Solubility Product
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Transcript of Solubility Product
Solubility Product Constant
A special case of equilibrium involving dissolving.Solid Positive Ion + Negative Ion
Mg(NO3)2 Mg2+ + 2NO3-
Keq = [Mg2+ ] [NO3-]2
Because the constant is a product of solubility, we call it the solubility product constant Ksp
Solubility Product Problems
•Given Ksp, find solubility
•Given solubility, find Ksp
•Find solubility in a solution with a common ion
•Predicting precipitation
Solubility Product Constant (Ksp)
BaF2(S) Ba+2(aq) + 2F-
(aq)
Write out the equilibrium law expression…
Ksp = [Ba+2][F-]2
Solubility Generalizations
•All nitrates are soluble
•All compounds of the alkali metals are soluble (Li, Na, K, etc.)
•All compounds of the ammonium (NH4
+) are soluble
Given Ksp, Find Solubility
What is the solubility of Silver Bromide (Ksp = 5.2 x 10-13)
AgBr Ag+ + Br-
Ksp = [Ag+][Br-] = 5.2 x 10-13
Let x = the solubility
(x)(x) = 5.2 x 10-13
X2 = 5.2 x 10-13 X = 7.2 x 10-7
Another Example
What is the solubility of PbI2 (Ksp = 7.1 x 10-9)
PbI2(s) Pb+2 + 2I-
Ksp = [Pb+2][I-]2 = 7.1 x 10-9
Let x = the solubility
(x)(2x)2 = 7.1 x 10-9 (x)(4x2) = 7.1 x 10-
94x3 = 7.1 x 10-9X = 1.2x10-3M
Find Ksp Given Solubility
What is the Ksp of Boric Acid, given its solubility of 2.15 x 10-3 Moles/liter?
H3BO3 3H+ + BO3-3
Ksp = [H+]3[BO3-3]
[3(2.15x10-3)]3 [2.15x10-3]
= 5.77 x 10-10
Solubility with a Common Ion
What is the solubility of lead iodide (PbI2) in a .15M solution of KI ?
PbI2 Pb+2 + 2I-
KI K+ + I-
Ksp = [Pb+2][I-]2 = 7.1 x 10-9
Solubility with a Common Ion
What is the solubility of lead iodide (PbI2) in a .15M solution of KI ?
PbI2 Pb+2 + 2I-
KI K+ + I-Ksp = [Pb+2][I-]2
Let x = solubilityThen: [Pb+] = x [I-] = .15+2x
Ksp = [Pb+2][I-]2 = 7.1 x 10-9
Let x = solubility of PbI2 in the solution
Then, [Pb+2] = x [I-] = .15 + 2x
Ksp = (x)(.15+2x)2 = 7.1 x 10-9
Ksp = [Pb+2][I-]2 = 7.1 • 10-9
Ksp = (x)(.15+2x)2 = 7.1 • 10-9
(x)(4x2 + .6x +.152) = 7.1 • 10-9
4x3 + .6x2 + .152x = 7.1 • 10-9
4x3 + .6x2 + .0225 – 7.1 • 10-9 = 0
Ksp = [Pb+2][I-]2 = 7.1 x 10-9
Ksp = (x)(.15+2x)2 = 7.1 x 10-9
Assume .15>>2x
then, .15+2x .15
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Ksp = [Pb+2][I-]2 = 7.1 x 10-9
Ksp = (x)(.15+2x)2 = 7.1 x 10-9
Assume .15>>2x
then, .15+2x .15
Ksp = (x)(.15)2 = 7.1 x 10-9
X = 3.16 x 10-7 moles/liters
Predicting Precipitation
A student mixes 0.010 mole Ca(NO3)2 in 2 liters of 0.10M Na2CO3 solution. Will a precipitate form?
Step 1: Write out the dissolving equations
Ca(NO3)2 (s) Ca2+ (aq) + 2NO3- (aq)
Na2CO3 (s) 2Na+ (aq) + CO32- (aq)
Step 2: Determine the most likely precipitate & write out it’s equation.
Ca(NO3)2 Ca2+ + 2NO3-
Na2CO3 2Na+ + CO32-
Recall the solubility generalizations…
•All nitrates are soluble
•All compounds of the alkali metals are soluble (Li, Na, K, etc.)
•All compounds of the ammonium (NH4
+) are soluble
Step 2: Determine the most likely precipitate & write out it’s equation.
Ca(NO3)2 Ca2+ + 2NO3-
Na2CO3 2Na+ + CO32-
CaCO3 Ca2+ + CO32-
CaCO3 Ca2+ + CO32-
Ksp = [Ca2+ ][CO32-] = 4.7 x 10-9
Step 3: Determine the molar concentrations & calculate the reaction quotient (Q).
[Ca2+] = .01 mole/2 liters = .005M
[CO32-] = 0.10 M (given)
Reaction Quotient (Q)
The product of the Ksp equation using the ion concentration before any reaction interaction.
If Q > Ksp Then a precipitate will form.
Q = [Ca2+ ][CO32-]
[Ca2+] = .01mole/2 liters = .005M
[CO32-] = 0.10 M (given)
Q = (.005)(0.10) = .0005
Q > Ksp a precipitate will form.
You try one….
.015 moles of AgNO3 is mixed with 5 liters of .02M NaCl solution. What is the most likely precipitate and will it form?
You try one….
.015 moles of AgNO3 is mixed with 5 liters
of .02M NaCl solution. What is the most likely precipitate and will it form?
AgNO3 Ag+ + NO3-
NaCl Na+ + Cl-
AgCl Ag+ + Cl-
AgCl Ag+ + Cl-
Q = [Ag+][Cl-]
Q = (.015/5)(.02) = .00006
Look up the Ksp (1.0 x 10-10)
Q > Ksp So a precipitate will form!