with the correct concentration, you'll need to calculate the amount of solute required. This often involves using the formula:
\[ Molar \thinspace Concentration = \frac{moles \thinspace of \thinspace solute}{volume \thinspace of \thinspace solution \thinspace in \thinspace Liters} \]
In the given example, we dissolved 0.250 moles of \(\mathrm{Ba}(\mathrm{OH})_2\) to make a 4.00 L solution. The preparation would involve measuring 0.250 moles of barium hydroxide and gradually mixing it into water until the total volume reaches 4.00 L. The solution's concentration is a crucial starting point for further stoichiometric and pH calculations.

Stoichiometry of Dissolution

Stoichiometry of dissolution is the calculation of relative quantities of reactants and products in a chemical reaction that occurs when a substance dissolves in a solvent. It's critical to understand the ratios in the chemical formula to predict the outcome of the dissolution.

For instance, barium hydroxide, \(\mathrm{Ba}(\mathrm{OH})_2\), dissociates into one barium ion (Ba²⁺) and two hydroxide ions (OH⁻) for each formula unit that dissolves. The stoichiometry for this reaction is:
\[ \mathrm{Ba}(\mathrm{OH})_2 \rightarrow \mathrm{Ba}^{2+} + 2\mathrm{OH}^- \]
Knowing this, we calculate that for every mole of \(\mathrm{Ba}(\mathrm{OH})_2\) dissolved, there will be twice as many moles of hydroxide ions produced. This stoichiometric relation is crucial for accurately determining the concentration of ions in the solution, a vital component for many quantitative chemical analyses.

pH and pOH Calculations

The pH and pOH of a solution are measures of its acidity and basicity, respectively. They are inversely related and based on the molar concentrations of hydrogen ions (H₃O⁺) and hydroxide ions (OH⁻). The pH is defined as the negative logarithm of the hydrogen ion concentration, while pOH is the negative logarithm of the hydroxide ion concentration. The calculations are simplified by the following formulas:
\[ pH = -\log[H_3O^+] \]
\[ pOH = -\log[OH^-] \]
For neutral water at 25°C, the pH and pOH both equal 7, as the concentrations of H₃O⁺ and OH⁻ each equal 1 × 10⁻⁷ M. In our example, once we've found the molar concentration of OH⁻ ions, we can calculate the pOH and then the pH, as they must sum to 14. This sum is a reflection of the water's ion product, Kw:
\[ Kw = [OH^-] \times [H_3O^+] = 1 \times 10^{-14} \]
By knowing the concentration of one of the ions, we can determine the other and then use these to calculate the pH and pOH, vital for understanding a solution's chemical properties.