Common Ion Effect

Common Ion Effect (explained by Le Chatelier)

· When a salt with an anion of a weak acid is added to that weak acid.

1.0 HF reacts with 1.0 M NaF

· More F- ions will shift equilibrium to left.

· Reverses the dissociation of acid and lowers percent dissociation.

· Same thing applies to salts with cations of a weak base added to that weak base.

Calculations are the same as last chapter!

· Initial [A-] is not zero because the solution also contains the salt of A-).

Example

Calculate [H+] and % dissociation of HF in a solution containing 1.0 M HF (Ka =7.2 x 10-4) and 1.0 M NaF.

· Compare with results of just [HF] = 1.0 where [H+] = 2.7 x 10-2 or 2.7%

Buffer Solutions

· One that resists change in the pH when a small amount of strong acid or strong base is added to it.

Usually made from a weak acid and one of its salts. Or can be made from weak base and one of its salts

Examples: HC2H3O2 and NaC2H3O2 OR NH3 and NH4Cl

· Major species Na+, C2H3O2-1 , HC2H3O2 and H2O

C2H3O2-1 : Absorbs excess acid: C2H3O2-1 + H3O+ ® HC2H3O2 +H2O

HC2H3O2 : Absorbs excess base: HC2H3O2 + OH- ® C2H3O2- + H2O

Þ “Mopping Up” ® pH remain unchanged

Example:

Calculate the pH of a Solution that is .50M HC2H3O2 and .25M NaC2H3O2

(Ka= 1.8 x 10-5).

Adding a Strong Acid à decrease pH or Strong base à increase pH

1. Assume the reaction goes to completion(all acid or base is used up)

2. Do stoichiometry first- use neutralization reaction

3. Strong base will grab H+ from weak acid, reducing initial concentration. [HA]o

4. Strong acid will add H+ to anion of the salt reducing initial concentration. [A-]o

5. Do equilibrium problem – use Ka or Kb and new concentrations

Example:

E HA + OH- ® A- + H2O } rxn is complete

M weak acid strong base

E A- + H+ ® HA } reaction is complete

R weak base strong acid

Calculate the pH that occurs when .010 mol NaOH is added to 1.0L of .50M HC2H3O2 / .50M NaC2H3O2 buffer. (Ka [HC2H3O2 ]= 1.8 x 10-5)

Another method: Henderson Hasselbach

· General equation: Ka = [H+][A-] so [H+] = Ka [HA]

[HA] [A-]

· [H+] depends on ratio Ka and [HA] if you take the (–) log of both sides

[A-]

if working with an acid if working with a base

· pH = pKa + log [A-] / [HA] pOH = pKb + log [HB+] / [B]

pH = pKa + log (base / acid) pOH = pKb + log (acid / base)

Example:

Calculate the pH of a solution containing .75 M lactic acid , HC3H5O3 (Ka = 1.4 x 10-4) and .25 M sodium lactate, Na C3H5O3

Another example:

A buffered solution contains .25 M NH3(Kb = 1.8 x 10-5) and .40 M NH4Cl. Calculate the pH.

Buffer Capacity

· the amount of acid or base the buffer can neutralize before pH changes to an appreciable amount

· the pH of a buffered solution is determined by the ratio [A-] / [HA] ; as long as ratio is large compared to amount of OH- added, the ratio will not change much

· as long as this ratio does not change much, the pH does not change much

· the more concentrated these two are, the more H+ and OH- the solution will be able to absorb

· large concentrations- big buffer capacity

Example:

Calculate the change in pH that occurs when .010 mol HCl is added to 1.0 L of each of the following: (Ka= 1.8 x 10-5)

a) 5.00 M HC2H3O2 and 5.00 M NaC2H3O2

b) .050 M HC2H3O2 and .050 M NaC2H3O2

· Best Buffers have a ratio of [A-] / [HA] = 1

· This is true when [A-] = [HA]

· This makes the pH = pKa since log 1 = 0

à pKa should be as close to desired pH as possible

Example: pH = pKa + log [A-] / [HA]

pH of 4.0 is wanted so 4.00 = pKa and thus Ka = 1.0 x 10-4

Solubility Equilibria

· Dissolving a solute into a solvent will eventually cause saturation

· Solid ↔ Dissolved solid

Equilibrium ( solids precipitate as fast as it dissolves)

Ionic Compound

Ma NMb (s) à aM+ (aq) + bNM-(aq)

CaCl2à Ca+2(aq) + 2 Cl-(aq)

Ksp = [Ca+2]1 [Cl-]2

· Ksp = Solubility product (equilibrium constant) - only changes with temperature change

How to calculate Ksp?

Example:

Calculate Ksp of copper(I) bromide, which has a measured solubility of 2.0 x 10-4 M @25C.

Calculate Ksp for bismuth sulfide (Bi2S3), which has a solubility of 1.0 x 10-154 M @25C.

· WATCH OUT! Solubility is not the same as the solubility product constant. It is the equilibrium position for how much can dissolve

How to calculate solubility?

Example:

Ksp for copper iodate Cu(IO3)2 is 1.4 x 10-7 @ 25 C. Calculate its solubility?

· Ksp will only allow you to compare the solubility of solids that fall apart into the same number of ions (NaCl, KF…)

· The larger the Ksp, the more soluble!

· If they fall into different number of ions- must do the math!

Example: Put in order from greatest to least solubility!

