Respuesta :

Willchemistry
first converts g/L in M :

Molar mass Silver Bromide ( AgBr) = 187.77 g/mol

M = C / molar mass

M = 13.5 x 10⁻⁵ / 187.77

= 7.189 x 10⁻⁷ M

AgBr ----- > Ag⁺    + Br ⁻

Ksp = [ Ag⁺] [ Br ⁻] = x₂

x =  7.189 x 10⁻⁷ * 7.189 x 10⁻⁷

Ksp = 5.16 x 10⁻¹³ M

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