a. Calculate the molar solubility of AgBr in pure water. \(K_{\text {sp }}\) for 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 \(\mathrm{AgBr}\) 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 \(\mathrm{b}\) ?

We know that \(AgBr \rightleftharpoons Ag^+ + Br^-\). We can represent the solubility of AgBr as 's'. The Ksp expression for this reaction will be: \[ K_{sp} = [Ag^+][Br^-] \] since Ksp is given as \(5.0 × 10^{-13}\), \[ 5.0 × 10^{-13} = (s)(s) \] Solving this equation for 's' (the molar solubility of AgBr in water) will give us the answer. \[s = \sqrt{5.0 × 10^{-13}} \] \[s = 2.24 × 10^{-7} M\]

In the presence of NH3, the reaction between Ag+ and NH3 occurs which forms the complex ion Ag(NH3)2+ with a formation constant, K, given as \(1.7 × 10^7\). The reaction can be written as : \[Ag^+ + 2NH_3 \rightleftharpoons Ag(NH_3)_2^+\] Let's assume the molar solubility of AgBr in 3.0 M NH3 is 'x'. Then, \[ K = \frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2} \Rightarrow [Ag^+] = \frac{[Ag(NH_3)_2^+]}{K[NH_3]^2} \] Now, using the Ksp expression we derived in Part a, \[K_{sp} = [(x)][Br^-] = x^2\] We know that \([Br^-] = [Ag^+]\), so we can replace the Ksp expression with the information about the complex ion and NH3: \[5.0 × 10^{-13} = \frac{[Ag(NH_3)_2^+]^2}{[1.7 × 10^7][3.0]^2} \] Solve for \([Ag(NH_3)_2^+]\), as it's equal to the molar solubility of AgBr in 3.0 M NH3. \[ Ag(NH_3)_2^+ = \sqrt{5.0 × 10^{-13} × [1.7 × 10^7] × [3.0]^2} = 2.49 × 10^{-4} M\]

Comparing the solubilities from parts a and b: \[ s = 2.24 × 10^{-7} M \ \ (Pure \ water) \] \[ Ag(NH_3)_2^+ = 2.49 × 10^{-4} M \ \ (3.0 \ M \ NH_3) \] The molar solubility of AgBr in 3.0 M NH3 is significantly higher than in pure water. This difference is due to the complex ion formation (Ag(NH3)2+) in the presence of NH3, lowering Ag+ concentration and increasing the solubility of AgBr.

We already found that the molar solubility of AgBr in 3.0 M NH3 is 2.49 × 10^{-4 } M. We can now calculate the mass of AgBr that will dissolve in 250.0 mL (0.250 L) of 3.0 M NH3: \[ mass \ of \ AgBr = (molar \ solubility) \times (volume) \times (molar \ mass \ of \ AgBr) \] \[ mass \ of \ AgBr = (2.49 × 10^{-4} \ mol/L) \times (0.250 \ L) \times (187.77 \ g/mol) \] \[ mass \ of \ AgBr = 0.0117 \ g \]

Adding HNO3 to the solution will provide a source of H+ ions, which protonate the NH3, forming NH4+ and decreasing the concentration of NH3: \[HNO_3 \to H^+ + NO_3^-\] \[NH_3 + H^+ \rightleftharpoons NH_4^+\] With less NH3 available, the complex ion formation in Part b (Ag(NH3)2+) is reduced, which raises Ag+ concentration. This increases the Ksp of AgBr and thus decreases the solubility of AgBr in HNO3. In conclusion, the addition of HNO3 will decrease the solubility of AgBr in both pure water and the 3.0 M NH3 solution.