Chapter 15 Applications of Aqueous Equilibria

• • Some of what we do in this chapter will be the same as what we did in Chapter 14. Let’s start with common ions.

What do the weak acid HF and the salt NaF have in common?

The common ion effect

• • When the salt with the anion of a weak acid is added to that acid: –

it reverses the dissociation of the acid

–

Lowers the % dissociation of the acid

This same principle applies to salts with the cation of a weak base as well.

• Calculations are the same as last chapter

15.2 Buffered solutions

• • • • When an acid-base solution contains a common ion, it’s called a buffered solution.

A buffered solution is one that resists a change in pH with the addition of hydroxide ions or protons.

Often buffered solutions contain a weak acid and it’s salt (HF and NaF) OR a weak base and it’s salt (NH 3 and NH 4 Cl).

We can make a buffer of any pH by varying the concentrations of these solutions.

Finding the pH of a buffered solution • Calculate the pH of a 1L buffered solution of 0.20M acetic acid solution and 0.30M sodium acetate. K a =1.8x10

-5 1.8x10

 5   C H O 2 3 HC H O 2 3 2  2    0.2

 x x  1.8x10

 5  0.3x

0.2

pH   x  1.2x10

 5  4.92

pH changes in a buffered solution

• • We will use the solution from the last problem, but add 0.01M NaOH. Compare the pH change that occurs with the addition of the NaOH solid.

There are 2 major step to proceed with these types of problems!! – 1. The stoichiometry – 2. The equilibrium

The stoichiometry

1L of 0.20M acetic acid solution and 0.30M sodium acetate (K a =1.8x10

-5 ) [H + ]=1.2x10

-5 Buffered with .01M NaOH • Use moles not molarity Before reaction After reaction

HC 2 H 3 O 2

0.20 mol 0.20 – 0.01

=0.19mol

OH-

0.01

0

↔ H 2 O + C 2 H 3 O 2 -

0.30 mol 0.30+0.01

0.31mol

• We will assume that all of the NaOH will be consumed

Now for the equilibrium - ICE

•

HC 2 H 3 O 2 OH ↔ H 2 O

Before reaction After reaction 0.20 mol 0.20 – 0.01

=0.19mol

0.01

0 Since there is 1L of solution

+ C 2 H 3 O 2 -

0.30 mol 0.30+0.01

0.31mol

Concentration

Initial Change Equilibrium

HC 2 H 3 O 2

0.19M

-x 0.19-x

↔ H +

0 +x X

+ C 2 H 3 O 2 -

0.31M

+x 0.31+x

1.8x10

 5   C H O 2 3 2 HC H O 2 3 2     0.19

 x  1.8x10

 5  0.31x

0.19

pH   x  1.1x10

 5  4.96

pH change = 4.96 – 4.92 (from previous problem) Change in pH = .04 with the addition of a buffer.

• • • •

Think about what would have happened….

If base was added to just water K w H  OH  1x10  14 H   0.01

 1x10  12 That means the pH of 0.01M NaOH in water is 12 Minus the pH of water (7) means a change in pH of 5. Compare adding NaOH to a buffered solution vs. an unbuffered solution….5 vs 0.04…

Remember

• • Buffered solutions are simply solutions of weak acids and weak bases containing a common ion When a strong acid is added to a buffered soltuion…You know how to do these problems. Just remember the two steps… – Stoichiometry first – ICE (equilibrium) second

How does the buffer work?

• • Solve the equilibrium expression for [H + ] K a  H  A  a   A  This means the [H+] depends on the ratio of [HA]/[A ]…if you take the –log of both sides…(I won’t bore you with the math)… you get… pH  pK a  log   A   

•

It has a name!

pH  pK a  log   A      It’s called the Henderson-Hasselback equation!

• • It also works for an acid and its salt, like HNO 2 and NaNO 2 Or a base and its salt, like NH 3 and NH 4 Cl, but you must remember to convert K a to K b in the equation

• •

An assumption

pH  pK a  log   A    One assumption of this formula is that the initial concentrations and the equilibrium concentrations are equivalent. (<5% validity) It is a pretty safe assumption since the initial concentrations of HA and A are relatively large in a buffered system.

