Question

One handbook lists a value of 9.5 for \(\mathrm{p} \mathrm{K}_{\mathrm{b}}\) of quinoline, \(\mathrm{C}_{9} \mathrm{H}_{7} \mathrm{N},\) a weak base used as a preservative for anatomical specimens and to make dyes. Another handbook lists the solubility of quinoline in water at \(25^{\circ} \mathrm{C}\) as \(0.6 \mathrm{g} / 100 \mathrm{mL} .\) Use this information to calculate the \(\mathrm{pH}\) of a saturated solution of quinoline in water.

Short answer

To find the pH, apply the \(pOH\) to \(pH\) conversion after finding the concentration of the hydroxide ion using the provided \(K_b\) and solubility data for quinoline. The exact pH value will depend on the calculated values in each step.

Step-by-step solution

Calculate \(K_b\) from \(\mathrm{p}K_b\)

The relationship between \(K_b\) and \(\mathrm{p}K_b\) is given by \(\mathrm{p}K_b = -\log(K_b)\). Therefore, \(K_b\) can be found as \(K_b = 10^{-pK_b}\). Given that \(\mathrm{p}K_b = 9.5\), we substitute this value into the formula to find \(K_b = 10^{-9.5}\).

Find the concentration of quinoline

The solubility of quinoline at \(25°C\) is given as 0.6 g/100 mL. To convert this into molarity (M), we first find the molar mass of quinoline, \(C_9H_7N\), which is approximately 129.16 g/mol. The molarity, \(C\), can then be calculated using the formula \(C = \frac{mass}{volume \times molar \, mass}\) where mass is 0.6 g, volume is 100 mL (or 0.1 L), and molar mass is 129.16 g/mol, giving \(C = \frac{0.6 \, g}{0.1 \, L \times 129.16 \, g/mol}\).

Find the base ionization constant (\(K_b\))

Quinoline is a weak base, and its ionization in water can be represented as \(C_9H_7N + H_2O \leftrightharpoons C_9H_7NH^+ + OH^-\). Because the solution is saturated, it's safe to assume that all the quinoline ionizes, and the ion concentration is then equal to the initial concentration of the quinoline. Thus, we can define the \(K_b\) expression as \(K_b = \frac{[C_9H_7NH^+][OH^-]}{[C_9H_7N]}\) or simply \(K_b = [OH^-]\), as the concentrations of \(C_9H_7NH^+\) and \(OH^-\) will be equal and the denominator is 1. The concentration of \(OH^-\) can then be found by taking the square root of \(K_b\), as \(K_b = [OH^-]^2\).

Calculate the \(pOH\) and then the \(pH\)

Once the hydroxide ion concentration, [OH^-], has been determined, the \(pOH\) can be found by using the equation \(pOH = -\log([OH^-])\). The \(pH\) can then be found by using the relationship \(pH = 14 - pOH\).

Key concepts

Weak Base Equilibrium

Understanding the behavior of weak bases in solution is key to calculating their pH. Unlike strong bases that dissociate completely in water, weak bases only partially ionize. Quinoline, for example, is a weak base that establishes an equilibrium when dissolved in water, following the equation:
\[ C_{9}H_{7}N + H_{2}O \leftrightarpoons C_{9}H_{7}NH^+ + OH^- \]
In this equation, quinoline reacts with water to form its conjugate acid, quinolinium ion (\(C_{9}H_{7}NH^+\)) and hydroxide ions (\(OH^-\)). The extent to which this equilibrium is established is represented by the base ionization constant (Kb), which indicates the strength of the base. A low Kb value signifies a weak base, resulting in fewer hydroxide ions and a higher pH compared to strong bases.

pKb to Kb Conversion

For weak bases, it is often more convenient to deal with the pKb value rather than the direct Kb. pKb is the negative logarithm of the base ionization constant (Kb), represented by the equation:
\[ pKb = -\log(Kb) \]
To convert pKb to Kb, which is necessary for calculating the concentration of hydroxide ions, use the inverse of the logarithmic relationship:
\[ Kb = 10^{-pKb} \]
In the case of quinoline, with a given pKb of 9.5, the Kb is found by calculating \(10^{-9.5}\), which provides the value needed to figure out how much the base will ionize in solution.

Molarity Calculation

Molarity, symbolized by M, is a measure of the concentration of a substance in solution, defined as moles of solute per liter of solution. The steps to calculate the molarity of quinoline involve converting grams to moles and liters to solution volume:
\[ C = \frac{mass}{volume \times molar \, mass} \]
For our example, the mass of quinoline is 0.6 grams, and the volume of the solution is 0.1 liters (equivalent to 100 mL). Using quinoline's molar mass of approximately 129.16 g/mol, the molarity can be determined. This calculation of concentration is critical because it directly affects the pH of the solution.

pH and pOH Relationship

The pH and pOH of a solution are intricately linked through the pH scale, a measure of acidity or basicity of an aqueous solution. The pH is calculated as the negative logarithm of the hydrogen ion concentration, while pOH is similarly defined for hydroxide ions:
\[ pOH = -\log([OH^-]) \]
The sum of pH and pOH is always equal to 14, a constant value at 25 degrees Celsius due to the dissociation of water:
\[ pH + pOH = 14 \]
After determining pOH from the hydroxide ion concentration, this relationship allows us to find the pH. For weak bases, once the pOH is calculated, subtracting it from 14 gives the pH, which indicates how basic the solution is.