A student prepared an equilibrium solution by mixing the following solutions:A.2.00 mL of 0.00250 M Fe(NO3)3B.5.00 mL of 0.00250 M KSCNC.3.00 mL of 0.050 M HNO3Calculate the initial concentrations of all ions, after mixing, prior to the reaction occurring. The equilibrium concentration of Fe(NCS)2+ was determined using a spectrophotometer to be 3.6 x 10-5 M. Calculate the concentrations of all ions at equilibrium. Calculate the value for the equilibrium constant, K.

Explanation:

A. 2.00 mL of 0.00250 M

Moles of ferric nitrate = n

Volume of ferric nitrate = 2.00 ml = 0.002 L ( 1 mL=0.001 L)

Molarity of ferric nitrate = 0.00250 M

B. 5.00 mL of 0.00250 M

Moles of KSCN  = n'

Volume of KSCN  = 5.00 ml = 0.005 L ( 1 mL=0.001 L)

Molarity of KSCN = 0.00250 M

C. 3.00 mL of 0.050 M

Moles of nitric acid = n''

Volume of nitric acid = 3.00 ml = 0.003 L ( 1 mL=0.001 L)

Molarity of nitric acid = 0.050 M

After mixing A, B and C together and their respective initial concentration before reaction.

After mixing A, B and C together the volume of the solution becomes = V

V = 0.002 L=0.005 L+0.003 L= 0.010 L

Concentration of ferric nitrate :

Concentration of ferric ions :

Concentration of nitrate ions from ferric nitrate:

Concentration of KSCN :

Concentration of ions:

Concentration of potassium ions:

Concentration of nitric acid :

Concentration of hydrogen ion :

Concentration of nitrate ions from nitric acid  :

Concentration of nitrate ion in mixture = 0.0015 M + 0.0015 M = 0.0030 M

given concentration of at equilbrium =

initially :

0.0005 M     0.00125 M        0

At equilibrium

(0.0005-0.000036) M   (0.00125-0.000036) M      0.000036 M

0.000464 M     0.001214 M               0.000036 M

The expression of an equilibrium constant will be given as;

The value for the equilibrium constant is 63.91.

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