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x copy of Equilibrium Con × The Equilibrium Constar x Untitled doc https//docs.google.com/document/d/18gLAq DTBsiec0g48j8FUjzio5WtV-gz9D6lmJwrW4/edit brium Constant Determination Lab Notebook Template Help Last edit was made 6 days ago by William Wacholtz Insert Format Tools Table Add-ons 100%, Normal text Table 1. Raw data for generation of calibration graph. Cuvette Volume Volume Total AbsorbancSNFe" [FeSCN of Fe volume of KSCN Used Used 0.50mL | 24.50mL | 25.00 | 210 | 4,0x1o-5 | 40x10.5 .196 0.75mL 24.25mL 25.00 316 1.00mL 24.00mL. 25.00 415 8.0x10-5 .192 1.0x104-4 190 1.2x 10^-4 1.188 6.0x104-5 8.0x105 10x104-4 1.2x104 4 1.25ml 23.75ml. 25.00 80 1.50mL 23.50mL 25.00 655 Experimentally determined molar absorptivity (e) of [FeSCNr . Mcm 5770Mcm e to search

Explanation / Answer

Hi! I hope you´re doing nice!

I´m going to start from the next reactions:

(1) FeCl-3(ac) ----> Fe+3(ac) + 3Cl-(ac)

(2) KSCN(ac) ----> K+(ac) + SCN-(ac)

If the cation Fe+3(ac) and SCN-(ac) reacts results the followiong:

(3) Fe+3(ac) + SCN-(ac) <----> Fe(SCN)+2(ac) "This is the chemical reaction of the present problem"

The color of the solution obtained is red.

Once the concentration of [FeSCN]+2 in the equilibrium is determined and as the concentrations of Fe+3 and SCN- are known, it will be possible to determine the final or equilibrium concentrations of the reactants, according to the expressions following:

For trail 2 in table1:

[Fe+3]final = [Fe+3]i - [Fe(SCN)+2]eq = 0.194M - 6.0x10^-5M = 019394 M

[SCN-]final = [SCN-]i - [Fe(SCN)+2]eq = 6.0x10^-5M - 6.0x10^-5M = 0M ---> That is the situation I don´t want to have, I don´t want the [SCN-] to be zero

ICE TABLE: (Initial, Change and Equilibrium):

Reaction: Fe+3(ac) + SCN-(ac) <----> Fe(SCN)+2(ac)

Initial 0.194M 6.0x10^-5M 0.000M

Change -xM -xM +xM

Equilibrium (0.194-x)M (6.0x10^-5-x)M 6.0x10^-5M

The mathematical model for the equilibrium constant is the following:

Keq= [Fe(SCN)+2]eq / [Fe+3]eq * [SCN-]eq = 6.0x10^-5 / (0.194-x) * (6.0x10^-5-x)

If Keq is equal to 1.24x10^2 results the following:

1.24x10^2 = 6.0x10^-5 / (0.194-x) * (6.0x10^-5-x) (Equation N° 1)

I´m going to use Algebra for clear the variable "x" from the equation N° 1 and results the following quadratic equation:

x^2 - 0.19406x + 1.115613x10^-5 = 0

The quadratic formula gives two solutions (but only one physical solution) for x:

Intuition must be used in determining which solution is correct in this case. The resulting concentration can be more than the initial concentration of [SCN-](ac).

[Solutions: x = 0.1940, 5.7505x10^-5. x = 5.7505x10^-5 makes chemical sense and is therefore the correct answer.]

Let´s calculate the [SCN-] at equilibrium as following:

[SCN-]eq = 6.0x10^-5M - 5.7505x10^-5M = 2.495x10-6M

The total volume of solution is equal to 25.00 mL for trail 2 in table 1, using converstion factor I can have the total volumen in Liters (L)

1 mL = 10x10^-3 L, so 25.00 mL is equal to 25.00x10^-3 L

I´m goint to determine the SCN- equilibrium amount as following:

25.00x10^-3 L of solution * ( 2.496x10^-6 mol / 1 L of solution ) = 6.24x10^-8 mol of SCN- at equilibrum.

Answer: 6.24x10^-8 mol is the amount of SCN- that was not part of the complex ion at equilibrium for trail 2 in table 1.

I hope this explanation to be helpful to the question made. Regards

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