Equal quantities of 0.010\(M\) solutions of an acid \(\mathrm{HA}\) and a base \(\mathrm{B}\) are mixed. The \(\mathrm{pH}\) of the resulting solution is \(9.2 .\) (a) Write the chemical equation and equilibrium-constant expression for the reaction between \(\mathrm{HA}\) and \(\mathrm{B}\) . (b) If \(K_{a}\) for \(\mathrm{HA}\) is \(8.0 \times 10^{-5}\) , what is the value of the equilibrium constant for the reaction between \(\mathrm{HA}\) and \(\mathrm{B} ?\) (c) What is the value of \(K_{b}\) for \(\mathrm{B}\) ?

First, let's write the reaction between the acid HA and the base B (assuming B is a generic hydroxide ion donor): HA + B → A⁻ + HB Next, write the equilibrium-constant expression for this reaction: K = [A⁻][HB] / [HA][B]

We're given the pH of the resulting solution which is 9.2. To find the pOH of the solution, use the relationship between pH and pOH: pH + pOH = 14 9.2 + pOH = 14 pOH = 4.8 Now, we can find the OH⁻ concentration in the solution from the pOH value: \[pOH = - log [OH^-]\] \[4.8 = - log [OH^-]\] \[[OH^-] = 10^{-4.8}\] Since we're given the Ka value for HA as \(8.0 × 10^{-5}\), we can write the equilibrium-constant expression for HA as: \[K_a = [\mathrm{H^+}][\mathrm{A^-}]/[\mathrm{HA}]\] We're also given that equal quantities of 0.010 M solutions of HA and B are mixed. When HA reacts with B, it donates a proton to B. So the amount of H⁺ ions from HA is: \[[\mathrm{H^+}] = 0.010 - [OH^-]\] Now we can plug in all the values in the Ka expression: \[8.0 × 10^{-5} = (0.010 - [OH^-])[\mathrm{A^-}]/[\mathrm{HA}]\] We know that the concentration of A⁻ is equal to the concentration of HB, and the same for HA and B, we can find the value of the equilibrium constant K as follows: \[K = \frac{[\mathrm{A^-}][\mathrm{HB}]}{[\mathrm{HA}][\mathrm{B}]} = \frac{[\mathrm{A^-^2}]}{[\mathrm{HA}]^2} = \frac{(0.010 - [OH^-])^2 }{(8.0 × 10^{-5})}\] Now substitute the value of [OH⁻] in the above equation and calculate the value of K: \[K = \frac{(0.010 - 10^{-4.8})^2 }{(8.0 × 10^{-5})}\]

The relation between Ka, Kb, and Kw for an acid-base conjugate pair is given by: \[K_a × K_b = K_w\] Where Kw is the ion-product constant for water: \(K_w = 1.0 × 10^{-14}\) Using the given Ka value and the calculated K value, we can find the Kb value for the base B: \[K_b = \frac{K_w}{K_a} = \frac{1.0 × 10^{-14}}{8.0 × 10^{-5}}\] Calculate the final value for Kb by dividing the constants.