Solving Delta G from a Solubility constant

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The Solubility Constant for AgI(s) at 25 degrees Celsius is 8.3*10^-17
how do i find $$\Delta$$G(rxn) for [Ag+]=9.1*10^-9 & [I-]=9.1*10^-9

Using the equation:
ΔG=ΔGº+RTln(Q)

Let K=Our solubility constant:
0=ΔGº+RTlnK
ΔGº=-RTlnK
Now,
ΔG=-RTlnK+RTlnQ
ΔG=RTlnQ-RTlnK
ΔG=RT(ln(Q/K))

ΔG=-5.67KJ/mol

The answer in my practice exam claims to be 91.7kJ/mol.

Where did I go wrong?

Thanks!

 

[lingxiao sun]

hey, I just did the work, I got 91.73 kj/mol. I don't know why but I think u have to square the K value, try it.

 

[RandomMystery]

What did you plug in for Q and K?
T=25+273
R=8.31
K = Q/K = 1/([Ag]) = 1/((9.1*10^-9)^2) = 1/(9.1*10^-18)