AP Chemistry 7.11 Introduction to Solubility Equilibria Study Notes - New Syllabus Effective fall 2024

AP Chemistry 7.11 Introduction to Solubility Equilibria Study Notes- New syllabus

AP Chemistry 7.11 Introduction to Solubility Equilibria Study Notes – AP Chemistry – per latest AP Chemistry Syllabus.

LEARNING OBJECTIVE

Calculate the solubility of a salt based on the value of K_sp for the salt.

Key Concepts:

Solubility-Product Constant
Common-Ion Effect

AP Chemistry-Concise Summary Notes- All Topics

7.11.A.1 Reversible Dissolution and Ksp:

1. Dissolution and Reversible Equilibrium:

When a salt is dissolved in water, it disintegrates into ions. For example, NaCl dissolves according to the equation: NaCl (s)⇌Na+(aq)+Cl−(aq)

The salt dissolves at first, but when ion concentration increases, a portion of ions recombine and create solid salt, establishing dynamic equilibrium. The rate of dissolution and the rate of precipitation are the same at equilibrium and so ion concentration remains constant.

Factors influencing this equilibrium include temperature (which tend to increase solubility), concentration of ions (common ion effect), and the nature of the solvent. Solubility of a salt is defined in terms of its solubility product constant (Ksp), and the greater the Ksp value, the greater the solubility. The reaction is reversible, i.e., the addition of excess solid salt or removal of ions can reverse the dissolution-precipitation equilibrium.

2. Solubility Product (Ksp):

– Definition: Solubility product constant (Ksp) is an equilibrium constant that explains the solubility of a salt in water. It is the product of the concentration of the ions formed when a salt is dissolved, each raised to the power equal to its coefficient in the balanced dissolution equation.

– Calculation:

For a salt like , which dissociates as: AB2(s)⇌A+(aq)+2B−(aq)

The Ksp expression would be: Ksp=[A+][B−]2

where and are the molar concentrations of the ions at equilibrium.

– Relation to Ion Concentrations: Ksp value provides the maximum product of ion concentrations in a saturated solution. Higher Ksp indicates higher solubility of the salt, while lower Ksp indicates lower solubility. If the product of ion concentration is greater than Ksp, then precipitation takes place.

3. Factors Affecting Solubility:

i. Temperature:
– Solubility of solids generally increases with temperature.
– For gases, solubility goes down with an increase in temperature.

ii. Common-Ion Effect:
– Addition of an already dissolved ion reduces the solubility of a salt by shifting equilibrium towards the solid.

iii. pH:
– Solubility of salts of weak acid/base can increase or decrease with pH. Calcium carbonate dissolves greater in acidic solution as it acts with H⁺ ions.

7.11.A.2 Calculating Solubility from Ksp and Comparing Solubilities:

1. Solubility and Ksp Relationship:

– Ksp (Solubility Product Constant) is the balance of a solid salt and dissolved ions in solution. For a salt ( AB ): AB(s)⇌A+(aq)+B−(aq),

Ksp=[A+][B−]
– Calculating Solubility: To obtain molar solubility (S) from Ksp:
1. Write dissociation equation.
2. Assume salt solubility ( S ).
3. Substitute into the Ksp equation and solve for S.

– Sample Problem: For ( MgF2 ):

If is the solubility:
$K_{s p} = (S) (2 S)^{2} = 4 S^{3} Solve for S.$

Ksp helps determine how much salt will dissolve and calculate its solubility at equilibrium.

Solution:

Let’s walk through a step-by-step example of calculating the solubility of a salt from the Ksp value.

1. Write the dissociation equation:
MgF2(s)⇌Mg2+(aq)+2F−(aq)

2. Write Ksp:
Ksp=[Mg2+][F−]2

Let ( S ) be the molar solubility of ( MgF2 ). At equilibrium:
– ([Mg^{2+}] = S)
– ([F^-] = 2S)

3. Substitute these into the Ksp equation:

Let the molar solubility of be . At equilibrium: [F−]=2S

4. Solve for S:

If the Ksp for is
$6.4 \times 10^{- 9, then:}$ $6.4 \times 10^{-9} = 4S^3$ $S^3 = \frac{6.4 \times 10^{-9}}{4} = 1.6 \times 10^{-9}$ $S = \sqrt[3]{1.6 \times 10^{- 9}} = 1.17 \times 10^{- 3} M$

So, the molar solubility of is .

