I keep getting one as my equilibrium constant for each absorbance (#1-5). Please

Question

I keep getting one as my equilibrium constant for each absorbance (#1-5). Please take a look at my data. I need help calculating the equilibrium constant (K_eq). This data is from my triiodide lab.

# I2(mL) KI (mL) H2O(mL) Total Volume (mL) I2o Io^- A Temperature (°C) I3^- I2 I^- K_eq 1 2 1 1 4 5.00E-05 5.00E-04 0.235 23 1.40E-05 4.00E-03 3.50E-03 1.00 2 1 3 0 4 2.50E-05 1.00E-03 0.144 23 8.30E-06 3.40E-03 2.40E-03 1.00 3 2 1 1 4 5.00E-05 5.00E-04 0.216 23 1.30E-05 3.90E-03 3.40E-03 1.00 4 3 1 0 4 7.50E-05 5.00E-04 0.278 23 1.60E-05 4.20E-03 3.80E-03 1.00 5 1 3 0 4 2.50E-05 1.50E-03 0.162 23 9.40E-06 3.90E-03 2.40E-03 1.00

Explanation / Answer

I- (aq) +I2(aq)I3- (aq)

Keq=[I3-](aq)/[I-](aq)[I2](aq)

1)keq=1.4*10^-5/4*10^-3*3.5*10^-3=0.1*10=1

2)keq=8.3*10^-6/3.4*10^-3*2.4*10^-3=1.017

3)keq=1.3*10^-5/3.90*10^-3*3.4*10^-3=0.098

4)keq=1.60*10^-5/4.20*10^-3*3.80*10^-3=1.00

5)keq=9.40*10^-6/3.90*10^-3*2.4*10^-3=1.00

To calculate K, [I2(aq)], [I-(aq)], and [I3-(aq)] at equilibrium are needed.

Check the values of [I2(aq)], [I-(aq)], and [I3-(aq)] is calculated correctly.Approximate value of keq=2.5

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