Calculate the molar solubility of AgBr in a 0.25M solution of NH₃(aq) (K_sp (AgBr) = 7.7 x 10^-13 ; K_f (Ag(NH₃)₂⁺) = 1.5 x 10⁷.

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First, write out the equilibrium equation for the dissolution of AgBr in water: AgBr (s) ⇌ Ag+ (aq) + Br- (aq) Next, write out the expression for the solubility product constant, Ksp: Ksp = [Ag+][Br-] We can use this expression to write the equilibrium concentrations of Ag+ and Br- in terms of the molar solubility of AgBr, x: Ksp = (x)(x) = x^2 Now let's consider the effect of the 0.25M solution of NH₃ on the equilibrium. Since this is a weak acid (denoted by the "HS" in the formula), it will react with the Ag+ ion in solution to form the complex ion Ag(NH₃)₂⁺: Ag+ (aq) + NH₃ (aq) ⇌ Ag(NH₃)₂⁺ (aq) The formation constant for this reaction is given as Kf. Using the equilibrium constant expression for this reaction, we can write: Kf = [Ag(NH₃)₂⁺]/[Ag+][NH₃] At equilibrium, the product of the equilibrium concentrations of Ag+ and NH₃ must equal the equilibrium concentration of Ag(NH₃)₂⁺. We can express the equilibrium concentration of Ag+ in terms of x and the initial concentration of NH₃: [NH₃]eq = [NH₃]initial - [Ag(NH₃)₂⁺]eq [NH₃]eq = 0.25 - [Ag(NH₃)₂⁺]eq (since [NH₃]initial = 0.25M) Using the Kf expression and the equilibrium concentration of Ag+, we can solve for the equilibrium concentration of Ag(NH₃)₂⁺: Kf = [Ag(NH₃)₂⁺]/(x)(0.25 - [Ag(NH₃)₂⁺]eq) 1.5 x 10⁷ = [Ag(NH₃)₂⁺]/(x)(0.25 - [Ag(NH₃)₂⁺]eq) [Ag(NH₃)₂⁺]eq = (1.5 x 10⁷)(x)(0.25) / [1 + (1.5 x 10⁷)(x)] Substituting this expression back into the expression for [NH₃]eq and then into the Ksp expression, we get: [NH₃]eq = 0.25 - (1.5 x 10⁷)(x)(0.25) / [1 + (1.5 x 10⁷)(x)] Ksp = (x)(x + (1.5 x 10⁷)(0.25)(x)) = xS1S1U2S1U2S1U1S1U1 Solving for x gives a value of approximately 8.4 x 10^-4 M, which is closest to option C.

Calculate the Molar Solubility of AgBr in a 0