A test tube contains: . 5.00 mL of 0.002M iron (III) nitrate . 4.50 mL of 0.002M potassium thiocyanate . 0.50 mL of 0.500M nitric acid The reaction proceeds as Fe(NO3)3 + KSCN --> Fe(SCN)*2 + K* After the reaction is complete, the final concentration of iron (III) thiocyanate is calculated to be 5.2 x 10 M. Calculate the equilibrium constant for the reaction. 5.84 09.95 8.95 4.85

Chemistry by OpenStax (2015-05-04)
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Chapter15: Equilibria Of Other Reaction Classes
Section: Chapter Questions
Problem 65E: Sometimes equilibria fur complex ions are described in terms of dissociation constants. Kd . For the...
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A test tube contains:
• 5.00 mL of 0.002M iron (III) nitrate
.
• 4.50 mL of 0.002M potassium thiocyanate
. 0.50 mL of 0.500M nitric acid
The reaction proceeds as Fe(NO3)3 + KSCN --> Fe(SCN)*2 + K+
After the reaction is complete, the final concentration of iron (III) thiocyanate is calculated to be 5.2 x
10 M.
Calculate the equilibrium constant for the reaction.
O5.84
O 9.95
8.95
4.85
Transcribed Image Text:A test tube contains: • 5.00 mL of 0.002M iron (III) nitrate . • 4.50 mL of 0.002M potassium thiocyanate . 0.50 mL of 0.500M nitric acid The reaction proceeds as Fe(NO3)3 + KSCN --> Fe(SCN)*2 + K+ After the reaction is complete, the final concentration of iron (III) thiocyanate is calculated to be 5.2 x 10 M. Calculate the equilibrium constant for the reaction. O5.84 O 9.95 8.95 4.85
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