If the solubility product of $$\text{AB}_2$$ is $$3.20 \times 10^{-11}\,\text{M}^3$$, then the solubility of $$\text{AB}_2$$ in pure water is ______ $$\times 10^{-4}\,\text{mol L}^{-1}$$ [Assuming that neither kind of ion reacts with water]

First, we look at the dissolution equilibrium of the sparingly soluble salt $$\text{AB}_2$$ in pure water. The salt dissociates as

$$\text{AB}_2 (s) \;\rightleftharpoons\; \text{A}^{2+}(aq) + 2\,\text{B}^-(aq)$$

From this balanced equation, we see that one mole of solid $$\text{AB}_2$$ gives one mole of cation $$\text{A}^{2+}$$ and two moles of anion $$\text{B}^-$$ when it dissolves.

Let the molar solubility of $$\text{AB}_2$$ in water be $$s\; \text{mol L}^{-1}$$. That means

$$[\text{A}^{2+}] = s$$

because each formula unit produces exactly one cation, and

$$[\text{B}^-] = 2s$$

because each formula unit produces two anions.

The solubility product constant is defined by the general expression $$K_{sp} = [\text{A}^{2+}]\,[\text{B}^-]^2$$ for this salt, since the stoichiometric coefficients in the dissolution reaction are 1 and 2 respectively. Stating the formula clearly:

$$K_{sp} = [\text{A}^{2+}]\times([\text{B}^-])^2$$

We now substitute the concentrations expressed in terms of $$s$$:

$$K_{sp} = (s)\,\bigl(2s\bigr)^2$$

Carrying out the square first, we get

$$K_{sp} = s \times 4s^2$$

Multiplying the factors of $$s$$ together yields

$$K_{sp} = 4s^3$$

We are given that $$K_{sp} = 3.20 \times 10^{-11}\,\text{M}^3$$, so we write

$$3.20 \times 10^{-11} = 4s^3$$

To isolate $$s^3$$, we divide both sides by 4:

$$s^3 = \dfrac{3.20 \times 10^{-11}}{4}$$

Performing the division in the numerator, $$3.20 \div 4 = 0.80$$, so

$$s^3 = 0.80 \times 10^{-11}$$

It is often convenient to write the coefficient as an integer by adjusting the power of ten. Because $$0.80 = 8.0 \times 10^{-1}$$, we have

$$s^3 = 8.0 \times 10^{-12}$$

Now we take the cube root of both sides to solve for $$s$$. The cube root of the product can be taken separately for the coefficient and for the power of ten:

$$s = \sqrt[3]{8.0}\;\times\;\sqrt[3]{10^{-12}}$$

The cube root of 8.0 is exactly 2, because $$2^3 = 8$$. For the power of ten, we recall that

$$\sqrt[3]{10^{-12}} = 10^{-12/3} = 10^{-4}$$

Combining these two results gives

$$s = 2 \times 10^{-4}\,\text{mol L}^{-1}$$

The question asks for the numerical value that multiplies $$10^{-4}$$, so we read off that the solubility is $$2 \times 10^{-4}\,\text{mol L}^{-1}$$.

So, the answer is $$2$$.