91Ó°ÊÓ

The formula relating osmotic pressure, molarity, and temperature is: \(\pi = i M R T\) Where \(\pi\) is the osmotic pressure in atm, \(i\) is the van't Hoff factor (given as 1), \(M\) is the molarity in mol/L, \(R\) is the gas constant (0.0821 Latm/(mol.K)), and \(T\) is the temperature in Kelvin. We are given, \(\pi = 0.878\) atm and \(T = 25^{\circ} \mathrm{C} = 298 \mathrm{K}\). First, convert the temperature to Kelvin: \(T_K = T_{\mathrm{C}} + 273 = 25 + 273 = 298 \mathrm{K}\). Now, solve for the molarity, \(M\): \(M = \frac{\pi}{iRT} = \frac{0.878}{1 \times 0.0821 \times 298} = 0.0361 \mathrm{M}\)

Using the definition of pH, \([\mathrm{H^+}] = 10^{-\mathrm{pH}}\). We are given the pH as \(6.76\). Thus, \([\mathrm{H^+}] = 10^{-6.76} = 1.74 \times 10^{-7} \mathrm{M}\)

From the \(K_w\) expression, \( K_w = [\mathrm{H^+}] [\mathrm{OH^-}] = 1.0 \times 10^{-14}\). We have the value of \([\mathrm{H^+}]\). We can now find \([\mathrm{OH^-}]\) as follows: \([\mathrm{OH^-}] = \frac{K_w}{[\mathrm{H^+}]} = \frac{1.0 \times 10^{-14}}{1.74 \times 10^{-7}} = 5.75 \times 10^{-8} \mathrm{M}\)

We have the molarity of the weak acid, \(M\), and the concentration of \(\mathrm{OH^-}\) ions. We can assume that all \(\mathrm{OH^-}\) ions have come from the dissociation of the conjugate base. Therefore, the initial concentration of the conjugate base is \(0.0361 \mathrm{M}\). Now, we can use the expression for \(K_b\) in terms of conjugate base (\([\mathrm{B^-}]\)) and hydroxide ions (\([\mathrm{OH^-}]\)) concentrations: \(K_b = \frac{[\mathrm{B^-}] [\mathrm{OH^-}]}{[\mathrm{HB}]}\) In this case, we assume that all the weak acid has dissociated into its conjugate base. Therefore, \([\mathrm{B^-}] = 0.0361 \mathrm{M}\) and \([\mathrm{OH^-}] = 5.75 \times 10^{-8} \mathrm{M}\). Now, we can calculate \(K_b\) as follows: \(K_b = \frac{(0.0361)(5.75 \times 10^{-8})}{0.0361 - 5.75 \times 10^{-8}} \approx 5.75 \times 10^{-8}\) So, the value of \(K_b\) for the conjugate base is approximately \(5.75 \times 10^{-8}\).