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.

Chemistry
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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2 steps

Answer

600

Step 1: Write down the dissociation reactions and their equilibrium constants for the diprotic acid H2_2A. A diprotic acid dissociates in two steps:

  1. First dissociation: H2A⇌H++HA−H_2A \rightleftharpoons H^+ + HA^- The acid dissociation constant for this step is Ka1=2×10−6K_{a1} = 2 \times 10^{-6}.

  2. Second dissociation: HA−⇌H++A2−HA^- \rightleftharpoons H^+ + A^{2-} The acid dissociation constant for this step is Ka2=6×10−12K_{a2} = 6 \times 10^{-12}.

Step 2: Calculate the equilibrium concentrations from the first dissociation. Let the initial concentration of H2_2A be C=0.05C = 0.05 M. For the first dissociation: Initial: [H2A]=0.05[H_2A] = 0.05 M, [H+]=0[H^+] = 0, [HA−]=0[HA^-] = 0 Change: −x-x, +x+x, +x+x Equilibrium: 0.05−x0.05 - x, xx, xx

The expression for Ka1K_{a1} is: Ka1=[H+][HA−][H2A]=x⋅x0.05−xK_{a1} = \frac{[H^+][HA^-]}{[H_2A]} = \frac{x \cdot x}{0.05 - x} Given Ka1=2×10−6K_{a1} = 2 \times 10^{-6}: 2×10−6=x20.05−x2 \times 10^{-6} = \frac{x^2}{0.05 - x} Since Ka1K_{a1} is small, we can assume x≪0.05x \ll 0.05, so 0.05−x≈0.050.05 - x \approx 0.05. 2×10−6≈x20.052 \times 10^{-6} \approx \frac{x^2}{0.05} x2=2×10−6×0.05=1×10−7x^2 = 2 \times 10^{-6} \times 0.05 = 1 \times 10^{-7} x=1×10−7=10×10−8=10×10−4≈3.16×10−4Mx = \sqrt{1 \times 10^{-7}} = \sqrt{10 \times 10^{-8}} = \sqrt{10} \times 10^{-4} \approx 3.16 \times 10^{-4} M This value of xx is indeed much smaller than 0.05, so the approximation is valid. Thus, at equilibrium from the first dissociation: [H+]≈3.16×10−4M[H^+] \approx 3.16 \times 10^{-4} M [HA−]≈3.16×10−4M[HA^-] \approx 3.16 \times 10^{-4} M

Step 3: Calculate the concentration of A2−^{2-} from the second dissociation. Now consider the second dissociation, using the concentrations from the first step: HA−⇌H++A2−HA^- \rightleftharpoons H^+ + A^{2-} Initial: [HA−]≈3.16×10−4[HA^-] \approx 3.16 \times 10^{-4}, [H+]≈3.16×10−4[H^+] \approx 3.16 \times 10^{-4}, [A2−]=0[A^{2-}] = 0 Change: −y-y, +y+y, +y+y Equilibrium: 3.16×10−4−y3.16 \times 10^{-4} - y, 3.16×10−4+y3.16 \times 10^{-4} + y, yy

The expression for Ka2K_{a2} is: Ka2=[H+][A2−][HA−]=(3.16×10−4+y)⋅y(3.16×10−4−y)K_{a2} = \frac{[H^+][A^{2-}]}{[HA^-]} = \frac{(3.16 \times 10^{-4} + y) \cdot y}{(3.16 \times 10^{-4} - y)} Given Ka2=6×10−12K_{a2} = 6 \times 10^{-12}. Since Ka2K_{a2} is very small compared to Ka1K_{a1} and the concentrations, we can assume y≪3.16×10−4y \ll 3.16 \times 10^{-4}. Therefore, 3.16×10−4+y≈3.16×10−43.16 \times 10^{-4} + y \approx 3.16 \times 10^{-4} and 3.16×10−4−y≈3.16×10−43.16 \times 10^{-4} - y \approx 3.16 \times 10^{-4}. Substituting these approximations: 6×10−12≈(3.16×10−4)⋅y(3.16×10−4)6 \times 10^{-12} \approx \frac{(3.16 \times 10^{-4}) \cdot y}{(3.16 \times 10^{-4})} 6×10−12≈y6 \times 10^{-12} \approx y So, the equilibrium concentration of the dianion A2−^{2-} is: [A2−]=6×10−12M[A^{2-}] = 6 \times 10^{-12} M

Step 4: Express the answer in the required format. The question asks to express the answer as Z where [A2−]=Z×10−14[A^{2-}] = Z \times 10^{-14} M. We have [A2−]=6×10−12[A^{2-}] = 6 \times 10^{-12} M. To convert this to the desired format: 6×10−12M=6×102×10−14M=600×10−14M6 \times 10^{-12} M = 6 \times 10^2 \times 10^{-14} M = 600 \times 10^{-14} M Comparing this with Z×10−14Z \times 10^{-14} M, we find Z=600Z = 600.

The final answer is 600\boxed{600}.

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Write down the dissociation reactions and their equilibrium constants for the diprotic acid H_2A.

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.
Chemistry

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.

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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.