Question 6 of 19 Map d Determine the concentrations of each zinc-containing spec

Question

Question 6 of 19 Map d Determine the concentrations of each zinc-containing species in a solution saturated with solid Zn(OH)2 and having a fxed OfH concentration of 6.0 x 10" M. The equilibrium constants for the equilibria that occur in the solution follow Zn(OH)s) Zn(OH)' Zn(OH)s Zn(OH) Number 2n21 8.33x 10 M Incorrect Number for Zn(OH)' is the equilibrium constant for the reaction 2n(OH)1 4.88x 10M Zn"(aq) + OH (aq) Zn(OH)+(aq) and the equilibrium constant has the form Number (Zn(OH),-1-11 5.06 × 10 Zn (OH) 7 | M OH ]-2.5x10. Number You have correctly found that [2n21-0.00083 M, and you PnOH)42-1.ju8× 10 MI knohavat correctly foundthat@n2+1-000083 M and you

Explanation / Answer

For the heterogeneous system equation,

K = equilibrium constant

K = [pE]^3/([A]^2.[pB]^3.[C]

feeding the given values,

K = (3.52 x 10^6)^3/(4.75 x 10^-2)^2.(9.72 x 10^3)^3.(10.09)

   = 2.10 x 10^9

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Selective precipitation

A) Precipitate forms first from solution has lowest Ksp value = BaCrO4

B) [Ba2+] needed for BaCrO4 precipitation = Ksp/[CrO4^2-]

                                                                     = 2.1 x 10^-10/0.0422 = 4.98 x 10^-9 M

C) 10% C2O4^2- = 0.010 x 0.0608 = 6.08 x 10^-4 M

remaining [C2O4^2-] = 0.0608 - 0.000608 = 0.0602 M

[Ba2+] needed = Ksp/[C2O4^2-]

                         = 1.3 x 10^-6/6.08 x 10^-4 = 2.14 x 10^-3 M

D) ratio [C2O4^2-]/[CrO4^2-] when [Ba2+] = 0.0050 M

[C2O4^2-] = 1.3 x 10^-6/0.005 = 2.6 x 10^-4 M

[CrO4^2-] = 2.1 x 10^-10/0.005 = 4.2 x 10^-8 M

so,

ratio [C2O4^2-]/[CrO4^2-] = 6190.5

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Titration,

NaI titrated with AgNO3

(a) 36.20 ml of AgNO3 added

initial moles NaI = 0.08510 M x 25 ml = 2.1275 mmol

moles AgNO3 added = 0.05160 M x 36.20 ml = 1.86792 mmol

[I-] remained = (2.1275 - 1.86792) mmol/(25 + 36.20) ml = 0.00424 M

[Ag+] = Ksp/[I-] = 8.3 x 10-17/0.00424 = 1.96 x 10^-14 M

pAg+ = -log[Ag+] = 13.71

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(b) at Ve

initial moles NaI = 0.08510 M x 25 ml = 2.1275 mmol

moles AgNO3 added = 2.1275 mmol

Volume of AgNO3 added = 2.1275 mmol/0.05160 M = 41.23 ml

[Ag+] = sq.rt.(Ksp) = sq.rt.(8.3 x 10-17) = 9.11 x 10^-9 M

pAg+ = -log[Ag+] = 8.04

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(c) 47.10 ml of AgNO3 added

initial moles NaI = 0.08510 M x 25 ml = 2.1275 mmol

moles AgNO3 added = 0.05160 M x 47.10 ml = 2.43036 mmol

[Ag+] = (2.43036 - 2.1275) mmol/(25 + 47.10) ml = 0.30286 M

pAg+ = -log[Ag+] = 0.52

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