Answer
a) $K=14.5$
b) $1.68\ g$
Work Step by Step
a) $AgBr(s)\leftrightarrow Ag^+(aq)+Br^-(aq)\ Ksp = 5\times10^{-13}$
$Ag^+(aq)+2S_2O_3^{2-}(aq)\leftrightarrow [Ag(S_2O_3)_2]^{3-}(aq) \ K_f=2.9\times10^{13}$
$AgBr(s)+2S_2O_3^{2-}(aq)\leftrightarrow[Ag(S_2O_3)_2]^{3-}(aq)+Br^-(aq)$
$K=Kf\times Ksp=14.5$
b) Since K > 1, assume that all thiosulfate is consumed.
$1\ g\div 187.77\ g/mol=0.0053\ mol$
By stoichiometry: $0.011\ mol$
Mass: $0.011\ mol\times 158.11\ g/mol=1.68\ g$
