General expression of solubility product -
M_{x}X_{y}\rightleftharpoons xM^{p+}(aq)+yX^{2-}(aq)
(x.p^{+}=y.\bar{q})
- wherein
Its solubility product is
K_{sp}=[M^{p+}]^{x}[X^{q-}]^{y}
AgBr \rightleftharpoons Ag ^++ Br^-
K_{sp} \; \; of \; \; AgBr = [ Ag^+][Br^-]
5 \times 10 ^{-13 }= 0.05 [Br ^- ]
[Br ^-] = \frac{5 \times 10^{13}}{0.05}= 1 \times 10^{-11}M
Moles of KBr = = 1 \times 1 0^{-11}
Weight of KBr = = 1.2 \times 1 0^{-9}g