Question

Consider a weak organic base (nonelectrolyte) with molar mass \(281 \mathrm{~g} / \mathrm{mol}\). An aqueous solution of the base has a pH of 8.73 and an osmotic pressure of \(55 \mathrm{~mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C}\). What is \(K_{\mathrm{b}}\) for the weak base?

Short answer

The base ionization constant (\(K_b\)) for the weak organic base is approximately \(5.37 \times 10^{-6}\).

Step-by-step solution

Calculate the concentration of the weak base

To begin, we must find the concentration of the weak base in the aqueous solution. We will use the Van't Hoff formula to relate osmotic pressure, concentration, and temperature: $$ \Pi = \mathrm{cRT} $$ Where \(\Pi\) is the osmotic pressure, c is the concentration (in mol/L), R is the ideal gas constant (\(0.0821 \frac{\mathrm{L} \cdot \mathrm{atm}}{\mathrm{mol} \cdot \mathrm{K}}\)), and T is the temperature in Kelvin. First, let's convert the osmotic pressure and temperature to the appropriate units: $$ \Pi = 55 \frac{\mathrm{mm Hg}}{760 \frac{\mathrm{mm Hg}}{\mathrm{atm}}} = 0.0724 \mathrm{~atm} $$ $$ T = 25^{\circ} \mathrm{C} + 273.15 = 298.15 \mathrm{~K} $$ Now we can solve for the concentration of the weak base: $$ \mathrm{c} = \frac{\Pi}{\mathrm{RT}} = \frac{0.0724 \mathrm{~atm}}{0.0821 \frac{\mathrm{L} \cdot \mathrm{atm}}{\mathrm{mol} \cdot \mathrm{K}} \cdot 298.15 \mathrm{~K}} = 0.00300 \mathrm{~M} $$

Calculate the hydroxide ion concentration using the pH value

Next, we need to find the hydroxide ion concentration, \([\mathrm{OH}^-]\). We can do this using the given pH value and the relation between pH and pOH: $$ \mathrm{pOH} = 14 - \mathrm{pH} $$ $$ [\mathrm{OH}^-] = 10^{-\mathrm{pOH}} $$ Substituting the given pH value, we have: $$ \mathrm{pOH} = 14 - 8.73 = 5.27 $$ $$ [\mathrm{OH}^-] = 10^{-5.27} = 5.37 \times 10^{-6} \mathrm{~M} $$

Calculate the change in concentration and equilibrium concentrations for the weak base and hydroxide ions

Let x be the change in concentration of the weak base that has reacted and formed hydroxide ions. We can set up an ICE (Initial, Change, Equilibrium) table: $$ \begin{array}{c | c c} & \mathrm{Weak~base} & \mathrm{OH^-} \\ \hline \mathrm{Initial} & 0.00300 \mathrm{~M} & 0 \\ \mathrm{Change} & -x & +x \\ \mathrm{Equilibrium} & 0.00300 - x & x \\ \end{array} $$ Since \([\mathrm{OH}^-]\) at equilibrium is already known (\(5.37 \times 10^{-6} \mathrm{~M}\)), we can find the value of x and the equilibrium concentration of the weak base: $$ x = [\mathrm{OH}^-]_{\mathrm{eq}} = 5.37 \times 10^{-6} \mathrm{~M} $$ $$ [0.00300 - x] = 0.00300 - 5.37 \times 10^{-6} \approx 0.00300 \mathrm{~M} $$

Calculate \(K_b\) using the equilibrium concentrations

Finally, we can use the definition of \(K_b\) for the weak base: $$ K_b = \frac{[\mathrm{OH}^-][\mathrm{weak~base}]}{[\mathrm{unreacted~weak~base}]} $$ Substituting the equilibrium concentrations from the ICE table, we have: $$ K_b = \frac{(5.37 \times 10^{-6})(0.00300)}{0.00300 - 5.37 \times 10^{-6}} \approx \frac{(5.37 \times 10^{-6})(0.00300)}{0.00300} = 5.37 \times 10^{-6} $$ Thus, the base ionization constant (\(K_b\)) for the weak organic base is approximately \(5.37 \times 10^{-6}\).