"calculate the concentrations of fe31, scn2, and fescn21 at equilibrium if 0.020 mole of fe(no3)3 is added to 1.0 l of 0.10 m kscn. (neglect any volume change.)"

The reaction we are concerned with here is as follows:

Fe³⁺ + SCN⁻ ⇄ FeSCN²⁺

The equilibrium constant is K = 1.1 x 10³

We are told we have 0.020 mole of Fe(NO₃)₃ is added to a 1.0 L solution of 0.10 M KSN. We can determine our concentration of Fe(NO₃)₃:

0.020 mole Fe(NO₃)₃ / 1.0 L = 0.020 M Fe(NO₃)₃

We can also assume that both Fe(NO₃)₃ and KSCN are completely dissociated like so:

Fe(NO₃)₃ → Fe³⁺ + 3 NO₃⁻
KSCN → K⁺ + SCN⁻

Therefore, the concentration [Fe³⁺] = [Fe(NO₃)₃] = 0.020 M and [KSCN] = [SCN⁻] = 0.10 M

Now we can determine the equilibrium concentrations using an ICE table, which shows the initial, changes, and equilibrium concentrations.

       Fe³⁺   +   SCN⁻   ⇄   FeSCN²⁺
I     0.020        0.10              0
C      -x            -x                +x
E    0.020-x     0.10-x           x

K = [FeSCN²⁺] / [Fe³⁺][SCN⁻]
K = x / [0.020-x][0.10-x] = 1.1 x 10³
x = (x² - 0.12x + 0.002)(1100)
0 = 1100x² - 133x + 2.2

We now solve for x using the quadratic formula:

x = (-b +/- √(b²-4ac)) / 2a
x = (133 +/- √(133² - 4(1100)(2.2))) / 2200
x = (133 +/- 89.5) / 2200
x = 0.101 and x = 0.0198

The only answer that can work for this example is x = 0.0198, now we plug in x into the equilibrium concentrations to get the final concentrations:

[Fe³⁺] = 0.0002 M
[SCN⁻] = 0.0802 M
[FeSCN²⁺] = 0.0198 M 

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