Additional

CHAPTE Aspects of
R 17 Aqueous
Equilibria
17.1 The Common Ion Effect

We have looked at weak acid solutions:

For example:

acetic acid: CH3COOH ↔ CH3COO- + H+

Recall:
Calculate the pH of a 0.30 M solution of acetic acid? (Ka = 1.8 x 10-5)
1.8 x 10-5 = (X)(X) X = 2.32 x 10-3 M
CH3COOH ↔ CH3COO + H
- +
(0.30 – X)
0.30 M 0 0
X X X pH = -log(2.3x10-3M)
Omit X – then do a
0.30 – X X X = 2.63
validity check.
What happens when the solution now contains a
soluble salt of that same acid?

acetic acid: CH3COOH ↔ CH3COO- + H+

Adding CH3COONa would add Na+ and CH3COO-

Acetate is known as a “common ion”.

 pH will
shift increase

CH3COOH ↔ CH3COO- + H+
decreases

addition of salt
(CH3COONa)

Whenever a weak electrolyte and a strong electrolyte contain a

common ion, the weak electrolyte ionizes less than it would if it
were alone in solution.
Sample Problem:
What is the pH of a solution made by adding 0.30 mol of acetic
acid and 0.30 mol of sodium acetate to enough water to make
1.0 L of solution? (Ka for acetic acid is 1.8 x 10-5)
1. List species: CH3COOH, Na+, CH3COO-

2. Identify equilibrium providing H+ CH3COOH ↔ CH3COO- + H+

3. Use an ICE chart to find [E] and plug into MAE to find [H+] and pH
CH3COOH ↔ CH3COO- + H+ 1.8 x 10-5 = (0.30 + X)(X)
(0.30 – X)
[I] 0.30 mol/L 0.30 mol/L 0
small Ka – omit x
[C] -X +X +X 1.8 x 10-5 = (0.30 )(X)
(0.30 )
[E] 0.30 - X 0.30 + X X
X = 1.8 x 10-5 M = [H+]

pH = -log(1.8 x 10-5 ) = 4.74

Problem:
a) What is the pH of a 0.25 M solution of formic acid (HCOOH)?
(Ka = 1.7 x 10-4)
1.7x10-4 = X2 / 0.25 pH = 2.19
X = 0.0065 M

b) What is the pH of a solution containing 0.35 M HCOOH and

0.52 M HCOOK?
pH = 3.94
1.7 x 10 = (0.52+x)(x)
-4

(0.35 – x)
Omit x (Ka small)
 17.2: Buffered Solutions
In many chemical processes and in many reactions that
occur in living systems, the pH must remain at a constant
value.

For example; -sewage treatment

- in electroplating
- manufacture of photographic materials
- pH of blood has to be close to 7.4,
- saliva close to 6.8
-enzymes in the stomach function only
at pH’s close to 1.5.
*A buffer is a solution that maintains a nearly constant pH
when moderate amounts of acid and base are added to it.
Buffer solutions contain either:

 a weak acid and a salt containing its conjugate base,

(e.g. HF and NaF)
or
 a weak base and it’s conjugate acid (eg. NH3 / NH4Cl)
(conjugate acid-base pairs).

A buffer is most efficient in resisting changes in pH

when both the acid and its conjugate base are
present in approximately equal amounts.
To understand what happens, consider the following:

Reaction 1: A solution of just acetic acid.

CH3COOH  CH3COO- + H+
acetic acid shiftion
acetate

If some acid (HCl) is added to this mixture, the reaction will try to shift in the reverse.

LOTS LITTLE

However, since acetic acid is a weak acid, there is not a lot

of the acetate ion present in the solution, and it quickly runs
out.

Without the CH3COO-, the H+ can’t react, so it builds in the

solution. pH decreases.
Reaction 2: A solution of acetic acid and a salt
containing its conjugate base.
(CH3COOH and CH3COONa)

CH3COOH  CH3COO- + H+ (+ Na+ spectator)

acetic acid
shift
acetate ion

In this case if HCl is added, the reaction tries to shift in the reverse
direction.LOTS LITTLE
LOTS

Since acetic acid is a weak acid, there initially is not a lot of

acetate ion present. The addition of sodium acetate
increases the amount of CH3COO- available.
With the added CH3COO-, the H+ can react, so H+ gets used up
as it is added: pH remains the same.
 Reaction 3: A mixture of acetic acid and its salt.
shift
CH3COOH  CH3COO- + H+ (+ Na+ spectator)
acetic acidacetate ion
+ OH- → H2O

In this case a base, NaOH, is added to the mixture.

What will happen?
The presence of NaOH adds OH- to the mixture. This will
react with the H+, causing the reaction to shift forward.

