Ok so i have been told (now bear in mind im trying to translate this from danish to english) that:
25,0 ml 2,00 mM SrCL2 is mixed at a given temperature with 15,0 mL 4,00 mM Na2SO4. It creates a deposit of strontiumsulfate. The deposit is filtered away, and the actual concentration of SR2+ in the fitered liquid is 7,5*10^(-4) M.
Calculate the value of the solubilityproduct of strontiumsulfate at the given temperature.
Calculate the solubility (in g/100 mL) of strontiumsulfate at the given temperature.
Now i am not certain if i did it right but here are my results:
5,625*10^(-7)
and
1,37895 g/100ml
25,0 ml 2,00 mM SrCL2 is mixed at a given temperature with 15,0 mL 4,00 mM Na2SO4. It creates a deposit of strontiumsulfate. The deposit is filtered away, and the actual concentration of SR2+ in the fitered liquid is 7,5*10^(-4) M.
Calculate the value of the solubilityproduct of strontiumsulfate at the given temperature.
Calculate the solubility (in g/100 mL) of strontiumsulfate at the given temperature.
Now i am not certain if i did it right but here are my results:
5,625*10^(-7)
and
1,37895 g/100ml
-
Use 3 significant figures in your answers. Your decimal place in the second part went the wrong way!
Solubility = .00075 mole/L x 184 g/mole = 0.138 g/L take 1/10 that for 100 mL (0.100L)
0.138 g/L x 0.100 = 0.0138 g/100 mL
Solubility = .00075 mole/L x 184 g/mole = 0.138 g/L take 1/10 that for 100 mL (0.100L)
0.138 g/L x 0.100 = 0.0138 g/100 mL