Additional Aspects of Aqueous Equilibria

Part 1

Ch. 17 in Textbook

Splash.

I) The Common-Ion Effect A) What is it?

Given: a weak acid solution Added: a salt of the acid Because the salt will ionize completely (strong electrolyte),

you are adding the conjugate base

The presence of this common ion is a stress to the equilibrium of the weak acid

The equilibrium will then shift towards the molecular form of the weak acid and thus _____________ its solubility

The pH thus ______________

Man, now I’m all wet.

HW: 17.2

B) Acid ExampleEx 1) What is the pH of a solution

consisting of 0.30 M acetic acid and 0.30 M sodium acetate? (Ka = 1.8 x 10-5)

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“The Sounds of Nature: Crashing

Waves.”

C) Base ExampleEx 2) 5.60 g of solid NH4Cl are added to

a 2.0 M solution of NH3 with a total volume of 100.0 mL. What is the pH if the Kb value for ammonia is 1.8 x 10-5?

These seagulls are driving me nuts, I tell ya!

HW: 17.4 (a) – (c), 17.6 (use Appendix D), 17.8

II) Buffered SolutionsA) What are they?

Can resist changes in pH (like blood)Based on common ion effectContain both an acid to neutralize base added and a

base to neutralize acid addedPrepared by pairing a weak acid/weak base with its

conjugate saltMost effective when the concentration of the weak

acid/weak base is equal to that of its conjugate saltLinkLink

This surfer dude’s goin’ down…

LIKE TOTALLY!

B) Buffer Capacity

Buffer Capacity: the degree to which a buffer can resist changes in pH; the amount of acid or base that can be added to a buffer before a significant pH change occurs

Depends on the concentrations of weak acid/weak base and conjugate salt

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manatee. Ugh! I ALWAYS DO

THAT!

C) Calculating pH for Buffers

Using an ICE chart is the LONG way; however, there is a shortcut…

Henderson-Hasselbalch Equation: used to calculate the pH of a buffer

pH = pKa + log [base] [acid]

pOH = pKb + log [acid] [base]

Don’t pretend like I don’t see you getting

into that lifeboat, captain! What

happened to going down with your ship!?

Ex 3) What is the pH of a buffer that is 0.12 M lactic acid, HC3H5O3, and 0.10 M sodium lactate? (Ka = 1.4 x 10-4)

Mommy, does water REALLY go to live in the clouds when

it evaporates?

HW: 17.12, 17.14

D) Addition of Strong Acids or Bases to Buffers

If a strong acid or base is added to the buffer, we want to see how the pH is (slightly) changed.

Write the neutralization equation, do an ICE chart to find the new equilibrium concentrations, then apply the H-H equation…

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boooooey…

Ex 4) A buffer consisting of 0.300 M acetic acid and 0.300 M sodium acetate has a pH of 4.74. After the addition of 0.020 M NaOH (assume no volume changes), what is the new pH?

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Please stick to the rivers and the lakes

that you're used to…”

Ex 5) A buffer consisting of 0.300 M acetic acid and 0.300 M sodium acetate has a pH of 4.74. After the addition of 0.020 M HCl (assume no volume changes), what is the new pH?

This coastline is so ugly. I think

I feel a SCHUNAMI coming on!

HW: 17.18

III) Acid-Base Titrations A) What are they?

A known concentration of acid (or base) is added to a basic (or acidic solution) of unknown concentration

Equivalence Point: the point at which stoichiometrically equal amounts of acid and base have reacted with one another

End Point: where indicator color change occurs (usually before or after equivalence point, depending on where the indicator changes color)

Amazing…so much water and yet you’d die of thirst

if you kept drinking me!

A pH titration curve shows the pH as a function of the volume of titrant added

LinkAllows us to determine

the equivalence point which is not always a pH of 7!

Dspace.mit.edu

So seriously, be honest, am I blue or green? Clear? I’m colorblind, so I’m

never sure.

B) Strong Base added to Strong Acid

A: pH found from initial concentration of acid

B: pH found from concentration of acid that hasn’t been neutralized

C: pH= 7.00D: pH found from

excess baseA

B

D

C

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I taste salty for some reason…

Ex 6) Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.00 mL of 0.100 M HCl solution:

(a) 49.00 mL

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(b) 51.00 mLI don’t swim in your toilet, so

don’t pee in me!

HW: 17.28

Notice how rapidly the pH rises close to the equivalence point (even the addition of a single drop of base can cause a change of more than one pH unit)

An indicator such as methyl red can be used to change from yellow to red from pH 4.2-6.0 (an endpoint that occurs before the equivalence point)

An indicator such as phenolphthalein can be used to change from colorless to pink from pH 8.3 -10 (an endpoint that occurs after the equivalence point)

A

B

D

C

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Seems like just yesterday that someone got

devoured…

Physchem.co.za

You’d think by now I’d be used to the sight of a

dead fish, but it still creeps me out. It just floats there…gross!

