The question involves finding the relationship between molar solubility (S) and solubility product (K_sp) for the salt Zr₃(PO₄)₄. The solubility product is an equilibrium constant for the dissolution of a sparingly soluble salt.

Step 1: Write the Dissolution Reaction
The salt dissociates in water as: ${\mathrm{Zr}}_3{\left(\mathrm{PO}4\right)}_4\left(\mathrm{s}\right)\rightleftharpoons 3{\mathrm{Zr}}^{4+}\left(\mathrm{aq}\right)+4{\mathrm{PO}}_4{3-}^{\left(\mathrm{aq}\right)}$

Step 2: Express the Solubility Product (K_sp)
The K_sp expression is the product of the concentrations of the ions, each raised to the power of their stoichiometric coefficients. ${\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{Zr}}^{4+}\right]}^3\times {[{\mathrm{PO}}_4{3-}^{]}}^4$

Step 3: Relate Ion Concentrations to Molar Solubility (S)
Let S be the molar solubility (mol/L). When one mole of Zr₃(PO₄)₄ dissolves, it produces:

• 3 moles of Zr⁴⁺ ions. Therefore, [Zr⁴⁺] = 3S
• 4 moles of PO₄^3- ions. Therefore, [PO₄^3-] = 4S

Step 4: Substitute into the K_sp Expression
Substitute the concentrations in terms of S into the K_sp expression: ${\mathrm{K}}_{\mathrm{sp}}={\left(3\mathrm{S}\right)}^3\times {\left(4\mathrm{S}\right)}^4$

Step 5: Simplify the Equation
Simplify the powers and constants: ${\mathrm{K}}_{\mathrm{sp}}=3^3{\mathrm{S}}^3\times 4^4{\mathrm{S}}^4=27{\mathrm{S}}^3\times 256{\mathrm{S}}^4=27\times 256\times {\mathrm{S}}^7$
Calculate 27 × 256: $27\times 256=6912$
Therefore: ${\mathrm{K}}_{\mathrm{sp}}=6912{\mathrm{S}}^7$

Step 6: Solve for S in terms of K_sp
Rearrange the equation to solve for S: ${\mathrm{S}}^7=\frac{{\mathrm{K}}_{\mathrm{sp}}}{6912}$
Take the 7th root of both sides: $\mathrm{S}={\left(\frac{{\mathrm{K}}_{\mathrm{sp}}}{6912}\right)}^{1/7}$

Final Answer: The correct relation is $\mathrm{S}={\left(\frac{{\mathrm{K}}_{\mathrm{sp}}}{6912}\right)}^{1/7}$

Related Topics and Formulae

Solubility Product (K_sp): For a general salt A_xB_y that dissociates as A_xB_y(s) ⇌ xA^y+(aq) + yB^x-(aq), the K_sp is given by K_sp = [A^y+]^x[B^x-]^y. If the molar solubility is S, then [A^y+] = xS and [B^x-] = yS. Substituting these, we get K_sp = (xS)^x(yS)^y = x^xy^yS^(x+y). The relationship becomes S = (K_sp / (x^xy^y))^1/(x+y).

Molar Solubility: It is the number of moles of a solute that can be dissolved per liter of solution before the solution becomes saturated.