AgI Ksp = 1.5 x 10-16

CuI Ksp = 5.0 x 10-12

CaSO4 Ksp = 6.1 x 10-5

Common Ion Effect

· Dissolving a solid into a solution will either the cation or anion already present will cause less to dissolve – Le Chatelier’s Principle

Example

Calculate the solubility of solid CaF2 (Ksp = 4.0 x 10-11) in a .025 M NaF solution.

Precipitation Predictions

· Use ion product, Q = [M+]a [NM-]b à use [initial]

· If Q> Ksp precipitation

· If Q < Ksp no precipitation

· If Q = Ksp at equilibrium

Example

A solution is prepared by adding 750.0 ml of 4.00 x 10-3 M Ce (NO3)3 to 300.0 ml of 2.00 x 10-2 M KIO3. Will Ce (IO3)3 with a Ksp = 1.9 x 10-10 precipitate from solution?

· If you want to calculate equilibrium concentrations in solution after precipitation occurs

1) must determine Q to see if its takes place

2) reverse dissociation equation – do stoichiometry- and assume it goes to completion

3) Adjust back to equilibrium

Example:

A solution is prepared by mixing 150.0 ml of 1.00 x 10-2 M Mg(NO3)2 and 250.00 ml of 1.00 x 10-1 M NaF. Calculate the concentration of Mg +2 and F -1 at equilibrium with solid MgF2 (Ksp = 6.4 x 10-9)

Titrations

· Technique used to determine the amount of acid or base in solution

· Add known concentration of titrant( from buret) until substance being tested (anylate) is consumed. (Equivalence point)

· Equivalence point is where stoichiometrically equivalent amounts have been added

· Equivalence point is often determined by color change of an indicator (end point)

· End Point (color change) is different than equivalence point( equal amounts have been added)

· Titration curve- graph of pH vs ml added(titrant)

Important for the following reasons:

1. shape determines equivalence point

2. used to select suitable indicators

3. determine Ka or Kb of weak acid/base being titrated

· PH meters monitor the acid base reaction.

· To make calculations easier, usually work with millimoles 1 mmol = 1/1000 th of a mole

Molarity = mol / L = mmol/ml

Strong Acid – Strong Base Titrations

· A strong acid is titrated with a strong base

· Net ionic: H+ + OH- à H2O

· Curve can be thought of as 4 regions:

1) Before any base has been added-pH depends on only strong acid

2) No buffer region- as base is added, pH slowly increases initially and

then rapidly near equivalence point

3) Endpoint (equivalence point) is 7.0 always- equal number of moles

Of acid and base have reacted, leaving a neutral salt

4) After equivalence point- dependent on left over strong base

Case study: Titration of 100.0 ml of .100M HCl with a .100M NaOH.

·No NaOH added

· 20.0 ml of NaOH added

Because vol. has been added in titration process:

Unreacted HCl = Molarity [H+]

total volume

· 50.0 ml of NaOH added

· 100.0 ml of NaOH

Þ Equivalence point has been reached pH = 7 ; only a neutral salt remains-no hydrolysis reaction occurs

· 110.0 ml of NaOH added

SEE TITRATION CURVE #1.

Weak Acid- Strong Base Titration

· A weak acid is titrated with a strong base

· Curve can be thought of as 4 regions:

1. Before any base is added ; pH depends on weak acid

2. After base is added- before equivalence pt ;

pH depends on weak acid/salt (buffer) – 2parts

(stoich part and equilibrium part)

3. At Equivalence point- equal amounts of acid and base

have been added leaving a salt whose conjugate

undergoes hydrolysis;

pH depends on anion.

4. At Equivalence point-excess strong base ;

pH depends on base.

Case study: Titration of 50.0 ml of .10 M HC2H3O2 ( Ka=1.8 x 10-5) with .10 M NaOH.

· No NaOH added (Calculate pH of weak acid.)

· 10 ml of NaOH added

Hint: Buffer mixture of C2H3O2-1 [conjugate base] and [weak acid.]

Stoich HC2H302 Na+ OH- H2O

OH- + HC2H3O2 à C2H3O2-1 + H2O

Equilibrium HC2H302 Na+ C2H3O2-1 H2O

HC2H3O2  H+ + C2H3O2-1

· 50 ml of NaOH added

stoich: HC2H302 Na+ OH- H2O

OH- + HC2H3O2 à C2H3O2-1 + H2O

Equilibrium Na+ C2H3O2-1 H2O

C2H3O2-1 + H2O à OH- + HC2H3O2

Þ Equivalence point has been reached (pH will always be >7)

· 60 ml of NaOH added

OH- + HC2H3O2 à C2H3O2-1 + H2O

SEE TITRATION CURVE #2

How are the 2 curves different?

· The weal acid solution has a higher initial pH than a strong acid

· The rapid-rise portion of the curve is smaller for a weak acid than a strong acid

· PH at equivalence point is greater than 7 for weak acid and exactly 7 for strong acid

Weak Base -Strong Acid Titration

· Similar to weak acid strong bases except inverted

· A weak base is titrated by a strong acid

Weak Base -Weak Acid Titration

· Change in pH is too gradual near equivalence point for indicators to be used

Indicators

· Dyes that are weak acids

· Used to identify equivalence point

· Changes color at the endpoint of a titration

· Must choose an indicator with a Ka near that of an acid being titrated and one that changes color over the pH range of the sharp vertical step on each graph

· Strong Acid-strong Base – most indicators

· Weak Acid/ strong base – phenolphthalein

· Strong Acid/weak base – methyl orange or congo red