C) Strong Base added to Weak Acid

A: pH found from ICE chart, Ka

expression, and solving for [H+] B (buffering region): pH found

by doing ICE chart for initial neutralization, then applying Henderson-Hasselbalch for remaining weak acid

C: pH≠7! pH found by assuming all base reacted, then writing Kb expression for conjugate base of weak acid, solving for [OH-], pOH, then finally pH

D: pH found from excess base Link

A

C

B

D

Dspace.mit.edu

Hey! Hi! Shoot, I don’t think she saw me wave…

Ex 7) (a) Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M HC2H3O2. (Ka = 1.8 x 10-5)

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once. Total jerk.

(b) Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M HC2H3O2 with 0.100 M NaOH.

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you look tough or something?

HW: 17.22, 17.24, 17.29

D) Relationship between Ka and the Shape of the Titration Curve

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expert, James Burchalewski!

E) Polyprotic Acids

Have multiple equivalence points due to multiple neutralization reactions

Bio.cmu.edu

Whoa, look at the time. Time for some low tide

action. See you suckas later!

Additional Aspects of Aqueous Equilibria

Part 2

Chapter 17 in Textbook

Zazzle.com

IV) Solubility Equilibria:A) What are they?

Heterogeneous equilibria between dissolving and precipitation (a saturated solution)

A quantitative means of determining how soluble a solid is in water

Not always as simple as soluble or insoluble like in net ionic equations

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B) Ksp

The equilibrium constant for the dissolving of a solid is designated Ksp for solubility-product

Ksp remains constant at a given temperature

The solid dissolving never appears in the equilibrium expression and is NOT affected by Le Chatelier’s Principle since it is pure

Link

Genchem.rutgers.edu

Ex 8) Write the Ksp expression for a saturated solution of CaF2(s).

Allaboutfrogs.org

C) Ksp and solubility

If a question asks for the molar solubility, then we can solve for the concentration in M using the equilibrium expression

Commons.wikipedia.org

Ex 9) If the Ksp for CaF2 is 3.9 x 10-11, what is its molar solubility?

                                       

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HW: 17.34, 17.38, 17.40

V) Factors that Affect Solubility:A) Common-Ion Effect

Adding a common ion to a solution equilibrium decreases the solubility according to Le Chatelier’s

Link

Greenfroginternet.com

B) pH

As the pH decreases, the solubilities of metal hydroxides increases; think of the increasing H+ as reacting and removing the OH-, thus shifting the equilibrium to the right (towards the ions)

This rule also applies to anions that are weak bases: F-, CO3

-2, PO4

-3, CN-, or S-2

LinkMicropig.tumblr.com

C) Complex Ions When a metal ion acts as a Lewis acid,

it accepts electrons from a Lewis base; this often results in the formation of a complex ion

Given equilibrium constant Kf: the greater the formation constant, the more stable the complex ion and the more soluble the metal

Rule of thumb: solubility of metal salts increase in the presence of NH3, CN-, and OH-

(ligands) due to the formation of complex ions Studiotota.com

Ex 10) The Ksp for AgCl is 1.8 x 10-10, but adding ammonia greatly increases the solubility. Why?

Woodka.com

Ex 11) Write the Kf for the complexation of silver ion with ammonia.

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D) Amphoterism

Al+3, Cr+3, Zn+2, Sn+2 when combined with OH- or O-2 are not just soluble in acidic solutions, but also basic ones; they are thus amphoteric

This is a result of complexes formed in water

Ex) Al(OH)3(s) + OH- ↔ Al(OH)4

-(aq)Evilscience.co.uk

VI) Precipitation and Separation of Ions

Given the following:BaSO4(s) ↔ Ba+2(aq) + SO4

-2(aq)

At any given time, Q = [Ba+2][SO4

-2]

If Q > Ksp, precipitation occurs until Q = Ksp

If Q = Ksp, equilibrium exists

If Q < Ksp, solid dissolves until Q = Ksp

Blog.greens.org.nz

Ex 12) Will a precipitate form when 0.10 L of 8.0 x 10-3 M Pb(NO3)2 is added to 0.40 L of 5.0 x 10-3 M Na2SO4?

                                       

Yankeeexposure.

blogspot.com

HW: 17.50

CuS (Ksp = 6 x 10-37) is less soluble than ZnS (Ksp = 2 x 10-25)

If H2S is added to the green solution, black CuS precipitates first

If this precipitate is removed and more H2S is added, white ZnS forms

This is called selective precipitation

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VII) Qualitative Analysis

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Media.weirdworm.com

HW: 17.56, 17.58

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