a) How many grams of sodium sulfate can be added to 500 mL of 3.100E-3 M barium

Question

a)

How many grams of sodium sulfate can be added to 500 mL of 3.100E-3M barium chloride before a precipitate forms? The Ksp for BaSO4 is 1.1E-10.

= _____g

b)

Solid KSCN was added to a 2.66M Co2+ solution so that it was also initially 2.66M SCN-. These ions then reacted to give the complex ion Co(SCN)+, whose formation constant was 1.0E+2. What is the concentration of Co2+(aq) at equilibrium?

Be sure to check any simplifying assumption you make.

= _____M

c)

Calculate the molar solubility of silver bromide, AgBr, in 6.8 M NH3.

Ksp(AgBr) = 7.7E-13
Kf(Ag(NH3)2)+ = 1.7E7

= _____mol/L

Explanation / Answer

How many grams of sodium sulfate can be added to 500 mL of 3.100 X 10^-3M barium chloride before a precipitate forms? The Ksp for BaSO4 is 1.1E-10.

solution:

BaSO4 ---> Ba+2 + SO4-2

Ksp = [Ba+2] [SO4-2]

1.1 X 10^-10 = [Ba+2] [SO4-2]

Molarity of Ba+2 = 3.1 X10^-3

Ksp = 1.1 X 10^-10 = 3.1 X10^-3 [SO4-2] =

[SO4-2] =1.1 X 10^-10 / 3.1 X10^-3 = 3.54 X 10^-8 Molar

Since the formula Na2SO4 has one SO4 2- in it, then [Na2SO4] = 3.54 X 10^-8 M

If we add solid Na2SO4 to the BaCl2 solution, its volume will be 500 mL (0.5 L).

moles Na2SO4 = Moalrity Na2SO4 x L Na2SO4 = (3.54 X 10^-8)(0.5) = 1.77 x 10^-8 moles Na2SO4

Molecular weight of Na2SO4 = 142 g

So mass of Na2SO4 = moles x molecular wt = 1.77 x 10^-8 X 142 = 2.51x 10^-6 g Na2SO4

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