A student chemist used 2.0 times 10^-3 M Fe(NO_3)_3 in 0.10 M HNO_3 solution ins

Question

A student chemist used 2.0 times 10^-3 M Fe(NO_3)_3 in 0.10 M HNO_3 solution instead of 2.0 times 10 M Fe(NO_3)_3 in 0.10 M HNO_3 solution in preparing the standard solutions. Explain how would this affect the experimental results. How would his/her value of K_eq compare to K_eq obtained from an experiment conducted correctly? If you were able to start with a sample of 0.2 M [FeNCS^2+] in 0.1 M HNO_3 solution (imagine that FeNCS^2+ and HNO_3 are the only solutes in the watery describe what will be present when you let this solution come to equilibrium.

Explanation / Answer

7) The equation will be, Fe+3 + HSCN <-------------> FeSCN2++ H+
Keq = [FeSCN2+] [ H+] / [Fe+3] [HSCN]
From the above expression it is clear that Keq   [HNO3] i.e [H+] and inversely proportional
   to [Fe(NO3)3] i.e [Fe+3] . Therefore we can say that Keq decreases as [Fe3+] increases.
In the given question the correct concentration of H+ remains the same but the concentration of Fe+3
changes . Since the student chemist considered value of [Fe+3] less than the correct one, an increase in
Keq can be observed.
Keq, wrong > Keq, actual   

8) After equilibrium you will end up having Fe+3 and HSCN

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