Pada suhu tertentu 0,350 gram BaF2 Mr=175 larut dalam air membentuk 1 L larutan jenuh.hasil kali kelarutan BaF2 pada suhu tersebut adlah.....
fatihahalhusna
![\displaystyle S=\frac{a}{Mr\times V}=\frac{0,350}{175\times 1}=2\times10^{-3}\ mol/L\\\\K_{sp}\ BaF_2=4S^3=4[2\times10^{-3}]^3=\boxed{\boxed{\bold{3,2\times10^{-8}}}}](https://tex.z-dn.net/?f=%5Cdisplaystyle+S%3D%5Cfrac%7Ba%7D%7BMr%5Ctimes+V%7D%3D%5Cfrac%7B0%2C350%7D%7B175%5Ctimes+1%7D%3D2%5Ctimes10%5E%7B-3%7D%5C+mol%2FL%5C%5C%5C%5CK_%7Bsp%7D%5C+BaF_2%3D4S%5E3%3D4%5B2%5Ctimes10%5E%7B-3%7D%5D%5E3%3D%5Cboxed%7B%5Cboxed%7B%5Cbold%7B3%2C2%5Ctimes10%5E%7B-8%7D%7D%7D%7D "\displaystyle S=\frac{a}{Mr\times V}=\frac{0,350}{175\times 1}=2\times10^{-3}\ mol/L\\\\K_{sp}\ BaF_2=4S^3=4[2\times10^{-3}]^3=\boxed{\boxed{\bold{3,2\times10^{-8}}}}")
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