A 50.0 mL sample of 0.0152 M Na2SO4(aq)is added to 50.0 mL 0.0125 M Ca(NO3)2(aq). What percentage of the Ca2 remains unprecipitated

38.4%

Explanation:

Step 1 Obtain the concentration of Ca^2+ and SO4^2-

[Ca^2+] = 0.05 L (0.0125 mol/L) (1/0.1 L) = 6.25 * 10^-3 M

[SO4^2-] = 0.05 L (0.0152mol/L) (1/0.1L) = 7.60 * 10^-3 M

Step 2 Obtain the concentration of [SO4^2-]  left

         Ca^2+(aq)             +       SO4^2-(aq)> CaSO4(s)

initial  6.25 * 10^-3 M             7.60 * 10^-3 M

final        0                               1.35 * 10^-3 M

Step 3 Determine the amount of Ca^2+ left from the common ion problem

                      CaSO4(s) > Ca^2+(aq)             +       SO4^2-(aq)

initial                                             0                                 1.35 * 10^-3 M

equilibrium                                    x                                (1.35 * 10^-3  + x) M

But Ksp of CaSO4 = 9.1 * 10^-6

Ksp = [Ca^2+] [SO4^2-]

9.1 * 10^-6= [x] [1.35 * 10^-3  + x]

9.1 * 10^-6= 1.35 * 10^-3 x + x^2

Rearranging:

x^2 + 1.35 * 10^-3 x + 9.1 * 10^-6 = 0

solving the quadratic yields x = 2.4 * 10^-3

Step 4 Calculate the amount of Ca^2+ remaining;

% Ca^2+ remaining = 2.4 * 10^-3/ 6.25 * 10^-3 M * 100

% Ca^2+ remaining = 38.4%

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.