A standard solution of FeSCN^2+ is prepared by combining 9.0 mL of 0.20 M Fe(NO_

Question

A standard solution of FeSCN^2+ is prepared by combining 9.0 mL of 0.20 M Fe(NO_3)3 with 1.0 mL of 0.0020 M KSCN. The equation for the reaction is as follows. Fe(NO_3)_3 + KSCN Right and Left arrow FeSCN^2+ + KNO_3 + 2NO_3^- What allows us to assume that the reaction goes essentially to completion? The reaction quotient Q is greater than K_c. Under the conditions given above, LeChatelier's principle dictates that the reaction shifts to the left. The concentration of Fe(NO_3)_3 is much higher than the concentration of KSCN. The equlibrium reaction has a very high K_c. The excess Fe^3+ prevents the formation of the neutral Fe(SCN)_3. Based on that assumption, what is the equilibrium concentration of FeSCN^2+?

Explanation / Answer

The relation between equilibrium consatnt K p and reaction quotient Q as follows:

Q = K

When Q=K, the system is at equilibrium and there is no shift to either the left or the right.

Q < K

When Q<K, there are more reactants than products. As a result, some of the reactants will become products, causing the reaction to shift to the right.

Q > K

When Q>K, there are more products than reactants. To decrease the amount of products, the reaction will shift to the left and produce more reactants. For Q>K:

Hence the correct answer is, the reaction quotient Q is greater than Kc.

First calculate the number of moles of Fe(NO3)3 and KSCN :

Number of moles of Fe(NO3)3:

9.0 mL * 0.20 moles /1000 mL

=1.8*10^-3 moles

Number of moles of KSCN:

1.0 mL * 0.0020 moles /1000 mL

=2.0*10^-6 moles

Here KSCN is a limiting agent.

Now number of moles of FeCN2+

2.0*10^-6 moles KSCN * 1 mole of FeCN2+/ 1mol KSCN

= 2.0*10^-6 moles of FeCN2+

Molarity = number of moles / total volume in L

9.0+1.0 mL = 10.0 mL = 0.010L

Molarity = number of moles / total volume in L

= 2.0*10^-6 moles of FeCN2+ / 0.010L

= 2.0*10^-4 M

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