I need to calculate the solability of CuI in a 1,0*10^⁻4 Mol/l K -solution, knowing that Ksp(CuI)=1,1*10^-12 and K_s2 of the reaction:
CuI(s)+I
CuI2−=7,9⋅10^⁻4
I have already figured out that the solid CuI will partially disolve by complex forming, which will shift the precipitation reaction to the right.
I learned to solve this with a lineair system. But I only got a first equation:
CuI2−CuI⋅104= 7,9⋅10^⁻4
According to my textbook the final answer is: 9,0*10^⁻8 Mol/l
Can someone please help me out