a. Calculate the molar solubility of AgBr in pure water. \(K_{\text {sp }}\) for \(\mathrm{AgBr}\) is \(5.0 \times 10^{-13}\) b. Calculate the molar solubility of AgBr in \(3.0 M \mathrm{NH}_{3}\). The overall formation constant for \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\) is \(1.7 \times 10^{7}\), that is, \(\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K=1.7 \times 10^{7}\) c. Compare the calculated solubilities from parts a and b. Explain any differences. d. What mass of AgBr will dissolve in \(250.0 \mathrm{~mL}\) of \(3.0 \mathrm{M} \mathrm{NH}_{3}\) ? e. What effect does adding \(\mathrm{HNO}_{3}\) have on the solubilities calculated in parts a and b?

First, we have to find the molar solubility of AgBr in pure water. AgBr's solubility product constant (\(K_{sp}\)) is given as \(5.0 \times 10^{-13}\). When AgBr dissolves in water, it dissociates as follows: \[ AgBr \rightleftharpoons Ag^+ + Br^-\] Since both AgBr and its ions are all in a 1:1 stoichiometry, we can set up the following equation, where \(s\) represents the molar solubility of AgBr in pure water: \[ K_{sp} = [Ag^+][Br^-] = (s)(s) = s^2 \] Now, solve for \(s\): \[ s^2 = 5.0 \times 10^{-13} \] \[ s = \sqrt{5.0 \times 10^{-13}} ≈ 2.24 \times 10^{-7} M\] The molar solubility of AgBr in pure water is around \(2.24 \times 10^{-7} M\).

Now, we'll find the molar solubility of AgBr in a \(3.0 M NH_3\) solution. Here, the formation of a complex ion (\(Ag(NH_{3})_{2}^+\)) affects the solubility. The complex formation reaction is given by: \[Ag^+ + 2NH_3 \rightleftharpoons Ag(NH_3)_2^+ \quad K = 1.7 \times 10^7 \] First, let's write the equilibrium expressions for the AgBr dissolution and formation of the complex ion: \[ K_{sp} = [Ag^+][Br^-] \] \[ K_{f} = \frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2} \] Now, we will assume that all the Ag+ is initially bound to the complex ion formed, and the total concentration of NH3 is not affected significantly by the complex formation: \[ K_{sp} = [Ag^+]([Br^-] + [Ag(NH_3)_2^+]) \] \[ K_{f} = 1.7 \times 10^7 = \frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2} \] Considering \(s\) as the molar solubility of AgBr in this scenario, we can establish the following relationships: \[ [Br^-] = s \] \[ [Ag(NH_3)_2^+] = s \] \[ [Ag^+] = \frac{s}{1.7 \times 10^7[NH_3]^2} \] \[ [NH_3] \approx 3.0 M\] Then: \[ K_{sp} = s \left( s + \frac{s}{1.7 \times 10^7 (3.0)^2} \right) \] Now, we need to solve for \(s\): \[ K_{sp} = 5.0 \times 10^{-13} = \frac{1.7 \times 10^7 s^3}{9(1.7 \times 10^7)^2} \] \[ s = \sqrt[3]{\frac{5.0 \times 10^{-13} \times 9(1.7 \times 10^7)^2}{1.7 \times 10^7}} ≈ 1.42 \times 10^{-4} M \] The molar solubility of AgBr in the presence of \(3.0 M NH_3\) is around \(1.42 \times 10^{-4} M\).

Comparing solubilities, the molar solubility of AgBr in the presence of \(3.0 M NH_3 (1.42 \times 10^{-4} M)\) is significantly higher than that in pure water (\(2.24 \times 10^{-7} M\)). This difference is due to the formation of the complex ion \(Ag(NH_3)_2^+\) in the presence of NH3. This complex ion reduces the concentration of free Ag+ in the solution, shifting the equilibrium of AgBr dissolution towards the right, and therefore increasing solubility.

To find the mass of AgBr dissolved in \(250.0 mL\) of \(3.0 M NH_3\), we will use the molar solubility obtained in step 2: \[ mass = molar\;solubility \times volume \times molar\;mass \] \[ mass = (1.42 \times 10^{-4}\; M)(0.25\; L)(AgBr\; molar\;mass) \] \[ mass ≈ (1.42 \times 10^{-4}\; M)(0.25\; L)(187.77\; g/mol) \approx 0.0067\; g \] Approximately \(0.0067 g\) of AgBr will dissolve in the \(250.0 mL\) of \(3.0 M NH_3\) solution.

Adding HNO3 will increase the concentration of H+ ions in the solution. Since NH3 acts as a base, it will react with H+ to form NH4+: \[ NH_3 + H^+ \rightleftharpoons NH_4^+ \] This reaction consumes NH3, and as a consequence, the formation of the complex ion \(Ag(NH_3)_2^+\) will also decrease. This will cause an increase in the concentration of free Ag+ ions, shifting the equilibrium of AgBr dissolution back towards the left, and therefore decreasing the solubility of AgBr in both scenarios (parts a and b).