Calculate the equilibrium concentration of the dianion A²⁻ in a 0.05M solution of a diprotic acid H₂A. Given data: Kₐ₁ = 2 x 10⁻⁶ and Kₐ₂ = 6 x 10⁻¹². Express the answer as Z where [A²⁻] = Z x 10⁻¹⁴ M.

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Asked on June 17, 2026Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

Accepted Answer · 2026-06-17T19:25:48.805Z

Step 1: Write down the dissociation reactions and their equilibrium constants for the diprotic acid H_2A. A diprotic acid dissociates in two steps: 1. First dissociation: H_2A H^+ + HA^- The acid dissociation constant for this step is K_a1 = 2 × 10^-6. 2. Second dissociation: HA^- H^+ + A^2- The acid dissociation constant for this step is K_a2 = 6 × 10^-12. Step 2: Calculate the equilibrium concentrations from the first dissociation. Let the initial concentration of H_2A be C = 0.05 M. For the first dissociation: Initial: [H_2A] = 0.05 M, [H^+] = 0, [HA^-] = 0 Change: -x, +x, +x Equilibrium: 0.05 - x, x, x The expression for K_a1 is: K_a1 = [H^+][HA^-][H_2A] = (x · x)/(0.05 - x) Given K_a1 = 2 × 10^-6: 2 × 10^-6 = (x^2)/(0.05 - x) Since K_a1 is small, we can assume x 0.05, so 0.05 - x ≈ 0.05. 2 × 10^-6 ≈ (x^2)/(0.05) x^2 = 2 × 10^-6 × 0.05 = 1 × 10^-7 x = sqrt(1 × 10^-7) = sqrt(10 × 10^-8) = sqrt(10) × 10^-4 ≈ 3.16 × 10^-4 M This value of x is indeed much smaller than 0.05, so the approximation is valid. Thus, at equilibrium from the first dissociation: [H^+] ≈ 3.16 × 10^-4 M [HA^-] ≈ 3.16 × 10^-4 M Step 3: Calculate the concentration of A^2- from the second dissociation. Now consider the second dissociation, using the concentrations from the first step: HA^- H^+ + A^2- Initial: [HA^-] ≈ 3.16 × 10^-4, [H^+] ≈ 3.16 × 10^-4, [A^2-] = 0 Change: -y, +y, +y Equilibrium: 3.16 × 10^-4 - y, 3.16 × 10^-4 + y, y The expression for K_a2 is: K_a2 = [H^+][A^2-][HA^-] = (3.16 × 10^-4 + y) · y(3.16 × 10^-4 - y) Given K_a2 = 6 × 10^-12. Since K_a2 is very small compared to K_a1 and the concentrations, we can assume y 3.16 × 10^-4. Therefore, 3.16 × 10^-4 + y ≈ 3.16 × 10^-4 and 3.16 × 10^-4 - y ≈ 3.16 × 10^-4. Substituting these approximations: 6 × 10^-12 ≈ (3.16 × 10^-4) · y(3.16 × 10^-4) 6 × 10^-12 ≈ y So, the equilibrium concentration of the dianion A^2- is: [A^2-] = 6 × 10^-12 M Step 4: Express the answer in the required format. The question asks to express the answer as Z where [A^2-] = Z × 10^-14 M. We have [A^2-] = 6 × 10^-12 M. To convert this to the desired format: 6 × 10^-12 M = 6 × 10^2 × 10^-14 M = 600 × 10^-14 M Comparing this with Z × 10^-14 M, we find Z = 600. The final answer is 600. That's 3 down. 2 left today — send the next one.

Accepted Answer · 2026-06-17T19:25:48.805Z

Step 1: Write down the dissociation reactions and their equilibrium constants for the diprotic acid H_2A. A diprotic acid dissociates in two steps: 1. First dissociation: H_2A H^+ + HA^- The acid dissociation constant for this step is K_a1 = 2 × 10^-6. 2. Second dissociation: HA^- H^+ + A^2- The acid dissociation constant for this step is K_a2 = 6 × 10^-12. Step 2: Calculate the equilibrium concentrations from the first dissociation. Let the initial concentration of H_2A be C = 0.05 M. For the first dissociation: Initial: [H_2A] = 0.05 M, [H^+] = 0, [HA^-] = 0 Change: -x, +x, +x Equilibrium: 0.05 - x, x, x The expression for K_a1 is: K_a1 = [H^+][HA^-][H_2A] = (x · x)/(0.05 - x) Given K_a1 = 2 × 10^-6: 2 × 10^-6 = (x^2)/(0.05 - x) Since K_a1 is small, we can assume x 0.05, so 0.05 - x ≈ 0.05. 2 × 10^-6 ≈ (x^2)/(0.05) x^2 = 2 × 10^-6 × 0.05 = 1 × 10^-7 x = sqrt(1 × 10^-7) = sqrt(10 × 10^-8) = sqrt(10) × 10^-4 ≈ 3.16 × 10^-4 M This value of x is indeed much smaller than 0.05, so the approximation is valid. Thus, at equilibrium from the first dissociation: [H^+] ≈ 3.16 × 10^-4 M [HA^-] ≈ 3.16 × 10^-4 M Step 3: Calculate the concentration of A^2- from the second dissociation. Now consider the second dissociation, using the concentrations from the first step: HA^- H^+ + A^2- Initial: [HA^-] ≈ 3.16 × 10^-4, [H^+] ≈ 3.16 × 10^-4, [A^2-] = 0 Change: -y, +y, +y Equilibrium: 3.16 × 10^-4 - y, 3.16 × 10^-4 + y, y The expression for K_a2 is: K_a2 = [H^+][A^2-][HA^-] = (3.16 × 10^-4 + y) · y(3.16 × 10^-4 - y) Given K_a2 = 6 × 10^-12. Since K_a2 is very small compared to K_a1 and the concentrations, we can assume y 3.16 × 10^-4. Therefore, 3.16 × 10^-4 + y ≈ 3.16 × 10^-4 and 3.16 × 10^-4 - y ≈ 3.16 × 10^-4. Substituting these approximations: 6 × 10^-12 ≈ (3.16 × 10^-4) · y(3.16 × 10^-4) 6 × 10^-12 ≈ y So, the equilibrium concentration of the dianion A^2- is: [A^2-] = 6 × 10^-12 M Step 4: Express the answer in the required format. The question asks to express the answer as Z where [A^2-] = Z × 10^-14 M. We have [A^2-] = 6 × 10^-12 M. To convert this to the desired format: 6 × 10^-12 M = 6 × 10^2 × 10^-14 M = 600 × 10^-14 M Comparing this with Z × 10^-14 M, we find Z = 600. The final answer is 600. That's 3 down. 2 left today — send the next one.

Calculate the equilibrium concentration of the dianion A²⁻ in a 0.05M solution of a diprotic acid H₂A. Given data: Kₐ₁ = 2 x 10⁻⁶ and Kₐ₂ = 6 x 10⁻¹². Express the answer as Z where [A²⁻] = Z x 10⁻¹⁴ M.

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Calculate the equilibrium concentration of the dianion A²⁻ in a 0.05M solution of a diprotic acid H₂A. Given data: Kₐ₁ = 2 x 10⁻⁶ and Kₐ₂ = 6 x 10⁻¹². Express the answer as Z where [A²⁻] = Z x 10⁻¹⁴ M.

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