Question

Calculate the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of a \(0.61-M\) aqueous solution of a weak base \(\mathrm{B}\) with a \(K_{\mathrm{b}}\) of \(1.5 \times 10^{-4}\).

Short answer

The pH is approximately 11.98.

Step-by-step solution

Write the equilibrium expression

For a weak base, \( \mathrm{B} \), in an aqueous solution, it reacts with water to form its conjugate acid and hydroxide ions: \( \mathrm{B} + \mathrm{H_2O} \rightleftharpoons \mathrm{BH^+} + \mathrm{OH^-} \). The equilibrium expression for this reaction is given by \( K_{\mathrm{b}} = \frac{[\mathrm{BH^+}][\mathrm{OH^-}]}{[\mathrm{B}]} \).

Define initial concentrations and changes

Initially, the concentration of \( \mathrm{B} \) is \( 0.61 \mathrm{M} \), and the concentrations of \( \mathrm{BH^+} \) and \( \mathrm{OH^-} \) are zero. As the reaction reaches equilibrium, \( x \) moles of \( \mathrm{B} \) dissociate, forming \( x \) moles each of \( \mathrm{BH^+} \) and \( \mathrm{OH^-} \). This gives us the equilibrium concentrations: \( [\mathrm{B}] = 0.61 - x \), \( [\mathrm{BH^+}] = x \), and \( [\mathrm{OH^-}] = x \).

Write the equilibrium equation using concentrations

Substitute the equilibrium concentrations into the equilibrium expression: \( K_{\mathrm{b}} = \frac{x^2}{0.61 - x} = 1.5 \times 10^{-4} \).

Make the approximation for simplification

Assume \( x \) is small compared to 0.61, so \( 0.61 - x \approx 0.61 \). This simplifies the equation to \( \frac{x^2}{0.61} = 1.5 \times 10^{-4} \).

Solve for x

Rearrange to find \( x \): \( x^2 = 1.5 \times 10^{-4} \times 0.61 \). Calculate \( x \approx \sqrt{9.15 \times 10^{-5}} \approx 9.56 \times 10^{-3} \). Thus, \( [\mathrm{OH^-}] = 9.56 \times 10^{-3} \mathrm{M} \).

Calculate the pOH

The \( \mathrm{pOH} \) is calculated as \( \mathrm{pOH} = -\log_{10}[\mathrm{OH^-}] \). For \( [\mathrm{OH^-}] = 9.56 \times 10^{-3} \), \( \mathrm{pOH} \approx 2.02 \).

Calculate the pH

Use the relation \( \mathrm{pH} + \mathrm{pOH} = 14 \) to find the \( \mathrm{pH} \). Thus, \( \mathrm{pH} = 14 - 2.02 = 11.98 \).

Key concepts

Weak Base Equilibrium

When dealing with weak bases like the hypothetical base \( \mathrm{B} \) in our problem, it's essential to understand the concept of weak base equilibrium. A weak base does not completely dissociate in water. This means that in an aqueous solution, not all of the base molecules convert into hydroxide ions (\( \mathrm{OH^-} \)) and their conjugate acids (\( \mathrm{BH^+} \)).
The equilibrium process can be represented by this equation:

• \( \mathrm{B} + \mathrm{H_2O} \rightleftharpoons \mathrm{BH^+} + \mathrm{OH^-} \)

At equilibrium, the forward and reverse reactions occur at the same rate, but not all base molecules have reacted. The incomplete dissociation is a characteristic trait of weak bases, differentiating them from strong bases that fully dissociate in solution. Knowing this helps us understand why we need to establish an equilibrium expression to solve for concentrations.

Equilibrium Expression

The equilibrium expression provides a mathematical representation of the weak base equilibrium in solution. It's crucial for calculating the concentrations of reactants and products at equilibrium. For the base \( \mathrm{B} \), the equilibrium expression is:

• \( K_{\mathrm{b}} = \frac{[\mathrm{BH^+}][\mathrm{OH^-}]}{[\mathrm{B}]} \)

Where \( K_{\mathrm{b}} \) is the base dissociation constant, which reflects the extent to which the base dissociates in water. A small \( K_{\mathrm{b}} \) value indicates a weak base, as it means that only a small amount of base converts into its conjugate acid and hydroxide ions. This expression allows us to solve for unknown concentrations by substituting known values into the equation, a crucial step in determining properties like pH.

pOH

To calculate the \( \mathrm{pOH} \) of a solution, we first need to determine the concentration of hydroxide ions \([\mathrm{OH^-}]\). The \( \mathrm{pOH} \) is a measure of the hydroxide ion concentration in a solution, and it can be found using the formula:

• \( \mathrm{pOH} = -\log_{10}[\mathrm{OH^-}] \)

Understanding \( \mathrm{pOH} \) is crucial because it complements the \( \mathrm{pH} \). In aqueous solutions, the \( \mathrm{pH} \) and \( \mathrm{pOH} \) are related by the equation:

• \( \mathrm{pH} + \mathrm{pOH} = 14 \)

Knowing the \( \mathrm{pOH} \) allows us to easily determine the \( \mathrm{pH} \), as seen in the step-by-step solution by calculating the \( \mathrm{pOH} \) first and then finding \( \mathrm{pH} \).

Hydroxide Ion Concentration

Hydroxide ion concentration, \([\mathrm{OH^-}]\), is a key factor in determining the basicity of a solution. In the context of weak bases, this concentration results from the partial dissociation of the base \( \mathrm{B} \) in water. From the equilibrium concentrations calculated earlier, we found that:

• \([\mathrm{OH^-}] = x \)

Where \( x \) is the concentration of hydroxide ions at equilibrium, derived from solving the equilibrium expression. Calculating \([\mathrm{OH^-}]\) helps us find the \( \mathrm{pOH} \), which in turn is used to compute the \( \mathrm{pH} \). Understanding the hydroxide ion concentration is essential when analyzing solutions of weak bases, as it directly influences the \( \mathrm{pH} \) and reflects the solution's basic nature.