Let’s try

• What is the pH of a solution containing 0.30M HCOOH and 0.52M KCOOH (formic acid and potassium formate) K a =1.8x10

-4 pH    4     0.52

0.30

     4.01

Given base, salt, and K

b • • Calculate the pH of 0.25M NH 3 0.40M NH 4 Cl. (K b = 1.8 x 10 -5 ) and Don’t forget to find K a FIRST!! And the ratio is base over acid..

 14 K a  1x10 1.8x10

 5  5.6x10

 10 pH   log(5.6x10

 10     0.25

0.40

    pH  9.05

15.3 Buffer Capacity

• • • • The pH of a buffered solution is determined by the ratio of [A ]/[HA].

If the ratio doesn’t change much…then the pH won’t change much either The more concentrated these two are, the more H + and OH the solution will be able to absorb.

Larger concentrations = bigger buffer capacity

•

Try this problem.

Calculate the change in pH that occurs when 0.040 mol of HCl(g) is added to 1.0 L of each of the following: Ka= 1.8x10

-5 • 5.00 M HAc and 5.00 M NaAc 0.050 M HAc and 0.050 M NaAc Calculate initial pH for each solution (henderson-hasselbalch) pH    5          4.74

Now find the change after adding 0.04mol of HCl to 1.0L of solution

Before reaction After reaction

HC 2 H 3 O 2

5.0 mol 5.0 + 0.04

=5.04

↔ H +

0.04

0

+ C 2 H 3 O 2 -

5.0 mol 5.0 - 0.04

=4.96

• pH    5     4.96

5.04

     4.74

There’s virtually NO change. Now look at solution B. We already know the initial pH.

Solution B is 1L of 0.050 M HAc and 0.050 M NaAc

HC 2 H 3 O 2 ↔ H + + C 2 H 3 O 2 -

Before reaction 0.050 mol 0.04mol

0.050 mol • pH   After reaction 0.050 + 0.04

=0.09

 5 0 0.050 - 0.04

=0.01

    0.01

0.09

     3.79

Compared to 4.74 before the acid was added.

Solution A contains much larger quantities of buffering components and has a larger buffering capacity than solution B

•

Larger concentrations = bigger buffer capacity

Here’s why pH  pK a  log   A   

Buffer Capacity

• • • The best buffers have a ratio [A ]/[HA] = 1 This is the most resistant to change and true only when [A ] = [HA] Makes pH = pKa (since the log of 1=0)

10.5 Titrations and pH Curves

• • • Titration is commonly adding a solution of known concentration until the substance being tested is consumed.

This is often noted by a color change and is called the equivalence point. This is often observed in a graph of pH vs mL of titrant and is called a titration curve.

Strong acid with a strong base titration

• • • Net ionic equation H + + OH  H 2 O To know the amount of H solution.

+ at any point in the titration, we must determine the amount of H + remaining and divide by the total volume of the Let’s first consider a new unit for molarity that is smaller as most titrations usually deal with mL Millimole (mmol) = 1/1000 mol Molarity = mmol/mL = mol/L

The pH Curve for the Titration of 50.0 mL of 0.200 M HNO 3 with 0.100 M NaOH Where all the H + ions originally present, have reacted with all the OH ions added

• • • • • • • •

Things to note

You need to do the stoichiometry for each step mL x Molarity = mmol There is no equilibrium. Both the acid and base dissociate completely Use [H + ] or [OH ] to figure pH or pOH The equivalence point is when you have an equal number of moles of H + and OH In other words enough H + exactly with the OH is present to react Before the equivalence point H + is in excess After the equivalence point OH is in excess

The pH Curve for the Titration of 100.0 mL of 0.50 M NaOH with 1.0 M HCI

Strong base titrated with strong acid.

• Very similar to strong acid with a strong base except before the equivalence point the OH- is in excess and H+ is in excess after the equivalence point.

