Question #188595

 The Kc for the reaction of tin(II) ions with iron(III) ions

Sn2+(aq)     +     2Fe3+(aq)            Sn4+(aq)     +     2Fe2+(aq)

is 1.0  X  1010 at room temperature of 298K. In an experiment conducted by Jessica under the watchful eye of Christine, at 25oC, in which solutions of SnCl2 and Fe(NO3)3 were mixed, the equilibrium concentration of tin ions were found to be 

Sn2+(aq)  =  0.050mol dm-3   and Sn4+(aq) =   0.040mol dm-3  

Calculate the concentrations of iron(III) and iron(II) ions at equilibrium.


Expert's answer

Concentration of iron(III)=4.92M

Concentration of Iron (II) =3.74M


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