For the reaction $$ \mathrm{CdC}_{2} \mathrm{O}_{4}(s)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) $$ (a) calculate \(K .\left(K_{s p}\right.\) for \(\mathrm{CdC}_{2} \mathrm{O}_{4}\) is \(\left.1.5 \times 10^{-8} .\right)\) (b) calculate \(\left[\mathrm{NH}_{3}\right]\) at equilibrium when \(2.00 \mathrm{~g}\) of \(\mathrm{CdC}_{2} \mathrm{O}_{4}\) is dissolved in \(1.00 \mathrm{~L}\) of solution.

For the given reaction, we have a solid substance (CdC2O4) dissolving in water and forming a complex ion (Cd(NH3)4^(2+)) and an anion (C2O4^(2-)). We are provided with the solubility product constant, Ksp = 1.5 x 10^(-8), for CdC2O4. From this information, we can determine the equilibrium constant (K) of the given reaction. Remember that: $$ K = \frac{[Cd(NH_3)_4^{2+}][C_2O_4^{2-}]}{[CdC_2O_4][NH_3]^4} = K_{sp} $$ Since [CdC2O4] is a solid, its concentration can be considered constant, and it's not included in the equation. Therefore, K = Ksp.

Since Ksp = 1.5 x 10^(-8), we have: $$ K = 1.5 \times 10^{-8} $$ Step 2: Calculate the concentration of NH3 at equilibrium

In order to find the concentration of NH3 at equilibrium, we first need to convert the mass of CdC2O4 to moles. Given that we have 2.00 g of CdC2O4, the molar mass of CdC2O4 is 183.5 g/mol. Therefore: $$ \text{moles of CdC}_2\text{O}_4 = \frac{2.00 \text{ g}}{183.5 \text{ g/mol}} = 0.0109 \text{ mol} $$

Next, we will use an ICE (Initial, Change, Equilibrium) table to find the NH3 concentration at equilibrium. Let 'x' be the amount of CdC2O4 that dissociates. Initial concentrations: $$[Cd(NH_3)_4^{2+}]=0$$ $$[C_2O_4^{2-}]=0$$ $$[NH_3]= y$$ Changes: $$[Cd(NH_3)_4^{2+}]=+x$$ $$[C_2O_4^{2-}]=+x$$ $$[NH_3] = -4x$$ Equilibrium concentrations: $$[Cd(NH_3)_4^{2+}]=x$$ $$[C_2O_4^{2-}]=x$$ $$[NH_3] = y - 4x$$ Now we can use the K expression from Step 1 along with the information from the ICE table: $$ K = \frac{[Cd(NH_3)_4^{2+}][C_2O_4^{2-}]}{[NH_3]^4} = \frac{x \cdot x}{(y - 4x)^4} $$ Since x is very small compared to y, we can approximate and assume that 4x<<y so (y-4x) ≈ y. Substitute K = 1.5 x 10^(-8) and solve for x: $$ 1.5 \times 10^{-8} = \frac{x^2}{y^4} $$ We know that the number of moles of CdC2O4 that dissolved is equal to x, and since 1.00L of solution: $$ \frac{0.0109}{1.00} = x = 0.0109 \text{ mol/L} $$ Now substitute x back into our simplified equilibrium and solve for the initial concentration of NH3 (y) $$ 1.5 \times 10^{-8} = \frac{(0.0109)^2}{y^4} $$ $$ y^4 = \frac{(0.0109)^2}{1.5 \times 10^{-8}} $$ $$ y = 0.141 \text{ mol/L} $$ So, the concentration of NH3 at equilibrium is 0.141 mol/L.