The pH Curve for the Titration of 50.0 mL of 0.100 M HC 2 H 3 O 2 with 0.100 M NaOH weak acid – strong base At equivalence point (pH > 7): A is basic

• • • • •

Titrating a weak acid with a strong base

Again, there is no equilibrium Do the stoichiometry – the reaction of OH with the weak acid is assumed to run to completion & the concentrations of the acid remaining & conjugate base are determined.

Determine the major species Since HA is stronger than H 2 O it is the dominant equilibrium Then do the equilibrium (Henderson Hasselbach)

The pH Curve for the Titrations of 100.0mL of 0.050 M NH 3 with 0.10 M HCl weak base – strong acid At equivalence point (pH < 7): NH 4 + is acidic

In a titration curve

• Equivalence point is defined by the stoichiometry NOT by the pH.

• Remember it is where the mol of H + equal to the moles of OH are

• • • • •

Summary

Strong acid and base just stoichiometry.

Weak acid with 0 ml of base - K a Weak acid before equivalence point –Stoichiometry first –Then Henderson-Hasselbach Weak acid at equivalence point- K b -Calculate concentration Weak acid after equivalence - leftover strong base. -Calculate concentration

Summary

• • • • Weak base with 0 ml of acid - K b Weak base before equivalence point.

–Stoichiometry first –Then Henderson-Hasselbach Weak base at equivalence point K a . -Calculate concentration Weak base after equivalence – left over strong acid. -Calculate concentration

15.5 Acid-Base Indicators

• Two common methods for monitoring the pH – 1. pH meter – 2. acid-base indicator (this is not a good choice for finding the equivalence point.) Careful selection of indicator can help get results close.

Endpoint IS change in color Equivalence point IS moles acid = moles base

HIn • • Most indicators are themselves a weak acid. They are one color with the proton attached and another without the proton.

Phenolphthalein, the most common indicator, is colorless as an acid and pink as a base ( In )

Did the color change?

• Typically, 1/10 of the initial form must be converted for the human eye to see a new color. When titrating an acidic solution… K a     In     pH    In       1 10

If a basic solution is titrated

• Indicator will initially exist as In more HIn and will form when acid is added K a     In     pH    In       10 1

Which indicator to use?

• • • It is best to choose an indicator whose endpoint is closest to our equivalence point.

It is easier to choose an indictor if there is a large change in pH near the equivalence point (vertical area of pH curve) The weaker the acid, the smaller the vertical area around the equivalence point, giving less flexibility in indicator choice

15.6 Solubility Equilibria and K

sp • Will it dissolve, and if not, how much?

• • • solid dissolved If everything dissolves, it is an equilibrium position Partial dissolving -- The solid will precipitate as fast as it dissolves (equilibrium) Surface area changes the rate at which something dissolves, but NOT the amount (no change in equilibrium position)

Generic equation M Nm a b  aM  (aq)  bNm  (aq) • • • M + is the cation (usually metal) Nm is the anion (a nonmetal) The solubility product for each compound is found below K sp  M  Nm  b

Solubility vs. solubility product

• • • • • Solubility is NOT the same as solubility product.

Solubility product is an equilibrium constant.

It doesn’t change except with temperature.

Solubility is an equilibrium position for how much can dissolve.

A common ion can change this.

Let’s try a couple… first an easy one!

• What is the Ksp value of copper (I) bromide with a measured solubility of 2.0 x 10 -4

CuBr

mol/L.

 (s)

Cu

 (aq) 

Br

 (aq) I C E

CuBr

2x10 -4 solid solid

↔ Cu +

0 +2x10 -4 2x10 -4

+ Br -

0 +2x10 -4 2x10 -4 K sp  K sp  4x10  8

Now a bit more difficult

I C E • Calculate the K sp for bismuth sulfide, which has a solubility of 1.0x10

-15 M at 25 o C

Bi S

2 3(s) 

2Bi

3  (aq) 

3S

2  (aq)

Bi 2 S 3

1x10 -15 solid solid

↔ 2Bi +3 + S -2

0 0 +2(1x10 -15 ) +3(1x10 -15 ) K sp  (2x10  ) (3x10  ) K sp  1.1x10

 73 2x10 -15 3x10 -15

Solubility from K

sp • Find the solubility of Silver Chloride. K sp =1.6x10

-10 .

