First, let's understand the dissociation of the sparingly soluble salt $$\mathrm{A}_3 \mathrm{B}_4$$. When it dissolves in water, it dissociates into its ions as follows:

$$\mathrm{A}_3 \mathrm{B}_4 (s) \leftrightarrow 3 \mathrm{A}^{4+} (aq) + 4 \mathrm{B}^{3-} (aq)$$

If the solubility of $$\mathrm{A}_3 \mathrm{B}_4$$ is $$\mathrm{s}$$ grams per liter, and the molar mass of $$\mathrm{A}_3 \mathrm{B}_4$$ is $$\mathrm{M}$$ grams per mole, then the molar solubility (which is the concentration of the dissolved salt in moles per liter) is given by:

$$s_{\text{mol}} = \dfrac{s}{M}$$

Now, upon dissociation, one mole of $$\mathrm{A}_3 \mathrm{B}_4$$ produces 3 moles of $$\mathrm{A}^{4+}$$ ions and 4 moles of $$\mathrm{B}^{3-}$$ ions. Therefore, if the molar solubility is $$s_{\text{mol}}$$, the concentration of $$\mathrm{A}^{4+}$$ ions will be:

$$[\mathrm{A}^{4+}] = 3s_{\text{mol}} = 3 \dfrac{s}{M}$$

And the concentration of $$\mathrm{B}^{3-}$$ ions will be:

$$[\mathrm{B}^{3-}] = 4s_{\text{mol}} = 4 \dfrac{s}{M}$$

The solubility product constant $$\mathrm{K}_{\text{sp}}$$ is given by the product of the ion concentrations, each raised to the power of their respective coefficients in the balanced equation:

$$\mathrm{K}_{\text{sp}} = [\mathrm{A}^{4+}]^3 [\mathrm{B}^{3-}]^4$$

Substituting the values of the ion concentrations:

$$\mathrm{K}_{\text{sp}} = \left(3 \dfrac{s}{M}\right)^3 \left(4 \dfrac{s}{M}\right)^4$$

Now calculate the expression inside the parenthesis:

$$\mathrm{K}_{\text{sp}} = \left( \dfrac{3s}{M} \right)^3 \left( \dfrac{4s}{M} \right)^4$$

Which simplifies to:

$$\mathrm{K}_{\text{sp}} = \dfrac{27s^3}{M^3} \cdot \dfrac{256s^4}{M^4}$$

Multiplying these terms together gives:

$$\mathrm{K}_{\text{sp}} = \dfrac{27 \cdot 256 \cdot s^7}{M^7}$$

Further simplification yields:

$$\mathrm{K}_{\text{sp}} = \dfrac{6912s^7}{M^7}$$

Therefore, the correct expression for $$\mathrm{K}_{\text{sp}}$$ is:

$$\mathrm{K}_{\text{sp}} = 6912 \left(\dfrac{s}{M}\right)^7$$

The correct option is:

Option A

Ionic Equilibrum

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