The Solubility Constant Of Stronyium Sulfate, Srso4, Is 2.8 X 10-7. How Many Grams Of Srso4 Must Be Dissolved In Water To Produce 1 L Saturated Soluti

The solubility constant of stronyium sulfate, srso4, is 2.8 x 10-7. how many grams of srso4 must be dissolved in water to produce 1 l saturated solution?

 

Answer:

The answer is 0.1 g of Strontium Sulfate.

Step-by-step explanation:

Solubility constant k_{sp} is solved using the equation below

k_{sp}= A^+ B^-   (Eqn 1)

Where A^+ is the molar solubility of the cation and  B^- is the molar solubility of the anion. Based from our problem our solution goes like this

k_{sp}= Sr^2+ SO_4^2-\\k_{sp} = s*s\\k_{sp} = s^2\\2.8*10^{-7} = s^2\\s = 5.29*10^{-4} mol/L

Take note that s= Sr^2+ = SO_4^2- because their stoichiometric ratio is 1:1.

Next, we need to convert the molar solubility to mass solubility

Using the molar mass of SrSO_4= 183.68 g/mol, we have


s = 5.29*10^{-4} \frac{mol}{L} * \frac{183.68g}{1 mol} = 0.0972 g/L


Thus in order to  saturate 1 L of solution we need 0.0972 g of  SrSO_4 or approximately 0.1 g.

If you want to know more about solubility, you may visit the following links:

brainly.ph/question/1289331

brainly.ph/question/1113977

brainly.ph/question/1113850


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