Finding the value of Ksp.

Question

I have 50 of these problems and am asking for help on three to use as examples. They are as follows:

 

1) The solubility of MgF2 is measured and found to be 7.47×10-2 g/L. Use this information to calculate a Ksp value for magnesium fluoride.

2) The solubility of Ag2S is measured and found to be 7.09×10-15 g/L. Use this information to calculate a Ksp value for silver sulfide.

3)The solubility of CaSO3 is measured and found to be 1.41×10-2 g/L. Use this information to calculate a Ksp value for calcium sulfite

Answer 1

Note: K_sp is expressed in terms of molar concentrations, therefore convert the solubility to mol/L.

 

mol/L (MgF₂) = (7.47x10^-2 g/L)/(62.3018 g/mol)

                    = 0.00119 mol/L

 

Write the chemical equation for solubility of MgF_2.

 

MgF₂ (s)   →  Mg²⁺(aq)  +  2F^-(aq)

 

Now, write the solubility product expression.

 

K_sp  =  [Mg²⁺][F^-]²

 

1 mol MgF₂ dissolve to give 1 mol Mg²⁺ and 2 mol F^-

 

Therefore, [Mg²⁺]  =  0.00119 M and [F^-]  =  2x0.00119 M

 

Therefore K_sp(MgF₂)  = (0.00119)(2*0.00119)²

                               =7x10^-9 M^-3

Answer 2

(I chose to answer the second problem)


2) 7.09*10^-15g/L *(1mole/171g) = 2.858*10^-17mol/L

Ksp = [Cation⁺][Anion⁺]/[Salt]
Let S = x; Let Ag = 2x
Ksp = [x][2x]²
ksp = 4x³
ksp = 4[2.858*10^-17]³

ksp = 9.34*10^-50