0 \times 10^{-10}$
2) For the first equilibrium reaction: $K_1 = \frac{[AgBr][Cl^-]}{[AgCl][Br^-]} = 200$
3) For the second equilibrium reaction: $K_2 = \frac{[Ag_2S][Br^-]^2}{[AgBr]^2[S^{2-}]} = 1.6 \times 10^{24}$
Now, we want to find the solubility product of
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