Question

The pH of a \(2.642 \mathrm{M}\) solution of a weak acid, \(\mathrm{HB}\), is 5.32. What is \(K_{\mathrm{a}}\) for the weak acid?

Short answer

Question: Calculate the acid dissociation constant (\(K_a\)) for a weak acid HB, given a concentration of 2.642 M and a pH of 5.32. Answer: The acid dissociation constant, \(K_a\), for the weak acid HB is approximately \(6.892 \times 10^{-12}\).

Step-by-step solution

Find the concentration of hydrogen ions [H+] using the pH

To find the [H+], we can use the following formula:$$\mathrm{pH} = -\log\,[\mathrm{H^+}]$$The given pH is 5.32, so we can solve for [H+]:$$[\mathrm{H^+}] = 10^{-\mathrm{pH}} = 10^{-5.32} = 4.786 \times 10^{-6}\, \mathrm{M}$$

Write the balanced chemical equation for the dissociation of the weak acid, HB

The dissociation of the weak acid, HB, can be represented as follows:$$\mathrm{HB} \rightleftharpoons \mathrm{H^+} + \mathrm{B^-}$$

Find the initial, change, and equilibrium concentrations using the reaction quotient (Qc) and the given concentration of the weak acid

Let's set up a reaction table showing the initial, change, and equilibrium concentrations for each species: - [HB]₀ = 2.642 M, [H+]₀ = 0, and [B⁻]₀ = 0 - At equilibrium: [HB] = 2.642 - x, [H+] = x, [B⁻] = x (where x is the change in[H⁺] due to the dissociation of HB) - Since we already know the equilibrium [H+], we can find x:$$x = [\mathrm{H^+}]_{eq} = 4.786 \times 10^{-6}\, \mathrm{M}$$Now, we can find the equilibrium concentrations of HB and B⁻: $$[\mathrm{HB}]_{eq} = [\mathrm{HB}]_{0} - x = 2.642 - 4.786 \times 10^{-6}$$ $$[\mathrm{B^-}]_{eq} = [\mathrm{B^-}]_{0} + x = 0 + 4.786 \times 10^{-6}$$

Substitute the equilibrium concentrations into the \(K_a\) expression and solve for \(K_a\)

The expression for the \(K_a\) is:$$K_a = \frac{[\mathrm{H^+}][\mathrm{B^-}]}{[\mathrm{HB}]}$$Substitute the equilibrium concentrations found in step 3 into the \(K_a\) expression:$$K_a = \frac{(4.786 \times 10^{-6})(4.786 \times 10^{-6})}{(2.642 - 4.786 \times 10^{-6})}$$Now, we can solve for \(K_a\):$$K_a = 6.892 \times 10^{-12}$$ The acid dissociation constant, \(K_a\), for the weak acid HB is approximately \(6.892 \times 10^{-12}\).