Since acetic acid is a weak acid, there is a lot of acid

molecule present to shift forward. pH is constant.
(Salt had no effect.)
Could you make a buffer using the following
combinations? If so, show how they would work if an
acid were added to the mixture.

a) HCl + NaCl b) HNO2 + KNO2

c) LiHCO3 + H2CO3 d) Mg(OH)2 + MgSO4

e) NH3 + NH4Cl
Buffers can also be made starting with:
 weak acid and then converting a portion of that weak
acid via neutralization with a strong base.
e.g. mixing same volumes of 1.0 M acetic acid with 0.5 M sodium
hydroxide (this makes 0.5 M acetate)
CH3COOH + NaOH → NaCH3COO + H2O
1.0 M 0.5 M

Weak base and reaction with a strong acid

NaCH3COO + HCl → CH3COOH + NaCl
1.0 M 0.5 M
HX ↔ H+ + X-
Calculating the pH of a buffer:
Calculations are similar to common-ion calculations
(ICE charts can be used).

Henderson-Hasselbalch Equation
This equation is useful for determining the pH of a buffer, or the
ratio of the concentrations of the conjugate acid-base pair needed
to prepare a buffer with a specific pH.

Weak acid, HA, and one of its salts, NaA.

HA  H+ + A-

(salt: Na+ and A-)

 Ka = [H+] [A-]
[HA]

Rearranging: [H+] = Ka [HA]

[A-]

Negative logarithms of both sides:

-log[H+] = -log Ka + -log [HA]

[A-]

Or -log[H+] = -log Ka + log [A-]

[HA]

pH = pKa + log [conjugate base]

[acid]
 Kb = [BH+] [OH-]
[B]

Rearranging: [OH-] = Kb [B]

[BH+]

Negative logarithms of both sides:

-log[OH-] = -log Kb + – log [B]
[BH+]

Or -log[OH-] = -log Kb + log [BH+]

[B]

pOH = pKb + log [conj. acid]

[base]
 Note:
• The stronger the acid, the smaller the pKa
• pKa + pKb = 14 (for conjugate acid/base pairs
• Moles OR concentration can be used in the HH equation.

Example:
What is the pH of a buffer that is 0.12 M in lactic acid, HC 3H5O3,
and 0.10 M in sodium lactate? For lactic acid, Ka = 1.4 x 10–4.

pH = pKa + log [conjugate base]

[acid]

pH = -log(1.4x10-4) + log [0.10]

[0.12]
= 3.85 + (0.08) = 3.77
This can also be solved using ICE charts as shown in 17.1.
Buffer Capacity: the amount (moles) of acid or base the
buffer can neutralize before the pH begins to significantly
change.
This is simply a matter of stoichiometry between acid and base.

 The maximum amount of strong acid that can be added is

equal to the amount of conjugate base present in the buffer.

 The maximum amount of base that can be added is equal to

the amount of weak acid present in the buffer.

*Capacity increases with larger amounts of the conjugate

acid-base pair.

The pH range of a buffer is the range over which the buffer

acts effectively.
 Problem:
Buffer system #1: 125 mL of 0.250 M HF / 0.400 M NaF
Buffer system #2: 175 mL of 0.200 M HF / 0.300 M NaF

Determine the buffer capacity of each system in terms of adding

a) NaOH and b) HNO3
HF + NaOH → NaF + H2O NaF + HNO3 → NaNO3 + HF

a) When we add NaOH – it is the HF that will neutralize it.

Therefore, determine the moles of HF available in each buffer.

Buffer #1: moles acid: C x V = (0.250M)(0.125L) = 0.0313 moles

moles of acid = moles of base…….0.0313 moles of NaOH can be neutralized.

Buffer #2: moles acid: C x V = (0.200M)(0.175L) = 0.0350 moles

moles of acid = moles of base…….0.0350 moles of NaOH can be neutralized.
Problem:
Buffer system #1: 125 mL of 0.250 M HF / 0.400 M NaF
Buffer system #2: 175 mL of 0.200 M HF / 0.300 M NaF
Determine the buffer capacity of each system in terms of adding
a) NaOH and b) HNO3
NaF + HNO3 → NaNO3 + HF
b) When we add HNO3 – it is the F- that will neutralize it.
Therefore, determine the moles of NaF available in each buffer.
Buffer #1: moles base: C x V = (0.40M)(0.125L) = 0.0500 moles
moles of base = moles of acid…….0.0500 moles of HNO3 can be
neutralized.

Buffer #2: moles base: C x V = (0.300M)(0.175L) = 0.0525 moles

moles of base = moles of acid…….0.0525moles of HNO3 can be
neutralized.
Buffer #2 is more effective in both cases.
A buffer is most efficient in resisting changes in pH when
both the acid and its conjugate base are in equal amounts.

pH = pKa + log [conjugate base]

[acid]

pH = pKa + log 1 = pKa

*Therefore, we can choose a buffer for a given pH range

based on the basis of the pKa value of the acid.

