Calculate the molar solubility of ag2 cro4 in water. use 1.10 x 10-12 as the solubility product constant of agzcro4
7.42 x 10-7m
1.05 x 10-6m
6.50 x 105m
1.03 x 10-4m

Explanation:

1) Write the chemical equation for the solubility equilibrium:

2) Write the expression for the solubility product constant, Ksp:

3) Relate the solubility with the concentrations:

4) Substitute in the Ksp equation:

5) Solve for s:

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"