Problem 44 - Chemistry

Question

An aqueous solution of a strong base has a pH of 9.78 at \(25^{\circ} \mathrm{C}\). Calculate the concentration of the base if the base is (a) \(\mathrm{LiOH}\) and (b) \(\mathrm{Ba}(\mathrm{OH})_{2}\).

Step-by-Step Solution

Answer

The concentration is \(6.03 \times 10^{-5}\) M for LiOH and \(3.015 \times 10^{-5}\) M for \(\mathrm{Ba(OH)}_2\).

Step 1: Understand the pH

The pH of the solution is given as 9.78, which means the solution is basic. The relationship between pH and pOH is given by the formula: \[ \text{pH} + \text{pOH} = 14 \]We need to find the pOH to calculate the hydroxide ion concentration.

Step 2: Calculate the pOH

Using the formula \(\text{pH} + \text{pOH} = 14\), substitute the known value of the pH:\[ 9.78 + \text{pOH} = 14 \]\[ \text{pOH} = 14 - 9.78 = 4.22 \]

Step 3: Find [OH⁻]

The concentration of hydroxide ions, \([\text{OH}^-]\), is related to pOH by the formula:\[ [\text{OH}^-] = 10^{-\text{pOH}} \]Substituting the value of pOH:\[ [\text{OH}^-] = 10^{-4.22} \approx 6.03 \times 10^{-5} \text{ M} \]

Step 4: Calculate concentration for LiOH

Since \(\mathrm{LiOH}\) is a strong base that dissociates fully, the concentration of \([\text{LiOH}]\) is equal to the \([\text{OH}^-]\) concentration:\[ [\text{LiOH}] = 6.03 \times 10^{-5} \text{ M} \]

Step 5: Calculate concentration for Ba(OH)₂

For \(\mathrm{Ba(OH)}_2\), which dissociates into one \(\text{Ba}^{2+}\) ion and two \(\text{OH}^-\) ions, the relationship is:\[ [\text{OH}^-] = 2[\mathrm{Ba(OH)}_2] \]So, the concentration of \(\mathrm{Ba(OH)}_2\) is:\[ [\mathrm{Ba(OH)}_2] = \frac{6.03 \times 10^{-5}}{2} = 3.015 \times 10^{-5} \text{ M} \]

Key Concepts

Hydroxide Ion Concentration

In a basic solution, the concentration of hydroxide ions \([\text{OH}^-]\) plays a crucial role in determining the solution's properties. The relationship between pOH and the hydroxide ion concentration is given by the formula: \[ [\text{OH}^-] = 10^{-\text{pOH}} \]This formula allows us to calculate the amount of hydroxide ions present, which is an essential step in understanding the overall behavior of the solution.
For example, if the pOH is 4.22, substituting this into the equation gives: \[ [\text{OH}^-] \approx 6.03 \times 10^{-5} \text{ M} \] This concentration reflects the strength of the base in the solution.
Remember, the hydroxide ion concentration is directly tied to the solution's basic nature.

Strong Base Properties

Strong bases, such as \(\text{LiOH}\) and \(\text{Ba(OH)}_2\), completely dissociate in solution. This means every molecule separates into its constituent ions.

• For \(\text{LiOH}\), it dissociates into \(\text{Li}^+\) and \(\text{OH}^-\)
• For \(\text{Ba(OH)}_2\), it forms one \(\text{Ba}^{2+}\) ion and two \(\text{OH}^-\) ions

This complete dissociation is characteristic of strong bases and makes calculating concentrations straightforward once the \(\text{OH}^-\) concentration is known.
In basic solutions, strong bases raise the hydroxide ion concentration significantly, thus increasing the pH of the solution.

Acid-Base Equilibrium

Understanding the pH and pOH relationship is crucial in acid-base chemistry. The formula \( \text{pH} + \text{pOH} = 14 \) is used to calculate one if you know the other. This equilibrium helps us understand:

• How acidic or basic a solution is
• The concentration of hydrogen ions \(\text{H}^+\) and hydroxide ions \(\text{OH}^-\)

In the given problem, knowing the pH allowed us to find the pOH, which is then converted to the hydroxide ion concentration.
This relationship remains constant at \(25^\circ \text{C}\), providing a reliable tool for calculations.

Ba(OH)₂ Dissociation

When \([\text{Ba(OH)}_2]\) dissolves in water, it dissociates into one \(\text{Ba}^{2+}\) ion and two \(\text{OH}^-\) ions. This results in the equation: \[ [\text{OH}^-] = 2[\text{Ba(OH)}_2] \]
This means the hydroxide ion concentration is double the concentration of \(\text{Ba(OH)}_2\). With a \[ [\text{OH}^-] \approx 6.03 \times 10^{-5} \text{ M} \], we find: \[ [\text{Ba(OH)}_2] = \frac{6.03 \times 10^{-5}}{2} \approx 3.015 \times 10^{-5} \text{ M} \]
Understanding this dissociation is key to solving problems related to strong bases like \(\text{Ba(OH)}_2\). This knowledge ensures precise calculations.