Consider a 0.10 M solution of a weak polyprotic acid (H_2

Question

Consider a 0.10 M solution of a weak polyprotic acid (H₂A) with the possible values of K_a1 and K_a2 given here.

a. K_a1 = 1.0 × 10^–4; K_a2 = 5.0 × 10^–5

Calculate the contributions to [H₃O⁺] from each ionization step. At what point can the contribution of the second step be neglected?

Accepted Answer

Step 1: Understand the ionization process of the polyprotic acid $$H_{2}A. $$The first ionization step is: $$H_{2}A$$ ⇌ $$H^{+}$$ + $$HA^{-}$$ with K_{a1} = 1.0 × $$10^{-4}. $$The second ionization step is: $$HA^{-}$$ ⇌ $$H^{+}$$ + $$A^{2-}$$ with K_{a2} = 5.0 × $$10^{-5}. $$Each step contributes to the [$$H_{3}$O^{+}$$].. Step 2: Set up the equilibrium expression for the first ionization step: K_{a1} = [$$H^{+}$$][$$HA^{-}$$]/[$$H_{2}A$$]. Assume initial concentration of $$H_{2}A is 0.10 M $$and initial [$$H^{+}$$] and [$$HA^{-}$$] are 0. Use an ICE table to express changes in concentration and solve for [$$H^{+}$$] from the first ionization.. Step 3: Calculate the [$$H^{+}$$] from the first ionization step using the approximation that x (the change in concentration) is small compared to the initial concentration of $$H_{2}A. $$This simplifies the expression to K_{a1} ≈ $$x^{2}/0.10.. $$Step 4: For the second ionization step, set up the equilibrium expression: K_{a2} = [$$H^{+}$$][$$A^{2-}$$]/[$$HA^{-}$$]. Use the [$$H^{+}$$] from the first ionization as the initial concentration for the second ionization. Assume [$$HA^{-}$$] is approximately equal to the [$$H^{+}$$] from the first ionization.. Step 5: Determine when the contribution of the second ionization step can be neglected. This occurs when the [$$H^{+}$$] from the second ionization is significantly smaller than that from the first ionization. Compare the calculated [$$H^{+}$$] from both steps to decide if the second step's contribution is negligible.

Consider a 0.10 M solution of a weak polyprotic acid (H₂A) with the possible values of K_a1 and K_a2 given here.
a. K_a1 = 1.0 × 10^–4; K_a2 = 5.0 × 10^–5
Calculate the contributions to [H₃O⁺] from each ionization step. At what point can the contribution of the second step be neglected?

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Key Concepts

Polyprotic Acids

Dissociation Constants (Ka)

Hydronium Ion Concentration ([H3O+])

Consider a 0.10 M solution of a weak polyprotic acid (H2A) with the possible values of Ka1 and Ka2 given here.a. Ka1 = 1.0 × 10–4; Ka2 = 5.0 × 10–5Calculate the contributions to [H3O+] from each ionization step. At what point can the contribution of the second step be neglected?