Question

A solution of 5.00 mL of 0.00300 M Fe(NO3), 4.00 mL of 0.00300 M KSCN, and 3.00 mL of 1.0 M HNO3 is mixed together and allowed to reach equilibrium. The concentration of Fe(SCN)2* is found to be 2.72x10Mat equilibrium. Calculate the equilibrium constant for this reaction?

Answer 1

The equilibrium constant for the reaction is 5.99 x 10^6.

Step-by-step explanation:

To calculate the equilibrium constant for the reaction, we need to use the concentrations of the reactants and products at equilibrium. The balanced equation for the reaction is:

Fe(NO3)3 + 3KSCN + 3HNO3 → Fe(SCN)2 + 3KNO3 + 3H2O

From the given information, we have 5.00 mL of 0.00300 M Fe(NO3)3, 4.00 mL of 0.00300 M KSCN, and 3.00 mL of 1.0 M HNO3. We can convert these volumes to moles and calculate the concentrations:

1. Fe(NO3)3: (5.00 mL / 1000) L * 0.00300 mol/L = 0.000015 mol
2. KSCN: (4.00 mL / 1000) L * 0.00300 mol/L = 0.000012 mol
3. HNO3: (3.00 mL / 1000) L * 1.0 mol/L = 0.0030 mol
4. Fe(SCN)2: 2.72 × 10-4 mol (given)

Using these concentrations, we can calculate the equilibrium constant:

K = ([Fe(SCN)2] / ([Fe(NO3)3] × [KSCN] × [HNO3]) = (2.72 × 10-4 mol) / (0.000015 mol × 0.000012 mol × 0.0030 mol) = 5.99 × 106