Question

The generic metal A forms an insoluble salt AB(s) and a complex AC₅(aq).

The equilibrium concentrations in a solution of AC₅ were found to be [A]=0.100 M[A]=0.100 M, [C]=0.0260 M[C]=0.0260 M, and [AC5]=0.100 M[AC5]=0.100 M. Determine the formation constant, K_f, of AC₅.

?f=

The solubility of AB(s) in a 1.000 M solution of C(aq) is found to be 0.1310.131 M. What is the K_sp of AB?

?sp=

Answer 1

(x) Finding the formation constant - The possible reaction is- A +5c - Als concentration The formation constant ky is a hatio of the concenthation of products to the ce of reactants to the concentration of products concentration of heactants [AC ] ita- I [A] [egs , We have been diven with the following data [A] = 0-100M 143 = 00260M, [AC] = 0.100M substitutes these values as ☺ Kp Pro too) Kf = 0100 0.100 (0.026075 - Kf = 1.1881 x 108 Kf = 0-8416 x 108 Kf = 8116 x 10 9 (approx)

Part B - Finding the Kep. The possible reaction occutang between Aband a is- AB (s) + 52 (aq) = Alglag) & Bear). To find the Kap, we need the concentrations of all the species in equilibrium for which we shall create an ICE table and them, we find the equilibrium constant K. This value is then divided by the value of kf obtained from earliete stap to find КЕР [AB] Given is = 0.131M [c] = IM AB (s) + 5 ccaq) Alglaq) & Beaq) Initial (I) 0.131 chandelt) -0.131 -5 (0-131) +0.131 +0-131 Equilibrium (E) [1-45 (0.131) 0.131 0.131 [NOTE = concentration of solids = 11 [AC] [B] [0]5 (0.131) (0.131) [1-560131)75 (0.017161) (0.345)5 : K

0.017 16/ 4.887x10-3 - 305 1156 x 1073 10-3 - 3. 51156 The value of ksp = If 3,51156 8.416 * 109 (Ksp = 4.1724x1000)