Problem 106 - Chemistry The Central Science

Question

Calculate the \(\mathrm{pH}\) of a solution made by adding \(2.50 \mathrm{~g}\) of lithium oxide \(\left(\mathrm{Li}_{2} \mathrm{O}\right)\) to enough water to make \(1.500 \mathrm{~L}\) of solution.

Step-by-Step Solution

Answer

The pH of the solution, made by adding $2.50\:g$ of lithium oxide $\mathrm{Li}_{2} \mathrm{O}$ to enough water to make $1.500\:L$ of solution, is approximately 13.03.

Step 1: Determine the moles of Li₂O

Before we can start, we need to convert the given mass of lithium oxide to moles. The molar mass of Li₂O can be calculated as follows: Molar mass of Li₂O = \(2\times\) molar mass of Li + molar mass of O = \(2\times 6.94\) g/mol + \(16.00\) g/mol = \(30.88\) g/mol Now, we can calculate the amount in moles: Moles of Li₂O = mass / molar mass = \(2.50\) g / \(30.88\) g/mol ≈ \(0.081\) mol

Step 2: Determine the concentration of LiOH

Lithium oxide (Li₂O) reacts with water to form lithium hydroxide (LiOH) in the following reaction: \[ \mathrm{Li}_{2} \mathrm{O} + \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{LiOH} \] From the stoichiometry of the equation, one mole of Li₂O reacts with one mole of water to produce two moles of LiOH. So, the moles of LiOH produced will be: Moles of LiOH = \(2 \times 0.081\) mol = \(0.162\) mol Since the volume of the solution is 1.500 L, we can now calculate the concentration of LiOH: \[ \mathrm{[LiOH]} = \frac{0.162 \, \text{mol}}{1.5 \, \text{L}} = 0.108 \, \text{M} \]

Step 3: Determine the concentration of OH⁻ ions

Region For every mole of LiOH, one mole of OH⁻ ions is produced, as LiOH is a strong base and dissociates completely in water: \[\mathrm{LiOH} \longrightarrow \mathrm{Li}^+ + \mathrm{OH}^-\] So, the concentration of OH⁻ ions is equal to the concentration of LiOH: \[[OH^-] = 0.108 \, \text{M}\]

Step 4: Calculate the pOH and pH of the solution

Now that we know the concentration of OH⁻ ions, we can calculate the pOH: \[ \mathrm{pOH} = -\log_{10}[\mathrm{OH}^-] = -\log_{10}(0.108) \approx 0.97 \] We then use the following relationship between pH and pOH to find the pH of the solution: \[ \mathrm{pH} + \mathrm{pOH} = 14\] Therefore, \[ \mathrm{pH} = 14 - \mathrm{pOH} = 14 - 0.97 \approx 13.03 \]

Step 5: Final Answer

The pH of the solution is approximately 13.03.

Key Concepts

Molar Mass Calculation

Calculating the molar mass is an essential part of understanding how many moles of a substance you have. The molar mass is the sum of the atomic masses of all atoms in a molecule. For lithium oxide (\(\text{Li}_2\text{O}\)), it consists of 2 lithium atoms and 1 oxygen atom.
Each lithium atom has a molar mass of 6.94 g/mol, and oxygen has a molar mass of 16.00 g/mol. Hence, the molar mass of lithium oxide can be calculated using the formula:

• Molar mass = \( 2 \times 6.94 \, \text{g/mol} + 16.00 \, \text{g/mol} = 30.88 \, \text{g/mol} \)

This calculation allows you to convert a given mass of compound into moles, which is crucial for stoichiometric calculations and further determining other properties of reactions.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It involves the use of coefficients from a balanced chemical equation to infer the relative amounts of reactants and products. In this exercise, lithium oxide (\(\text{Li}_2\text{O}\)) reacts with water to form lithium hydroxide (LiOH) according to the balanced equation:

• \( \text{Li}_2\text{O} + \text{H}_2\text{O} \longrightarrow 2 \text{LiOH} \)

From the stoichiometry, one mole of \(\text{Li}_2\text{O}\) produces two moles of \(\text{LiOH}\). This relationship allows us to determine how much of each substance is involved in the reaction, facilitating the conversion from moles of reactants to moles of products.

Acid-Base Chemistry

Understanding acid-base chemistry is crucial when dealing with reactions involving acids and bases. Lithium hydroxide (\(\text{LiOH}\)) is a strong base that dissociates completely in water:

• \( \text{LiOH} \longrightarrow \text{Li}^+ + \text{OH}^- \)

This complete dissociation is characteristic of strong bases. The concentration of hydroxide ions (\(\text{OH}^-\)) in the solution is directly related to the concentration of the \(\text{LiOH}\) calculated earlier. Understanding this chemistry helps predict the strength and nature (acidic or basic) of the solution.

Concentration Determination

Concentration determination is the process of finding the amount of solute present in a solution. Typically expressed in molarity (M), it is defined as moles of solute per liter of solution. This is calculated using the formula:

• \( \text{Concentration (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \)

In the exercise, after forming \(\text{LiOH}\) from \(\text{Li}_2\text{O}\) and water, the concentration of \(\text{LiOH}\) (and hence \(\text{OH}^-\)) can be calculated for a solution with 1.500 L of volume. This value is used to further calculate the pH of the solution. Accurate concentration determination is key for predicting the behavior of the solution.