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%

girl imma just put my comment here because i need a couple point okurrttt

substituting a hydrogen atom with a halogen in a hydrocarbon:

- the single bond remains single. a is wrong.

- the bond with hydrogen is as strong as with a halogen. c is wrong

- a hydrocarbon is saturated if there is no double bond. we do not know that so d is wrong.

- carbon will always have four valence electrons. e is wrong.

by elimination, the correct answer is b. the boiling point of the new compound increases.