please solve these problems in detail. thanks To a 250.0 mL volumetric flask are

Question

please solve these problems in detail. thanks

To a 250.0 mL volumetric flask are added 1.00 mL volumes of three solutions: 0.0100 M AgNO3, 0.235 M NaBr, and 0.100 M NaCN. The mixture is diluted with deionized water to the mark and shaken vigorously. What mass of AgBr would precipitate from this mixture? (Hint: The Ksp of AgBr is 5.4 times 10-13 and the Kf of Ag(CN)2- is 1.0 times 1021) Suppose a 500 mL solution contains 0.50 millimoles of Co(NO3)2, 100.0 millimoles of NH3, and 100.0 millimoles of ethylenediamine. What is the concentration of Co2+(aq) in the solution? (The Kf for the formation of Co(NH3)62+ is 7.7 times 104. The Kf for the formation of Co(en)32+ is 8.7 times 1013.)

Explanation / Answer

that the initial concentration of the CN- was equal to (.001 L x .100 mol/L) / .250L = 4E-4as was the initial conc. of the Br-.

Likewise, the intial concentration of Ag+ would be (.001L x .01mol/L)/.250L = 4E-5

CN                Ag(CN)2          Br
I 4E-4 0 0
C             -2x x x
E            4E-4 -2x x x

Keq = x 2/ (4E-4 - 2x) = 5.40E8

Solve this for x which gives us the equilibrium concentration of the Br-.

Once we know how much Br remains in solution, and the amount that we started with we can figure out how much Br- precipitated and from that the mass of the AgBr that precipictate.

b)

Initial concentration of Co2+ = 0.001 M

. concentration of NH3 = 0.2M and

concentration of en = 0.35M .

let concentration of Co(en)3 = x and concentration of Co(NH3)6 = y.

Hence final concentration of

Co2+ - 0.001 - x - y.

Now Kf = 8.7 X 10^13 = x/(0.001 - x - y)*(0.35)^3

and 7.7X10^4 = y/(0.001-x-y)*(0.2)^6 ,

Hence x + y = {8.7 * 10^13 * 0.35^3 + 7.7 X 10^4 * 0.2^6 } * 0.001 - x - y

Let 0.001 - x - y = z, hence x + y = 0.001 - z,

so we have 0.001 - z = {8.7 * 10^13 * 0.35^3 + 7.7 X 10^4 * 0.2^6 } * z. hence z = 0.001/{8.7 * 10^13 * 0.35^3 + 7.7 X 10^4 * 0.2^6 + 1}

So final concentration = 2.68 X 10^-15

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