A standard solution of FeSCN^2+ is prepared by combining 6.00 mL of 0.00150 M Fe

Question

A standard solution of FeSCN^2+ is prepared by combining 6.00 mL of 0.00150 M Fe(NO_3)_3, 1.030 g KSCN, and deionized water in a 250 mL volumetric flask. The net ionic equation for the reaction is below. Fe^3+ + SCN^- Rightwards Harpoon Over Leftwards Harpoon Fe(SCN)^2+ What allows us to assume that the reaction goes essentially to completion? The reaction quotient Q is greater than K_c. The equilibrium reaction has a very high K_c. The concentration of Fe(NO_3)_3 is much higher than the concentration of KSCN. The concentration of KSCN is much higher than that of Fe(NO_3)_3. Under the conditions given above, LeChatelier's principle dictates that the reaction shifts to the left. Based on that assumption, what is the equilibrium concentration of Fe(SCN)^2>? For clarification see the example calculation on page 10-7 of your lab manual. Once checked, the calculated molarity will be copied to the appropriate space on Data Sheet page 10-11.

Explanation / Answer

The reaction is Fe+3 + SCN- ------->. Fe(SCN)+2

6x 0.0015x10-3 1.03/97 0 initial moles

9.0 x10-6/250 1.03/(97x250) 0 initial concentrations

0 1.03/(97x250)   9.0 x10-6/250 equilibrium concentrations

Thus we can assume the reaction goes to completion as the concentration of SCN- >> Fe+3

Based on this assumption, we take all the Fe+3 is completely reacted adn the product concentration is equal to the concentration of Fe+3 reacted = 3.6 x 10-8M

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