Determine the molar solubility of AgBr (Ksp = 5.40e-13) in 2.502 M NH3 if the complex ion [Ag(NH3)2]+ forms with a Kf = 1.70e7
will I need to make an chemical equation using the compounds they give me? and then do an ICE table?
That's what I would do.
To determine the molar solubility of AgBr in 2.502 M NH3, we need to consider the equilibrium reaction involved and use the concepts of equilibrium and solubility product.
Firstly, let's write the equilibrium reaction for the formation of the complex ion [Ag(NH3)2]+.
AgBr (s) + 2 NH3 (aq) ⇌ [Ag(NH3)2]+ (aq) + Br- (aq)
Now, we will define the equilibrium constant for the formation of the complex ion, Kf = [Ag(NH3)2]+ / ([AgBr] * [NH3]^2).
Next, let's consider the solubility product, Ksp, which is given as 5.40e-13. The expression for Ksp for AgBr is given by Ksp = [Ag+] * [Br-], assuming that AgBr is a sparingly soluble salt.
Since we are interested in finding the molar solubility of AgBr, we can represent the solubility of AgBr as "x" (mol/L). Therefore, [Ag+] and [Br-] would both be equal to "x".
Now, we know that [Ag(NH3)2]+ = "x" and [NH3] = 2.502 M.
Using the definition of Kf, we can write:
Kf = [Ag(NH3)2]+ / ([AgBr] * [NH3]^2)
1.70e7 = x / (x * (2.502)^2)
1.70e7 = x / (x * 6.26)
Simplifying the equation:
1.70e7 = 1 / 6.26
From this equation, we can calculate the value of "x" or the molar solubility of AgBr in 2.502 M NH3.