7.11.A.3 Solubility Rules and ksp

$Relationship:$

1. Solubility Rules and

i. Solubility Rules:
– Soluble: Alkali metals, nitrates, ammonium compounds, and most halides except for (except with

$\text{Ag}^+, \text{Pb}^{2+}$

etc.).
– Insoluble: Most carbonates, phosphates, and sulfides (except alkali metals and ammonium).

ii. (Ksp) (Solubility Product):
– (Ksp) measures sparingly soluble compounds’ solubility. The larger (Ksp), the higher the solubility.
– Example: For example: For ,

$K_{sp} = [\text{A}^+][\text{B}^-]^2$

iii. Interpretation:
– Larger ( Ksp ): More soluble.
– Smaller (Ksp): Less soluble.

Conclusion:
A larger (Ksp) corresponds to a larger solubility, and solubility rules predict whether a compound will or will not dissolve.

2. Solubility Calculation:

Write Dissolution Equation: Example:

${BaSO}_{4} (s) ⇌ {Ba}^{2 +} (a q) + {SO}_{4}^{2 -} (a q)$

Set Up

$K_{sp}$

Expression:

$K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}]$

Assume Solubility (s):

$K_{s p} = s^{2}$

Solve for Solubility:

$s = \sqrt{K_{s p}}$

Apply Solubility Rules:
Low Ksp = insoluble; high Ksp = soluble.

3. Common Ion Effect:

The common ion effect decreases the solubility of a salt when a common ion is added. The greater ion concentration shifts the equilibrium to the solid phase, reducing dissociation.

Example:
For , adding (which provides

$\text{Cl}^-$

) reduces the solubility of by shifting the equilibrium to the left.

Conclusion:
Adding a common ion lowers the solubility of a salt.

7.11.A.4 Calculating Ksp from Molar Solubility:

1. Molar Solubility:

– Definition: Molar solubility is the number of moles of a solute that can dissolve in one liter of solution to form a saturated solution at a specified temperature.

– Calculation:
1. Write the dissociation equation for the solute.
2. Set up an ICE table (Initial, Change, Equilibrium) for ion concentrations.
3. Use the ( Ksp ) expression to state the solubility of the compound in terms of its ions at equilibrium.
4. Plug in the known ( Ksp ) to solve for molar solubility (S).

2. Dissolution Equilibrium:

– Dissociation: When an ionic compound is dissolved in water, it separates into its constituent ions. For example, for salt ( AB ) dissolving:

AB(s)⇌A+(aq)+B−(aq)

– ( Ksp ) Expression: The solubility product constant, ( Ksp ), is the equilibrium constant for the reaction of dissolution. It is expressed as the product of the ion concentrations, to the power of their coefficients in the equation for dissociation: Ksp=[A+][B−]

– Application: The solubility of the compound in a saturated solution can be determined using the value of ( Ksp ). The larger the ( Ksp ), the more soluble the substance.

3. Using ICE Tables: Set up Dissociation Equation: For AB :

$AB (s) \rightleftharpoons A^+ (aq) + B^- (aq)$

ICE Table:

Species	Initial (M)	Change (M)	Equilibrium (M)
$AB (s)$	–	+s	–
$A^+ (aq)$	0	+s	s
$B^- (aq)$	0	+s	s

ksp Expression:

$K_{sp} = [A^+][B^-] = s^2$

Solve for Ksp: Substitute the known solubility s to calculate ksp. For s =0.1M

$K s p = (0.1)square = 0.01$

$K_{sp} = (0.1)^2 = 0.01$

4. Stoichiometry and

– For salt ( AB2 ), dissociation is:

$AB_2 (s) \rightleftharpoons A^+ (aq) + 2B^- (aq)$

– The stoichiometric ratios (1 for A^+ , 2 for B^- ) affect the concentrations of the ions at equilibrium.

2. ICE Table:

Species	Initial (M)	Change (M)	Equilibrium (M)
	–	+s	–
$A^+ (aq)$	0	+s	s
$B^- (aq)$	0	+2s	2s

iii. ( Ksp ) Expression:
– The ( Ksp) expression accounts for the stoichiometric coefficients:

$K_{sp} = [A^+][B^-]^2 = s \times (2s)^2 = 4s^3$

iv. Influence of Stoichiometric Coefficients:
– The coefficients (e.g., 2 for ( B^- ) influence the concentration factors and the power to which they are raised in the ( Ksp) formula.

AP Chemistry 7.11 Introduction to Solubility Equilibria Study Notes

AP Chemistry 7.11 Introduction to Solubility Equilibria Study Notes - New Syllabus Effective fall 2024

AP Chemistry 7.11 Introduction to Solubility Equilibria Study Notes- New syllabus

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