The value of \(K_{s p}\) for \(\mathrm{Cd}(\mathrm{OH})_{2}\) is $2.5 \times 10^{-14} .(\mathbf{a})$ What is the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2} ?(\mathbf{b})\) The solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) can be increased through formation of the complex ion \(\mathrm{CdBr}_{4}^{2-}\left(K_{f}=5 \times 10^{3}\right) .\) If solid \(\mathrm{Cd}(\mathrm{OH})_{2}\) is added to a NaBr solution, what is the initial concentration of NaBr needed to increase the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) to $1.0 \times 10^{-3} \mathrm{~mol} / \mathrm{L} ?$

To find the molar solubility of Cd(OH)2, first, we start by writing a balanced dissociation equation: \[Cd(OH)_2 \xrightarrow{dissolution} Cd^{2+} + 2 OH^-\] Now, write the expression for the solubility product constant, Ksp, of Cd(OH)2: \[K_{sp} = [Cd^{2+}][OH^-]^2\] We are given the value of Ksp, and we can solve for the molar solubility, s, of Cd(OH)2. Let the molar solubility be denoted by s, thus the [Cd2+] = s and [OH-] = 2s \(K_{sp} = (s)(2s)^2\)

Now, insert the given Ksp value and solve for s: \(2.5 \times 10^{-14} = (s)(2s)^2\) \(2.5 \times 10^{-14} = 4s^3\) Divide both sides by 4: \(\frac{2.5 \times 10^{-14}}{4} = s^3\) Take the cube root of both sides: \(s = \sqrt[3]{\frac{2.5 \times 10^{-14}}{4}}\) Finally, calculate the value of s: \(s \approx 6.48 \times 10^{-6} \frac{mol}{L}\) Now, we move to the second part of the problem.

The complex formation reaction is: \[Cd^{2+} + 4Br^- \xrightarrow{formation} CdBr_4^{2-}\] The equilibrium constant for this reaction, Kf, is given as 5 x 10^3. The relationship between Ksp and Kf is derived as follows: \[K = K_{sp}K_{f} = (s)(2s)^2 \times \frac{[(s)(1.0 \times 10^{-3})]}{(4s)(1.0 \times 10^{-3})^4}\] We are given the value of K and solve for s.

First, calculate the value of the equilibrium constant K: \(K = K_{sp} K_{f}\) \(K = (2.5 \times 10^{-14})(5 \times 10^3)\) \(K = 1.25 \times 10^{-10}\) Now, insert the given molar solubility of 1.0 x 10^-3 mol/L and solve for the initial concentration [Br-]: \(1.25 \times 10^{-10} = \frac{[(s)(1.0 \times 10^{-3})]}{(4s)([Br^-])^4}\) Rearrange for [Br-]: \([[Br^-] = \sqrt[4]{\frac{(s)(1.0 \times 10^{-3})}{(4s)(1.25 \times 10^{-10})}}\) Substituting the value of s from step 2: \([[Br^-] = \sqrt[4]{\frac{(6.48 \times 10^{-6})(1.0 \times 10^{-3})}{(4)(6.48 \times 10^{-6})(1.25 \times 10^{-10})}}\) Calculate the value of [Br-]: \([Br^-] \approx 0.063 \frac{mol}{L}\) Therefore, the initial concentration of NaBr needed to increase the molar solubility of Cd(OH)2 to 1.0 x 10^-3 mol/L is 0.063 mol/L.