AgCl

(s) 

Ag

 (aq) 

Cl

 (aq) K sp Ag  1.6x10

 10    Cl     1.6x10

 10  1.3x10

 5  x x 2

One more…a bit tougher

• Find the solubility of copper (II) iodate, Cu(IO 3 ) 2 . The Ksp for Cu(IO 3 ) 2 = 1.4x10

-7

Cu(IO )



Cu

 2 3 2(s) K sp   Cu  1.4x10

 7  2(IO ) 3 (aq)   2    2 

2IO

 3 (aq) 1.4x10

 7  4x 3 3 3.5x10

 8  3 x 3 3.3x10

 3  x The solubility of copper (II) iodate is 3.3x10

-3

Comparing Ksp values

• Ksp values give info about solubility (relative solubility). 1. When salts have the SAME number of ions: the larger the Ksp the more soluble 2. When salts have different number of ions: you CANNOT compare directly.

Common ion effect and solubility

• The presence of a common ion DECREASES the solubility of the salt.

Let’s compare: What is the solubility of AgBr in both pure water & in 0.0010M NaBr. (pure H 2 O first) AgBr (s) K sp   Ag 7.7x10

  (aq) 13  Br  (aq) 7.7x10

 13  7.7x10

 13  8.8x10

 7  x x 2

Now with a common ion

What is the solubility of AgBr in 0.0010M NaBr NaBr (s)  Na  (aq)  Br  (aq) AgBr (s)  Ag  (aq)  Br  (aq)  Br  0.001M

I C E

AgBr

7.7x10

-13 solid solid

↔ Ag +

0 +x x

+ Br -

0.0010

+x .0010x

K sp 7.7x10

 13  7.7x10

 13    assume x is negligible  in x + 0.

0 010 7.7x10

 13  0.0010x

7.7

x 10  10  x

Compare the two results

K sp In Water  7.7x10

 13 7.7x10

 13  7.7x10

 13 8.8x10

 7   x x 2 Common ion solution K sp 7.7x10

 13  7.7x10

 13     assume x is negligible in x + 0.

0 010 7.7x10

 13  0.0010x

7.7

x 10  10  x Notice that the common ion reduces the solubility as stated earlier (the reverse reaction occurs faster: equilibrium lies to the left)

• •

pH and solubility

Presence of a common ion decreases solubility Insoluble bases dissolve in acidic solutions Insoluble acids dissolve in basic solutions Mg(OH) 2(s)  Mg 2  (aq)  add 2OH  (aq) K sp  1.2x10

 11 K sp Mg 2  OH  2 1.2x10

 11  s(2s) 2 1.2x10

 11  4s 3 s  1.4x10

 4 OH  pOH    2s  2.8x10

3.55 & pH   4 10.45

At pH less than 10.45

Lower [OH ] Increase solubility of Mg(OH) 2

At pH greater than 10.45

Raise [OH ] Decrease solubility of Mg(OH) 2

15.7 Precipitation – mixing two solutions to get a solid

• • • Let’s look at the reverse process of dissolving salts to form a solution.

b Ion product is Q  M  Nm  Initial concentrations are used

Q < K sp

Unsaturated solution No precipitate

Q = K sp

Saturated solution

Q > K sp

Supersaturated solution Precipitate

will form

Try this

• • • • A solution of 750.0 mL of 4.00 x 10 -3 M Ce(NO 3 ) 2.00 x 10 3 -2 is added to 300.0 mL of M KIO 3 . Will Ce(IO 3 ) 3 (K sp =1.9x10

-10 M) precipitate and if so, what is the concentration of the ions?

Step 1: find the concentration of each of the ions expected to precipitate Step 2: Find Q Step 3: compare to K sp