A standard solution of FeSCN^2+ is prepared by combining 3.00 mL of 0.00150 M Fe

Question

A standard solution of FeSCN^2+ is prepared by combining 3.00 mL of 0.00150 M Fe(NO_3)_3, 0.941 g KSCN and deionized water in a 100 mL volumetric flask. The net ionic equation for the reaction is below. Fe^3+ + SCN^Fe(SCN)^2+. What allows us to assume that the reaction goes essentially to completion? 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. The equilibrium reaction has a very high K_c. The reaction quotient Q is greater than K_c. The concentration of Fe(NO_3)_3 is much higher than the concentration of KSCN. 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. On checked, the calculated molarity will be copied to the appropriate space on Data Sheet page 10-11. Number M

Explanation / Answer

Q1.

From all data,

a) not really, it can't ensure total reaction

b) this is not true, even though it shift, we do not know up to howm uch extent

c) This is true, sice Kc will show us the relationship between productS:reactants

d) not really, since there could be pretty low Kc, a higher Q just ensures slightly shift to products, not high shigfts

e) Not really, both are important

Q2.

find equilibrium concentratino of Fe(SCN)2+

First, calculate concentrations

[Fe+3] = M*V1/(V1+V2) = 3*0.0015 / ((3+100)/1000) =0.0436 M

[SCN-] = mass / MW / (V1+V2) = 0.941/(97.181) / (103/1000) = 0.0940 M

ratio is 1:2 so

Fe+3 is limiting

now, assume

1 mol of Fe+3 --> 1 mol of Fe(SCN)+2

so

0.0436 M of Fe+3 --> 0.0436 M of Fe(SCN)+2

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