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# Solubility Product - PowerPoint PPT Presentation

Solubility Product. Solubility Product Constant. A special case of equilibrium involving dissolving. Solid  Positive Ion + Negative Ion. Mg(NO 3 ) 2  Mg 2+ + 2NO 3 -. Keq = [ Mg 2+ ] [NO 3 - ] 2.

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Product

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

• Given Ksp, find solubility

• Given solubility, find Ksp

• Find solubility in a solution with a common ion

• Predicting precipitation

BaF2(S) Ba+2(aq) + 2F-(aq)

Write out the equilibrium law expression…

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

• All nitrates are soluble

• All compounds of the alkali metals are soluble (Li, Na, K, etc.)

• All compounds of the ammonium (NH4+) are soluble

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

X = 7.2 x 10-7

X2 = 5.2 x 10-13

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 10-9

4x3 = 7.1 x 10-9

X = 1.2x10-3M

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

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

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 = solubility

Then: [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

A student mixes 0.010 mole Ca(NO3)2 in 2 liters of 0.10M Na2CO3solution. 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)

Ca(NO3)2  Ca2+ + 2NO3-

Na2CO3  2Na+ + CO32-

Recall the solubility generalizations… it’s equation.

• All nitrates are soluble

• All compounds of the alkali metals are soluble (Li, Na, K, etc.)

• All compounds of the ammonium (NH4+) are soluble

Ca(NO3)2  Ca2+ + 2NO3-

Na2CO3  2Na+ + CO32-

CaCO3 Ca2+ + CO32-

CaCO it’s equation.3 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) it’s equation.

The product of the Ksp equation using the ion concentration before any reaction interaction.

If Q > Ksp Then a precipitate will form.

Q = [ it’s equation.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…. it’s equation.

.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…. it’s equation.

.015 moles of AgNO3 is mixed with 5 liters of .02M NaCl solution. What is the most likelyprecipitate and will it form?

AgNO3 Ag+ + NO3-

NaCl  Na+ + Cl-

AgCl  Ag+ + Cl-

AgCl it’s equation. 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!