Answer
A)$K_{sp}$=[$Sr^{2+}$]$\times$[$CrO_{4}^{2-}$]= 3,48$\times$$10^{-5}$
B)$K_{sp}$=[$Bi^{3+}$]$\times$$[I^{-}]^{3}$=7,95$\times$$10^{-19}$
C)$K_{sp}$=[$Fe^{2+}$]$\times$$[OH^{-}]^{2}$=7,26$\times$$10^{-15}$
D)$K_{sp}$=[$Sn^{2+}$]$\times$$[I^{-}]^{2}$=1$\times$$10^{-4}$
Work Step by Step
A) The molar mass of $SrCr$$O_{4}$ is equal to:
molar mass= 203,62 g/mol
solubility=1,2 mg/mL=1,2 g/L=1,2 g/L$\div$ 203,62 g/mol= 5,9$\times$$10^{-3}$ mol/L= [$Sr^{2+}$]= [$CrO_{4}^{2-}$]
$K_{sp}$=[$Sr^{2+}$]$\times$[$CrO_{4}^{2-}$]= 3,48$\times$$10^{-5}$
B) molar mass of $Bi$$I_{3}$
molar mass=589,7 g/mol
solubility=7,7$\times$$10^{-3}$ g/L$\div$589,7 g/mol=1,31$\times$$\times$$10^{-5}$ mol/L=[$Bi^{3+}$]
[$I^{-}$]=3$\times$1,31$\times$$\times$$10^{-5}$ mol/L=3,93$\times$$10^{-5}$mol/L
$K_{sp}$=[$Bi^{3+}$]$\times$$[I^{-}]^{3}$=7,95$\times$$10^{-19}$
C)molar mass of $Fe$$(OH)_{2}$= 89,845 g/mol
solubility=1,1$\times$$10^{-3}$ g/L$\div$89,845 g/mol=1,22$\times$$\times$$10^{-5}$ mol/L=[$Fe^{2+}$]
[$OH^{-}$]=2$\times$1,22$\times$$\times$$10^{-5}$ mol/L=2,44$\times$$10^{-5}$mol/L
$K_{sp}$=[$Fe^{2+}$]$\times$$[OH^{-}]^{2}$=7,26$\times$$10^{-15}$
D)molar mass of $Sn$$I_{2}$=372,51 g/mol
solubility=10,9 g/L$\div$372,51 g/mol=0,0293 mol/L=[$Sn^{2+}$]
[$I^{-}$]=2$\times$0,0293 mol/L=0,0586 mol/L
$K_{sp}$=[$Sn^{2+}$]$\times$$[I^{-}]^{2}$=1$\times$$10^{-4}$
