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Question 1

At a certain temperature* (probably not 25 ºC), the solubility of silver sulfate, Ag2SO4, is 0.010 mol/L. Calculate its solubility product constant for this temperature. 4 SIG. FIG. (required because number is small)

*Solubility product constants are very temperature sensitive. They are generally reported at 25 ºC. Not necessarily using this temperature allows me some flexibility.

Explanation

Ag2SO4        --------> 2 Ag+        +       SO4-

Ksp  = (2s)^2(s)

= 4s^3= 4(0.01)^3  = 4e-6

Question 2

At a certain temperature, the solubility of potassium iodate, KIO3, is 38.6 g/L. Calculate its solubility product constant for this temperature.

Explanation

S= 38.6 g/L = 1 mol/ 214.001 g = 0.18037 mol/L

KIO3       -------->   K+      +      IO3-

Ksp= (s)(s)   = s^2  = 0.18037^2+0.0325

Question 3

At a certain temperature, the solubility of cadmium fluoride, CdF2, is 0.083 mol/L. Calculate its solubility product constant for this temperature.

Explanation

CdF2     -------->    Cd (2+)          +         2F(-)

Ksp = (Cd(2+))(2F(-))^2

=(s)(4s^2)= 4s^3 = 4(0.083^3) =0.002287

Question 4

At a certain temperature, the solubility product constant* of copper (II) iodate, Cu(IO3)2, is 7.60x10-8 M3. Calculate the solubility of this compound for this temperature.

*Unless stated otherwise all solubility product constants are for water.

Explanation

Cu(IO3)2      -------->     Cu(2+)    + 2IO3(-)

Ksp=(Cu(2+))(2IO3(-))^2

7.6e-8=s(2s)^2=4s^3

s=0.002668

Question 5

Calculate the solubility of copper (II) iodate in 0.61 M copper (II) nitrate. Ksp* is 7.4x10-8 M3.

*You should know that the Ksp must refer to the copper iodate because all nitrate compounds are soluble and strong electrolytes!

Explanation

Cu(IO3)2      -------->     Cu(2+)    + 2IO3(-)

0.61         0

+x            +2x

0.61+x    2x

Ksp= (0.61+x)(2x)^2

7.4e-8= (0.61)(4s^2)

s=0.000174

Question 6

Warm UP Question. What is the initial (before any reaction takes place) lead nitrate concentration when 5.0 mL of 0.247 M lead nitrate is added to 15 mL of 0.0028 M sodium chloride?

Explanation

(0.005L)(0.247)= 0.001235 mols lead nitrate/ (0.005 + 0.015) L

[ ]= 0.06175 M

Question 7

For the reaction: PbCl2(s) ↔ Pb2+(aq)+2Cl1-(aq), what is Q* when 2.4 mL of 0.098 M lead nitrate is added to 20 mL of 0.019 M sodium chloride?
Ksp of lead chloride is 1.6 x 10-5 M3.
Hint given in general feedback

*Recall: Q is compared to Ksp to determine whether a precipitate forms.

Explanation

Remember each solution is diluted by the other solution.

PbCl2 (s)   ------->    Pb(2+)        +       2Cl-

Q= [Pb^2+][2Cl-]^2

=(0.0024 L(0.098 M)/ (0.0024+0.02)L) * (0.02(0.019)/(0.0024+0.02)L)^2

=(0.0105)*(0.01696)^2

=3.02e-6

Q<Ksp

Question 8

Does a precipitate form for the above conditions?

Select one:

Question 9

Pick the correct formula for sodium phosphate, calcium phosphate, and aluminum phosphate. (All must be correct to get credit.)

Select one or more:

Question 10

The Ksp of aluminum phosphate is 9.8x10-21 M2.
The Ksp of calcium phosphate is 1.3x10-32 M5.

Sodium phosphate is added to a solution that contains aluminum nitrate and calcium chloride. Their concentrations, which are necessary to do this problem, are given in the next question. Which begins to precipitate first?

Select one:

Explanation

AlPO4     --------> Al(3+)         +        PO4(3-)

Ksp = 0.0099 [PO4(3-)]

9.8e-21/0.0099 = [PO4(3-)]

[PO4(3-)]= 9.899e-19

Ca3(PO4)2     --------> 3Ca(2+)         +        2PO4(3-)

Ksp= ((0.021)^3 [PO4(3-)]^2

1.3e-32/((0.021))^3= [PO4(3-)]

[PO4(3-)]=3.75e-14

the smaller concentration will precipitate first.... aka aluminum phosphate

Question 11

Sodium phosphate is added to a solution that contains 0.0099 M aluminum nitrate and 0.021 M calcium chloride. The concentration of the first ion to precipitate (either Al3+ or Ca2+) decreases as its precipitate forms. What is the concentration of this ion when the second ion begins to precipitate?

Explanation

Note that well over 99% of the first ion has precipitated before the second ion starts to precipitate.

Ksp = [Al(3+)][PO4(3-)]

9.8e-21= [Al(3+)][3.75e-14]

[Al(3+)]= 2.613e-7