please solve this question in detail. thanks To a 250.0 mL volumetric flask are
ID: 780526 • Letter: P
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please solve this question 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) 4101021_2Suppose 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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