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.

A diprotic acid dissociates in two steps:

1. First dissociation:

   $$\text{H}_2\text{A} \rightleftharpoons \text{H}^+ + \text{HA}^-$$

   The acid dissociation constant for this step is $K_{a1} = 2 \times 10^{-6}$.

2. Second dissociation:

   $$\text{HA}^- \rightleftharpoons \text{H}^+ + \text{A}^{2-}$$

   The acid dissociation constant for this step is $K_{a2} = 6 \times 10^{-12}$.

Let the initial concentration of H$_2$A be $C = 0.05$ M.
For the first dissociation:
Initial: $[\text{H}_2\text{A}] = 0.05$ M, $[\text{H}^+] = 0$, $[\text{HA}^-] = 0$
Change: $-x$, $+x$, $+x$
Equilibrium: $0.05 - x$, $x$, $x$

The expression for $K_{a1}$ is:

$$K_{a1} = \frac{[\text{H}^+][\text{HA}^-]}{[\text{H}_2\text{A}]} = \frac{x \cdot x}{0.05 - x}$$

Given $K_{a1} = 2 \times 10^{-6}$:

$$2 \times 10^{-6} = \frac{x^2}{0.05 - x}$$

Since $K_{a1}$ is small, we can assume $x \ll 0.05$, so $0.05 - x \approx 0.05$.

$$2 \times 10^{-6} \approx \frac{x^2}{0.05}$$ $$x^2 = 2 \times 10^{-6} \times 0.05 = 1 \times 10^{-7}$$ $$x = \sqrt{1 \times 10^{-7}} = \sqrt{10 \times 10^{-8}} = \sqrt{10} \times 10^{-4} \approx 3.16 \times 10^{-4} \text{ M}$$

This value of $x$ is indeed much smaller than 0.05, so the approximation is valid.
Thus, at equilibrium from the first dissociation:

$$[\text{H}^+] \approx 3.16 \times 10^{-4} \text{ M}$$ $$[\text{HA}^-] \approx 3.16 \times 10^{-4} \text{ M}$$

Now consider the second dissociation, using the concentrations from the first step:

$$\text{HA}^- \rightleftharpoons \text{H}^+ + \text{A}^{2-}$$

Initial: $[\text{HA}^-] \approx 3.16 \times 10^{-4}$, $[\text{H}^+] \approx 3.16 \times 10^{-4}$, $[\text{A}^{2-}] = 0$
Change: $-y$, $+y$, $+y$
Equilibrium: $3.16 \times 10^{-4} - y$, $3.16 \times 10^{-4} + y$, $y$

The expression for $K_{a2}$ is:

$$K_{a2} = \frac{[\text{H}^+][\text{A}^{2-}]}{[\text{HA}^-]} = \frac{(3.16 \times 10^{-4} + y) \cdot y}{(3.16 \times 10^{-4} - y)}$$

Given $K_{a2} = 6 \times 10^{-12}$. Since $K_{a2}$ is very small compared to $K_{a1}$ and the concentrations, we can assume $y \ll 3.16 \times 10^{-4}$.
Therefore, $3.16 \times 10^{-4} + y \approx 3.16 \times 10^{-4}$ and $3.16 \times 10^{-4} - y \approx 3.16 \times 10^{-4}$.
Substituting these approximations:

$$6 \times 10^{-12} \approx \frac{(3.16 \times 10^{-4}) \cdot y}{(3.16 \times 10^{-4})}$$ $$6 \times 10^{-12} \approx y$$

So, the equilibrium concentration of the dianion A$^{2-}$ is:

$$[\text{A}^{2-}] = 6 \times 10^{-12} \text{ M}$$

The question asks to express the answer as Z where $[\text{A}^{2-}] = Z \times 10^{-14}$ M.
We have $[\text{A}^{2-}] = 6 \times 10^{-12}$ M.
To convert this to the desired format:

$$6 \times 10^{-12} \text{ M} = 6 \times 10^2 \times 10^{-14} \text{ M} = 600 \times 10^{-14} \text{ M}$$

Comparing this with $Z \times 10^{-14}$ M, we find $Z = 600$.

The final answer is $\text{600}$.