E.g: For a buffer solution to operate around a pH of 9 we

might choose an NH3 – NH4+ buffer, since the pKa
for NH4+ is 9.3. (-logKa = -log(5.6x10-10) = 9.3)
Example:
Blood has a pH of 7.40. Use the H.H. Equation to calculate
the ratio of hydrogen carbonate ion, HCO3-, to carbonic acid,
H2CO3, in blood. pKa of H2CO3 = 6.37

pH = pKa + log [conjugate base]

[acid]
7.40 = 6.37 + log [conjugate base] / [acid]

7.40 – 6.37 = 1.03 = log [base] / [acid]

invlog 1.03 = [base] / [acid]

10.7 = [base] / [acid]
Mole Ratio of the base, HCO3-, to acid, H2CO3, is 10.7 to 1.
 Preparing a Buffer Solution with a Specific pH

1st – choose an acid whose pKa value is close to the desired pH.

2nd – plug desired pH and pKa into the Henderson-Hasselbalch

equation

3rd – Use the calculated ratio into an amount of solution that

you require
Ex: How would you prepare a phosphate buffer with a
pH of about 7.40?

There are 3 possibilities using phosphoric acid:

H3PO4  H+ + H2PO4- Ka = 7.5 x 10-3 pKa = 2.12

H2PO4-  H+ + HPO42- Ka = 6.2 x 10-8 pKa = 7.21
HPO4-  H+ + PO43- Ka = 2.2 x 10-13 pKa = 12.7

The most suitable substance would be the second

choice since the pKa value is closest to the desired pH
value.
H2PO4-  H+ + HPO42-
Could use NaH2PO4 (acid) and Na2HPO4 (base)
To find out the ratio of the two substances use the
H.H. equation:
pH = pKa + log ([conj. base] / [acid])

7.40 = -log(6.2 x 10-8) + log([HPO42-] / [H2PO4-])

7.40 = 7.21 + log ([HPO42-] / [H2PO4-])

7.40 – 7.21 = log([HPO42-] / [H2PO4-])
invlog(0.19) = [HPO42-] / [H2PO4-]
1.5/1 = [HPO42-] / [H2PO4-]

Therefore to make the buffer you use 1.5 parts of the

base (HPO42-) for every 1 part of the acid (H2PO4-)
used.
Problem:
How would you prepare a litre of “carbonate buffer” at a
pH of 10.10? (you have available H2CO3, NaHCO3, and
Na2CO3)

H2CO3 + H2O ↔ H3O+ + HCO3- Ka = 4.5 x 10-7

HCO3- + H2O ↔ H3O+ + CO32- Ka = 4.7 x 10-11

 Problem:
Calculate the pH of a solution containing
0.20 M CH3COOH and 0.30 M CH3COONa. Ka = 1.8 x 10-5
Two methods:
ICE chart:
CH3COOH CH3COO- + H+
[I] 0.20 M 0.30 M 0
[C] x x x
[E] 0.20 – x 0.30 + x x

Ka = 1.8 x 10-5 = (0.30 + x)(x) omit the ‘x’s

(0.20 –x) OR H.H.:
x = [H+] = 1.2 x 10-5 M pH=pKa+log[CH3COO-] / [CH3COOH]
pH = -log(1.2 x 10-5) = 4.92 pH = -log Ka + log (0.30) / (0.20)

= -log(1.8 x 10-5) + log (1.5)

= 4.74 + 0.18
= 4.92
Problem:
Calculate the pH of a solution containing
0.20 M CH3COOH and 0.30 M CH3COONa. Ka = 1.8 x 10-5

b) What would have been the pH of the same acetic

acid solution without the presence of the acetate ion
salt?

Ka = 1.8 x 10-5 = x2 / 0.20 – x (omit the x)

X = [H+] = 1.9 x 10-3 M
pH = 2.70
 Addition of Strong Acids or Bases to Buffers
E.g. What is the pH change when 0.010 mol of HCl is
added to a solution containing 0.050 mol of acetate ion
and 0.050 mol of acetic acid in 1.0 L of water?
(*assume no volume change)
Initially: CH3COOH  CH3COO- + H+
(0.05 mol/L) (0.05 mol/L) 0
x x x
(0.05 – x) (0.05 + x) x
Ka = 1.8 x 10-5 = (0.05 + x) x / (0.05 – x)
1.8 x 10-5 = 0.05 x / 0.05
x = 1.8 x 10-5 = [H+]
pH = -log(1.8 x 10-5) = 4.74

OR pH = -log(1.8x10-5) + log [0.050]

[0.050]
 After the addition of HCl: (HCl – 100% ionization)
(The H+ reacts with the acetate ion to form acetic acid)
This is a stoichiometry calculation.
CH3COOH  CH3COO- + H+
0.05 0.05 0.01 Before neutralization.
+ 0.01 -0.01 -0.01 Neutralization.
0.06 0.04 0 At new equilibrium.
Now plug these new concentrations into an ice chart or use
the H.H. equation to find out the new pH.
CH3COOH  CH3COO- + H+ 1.8 x 10-5 = (0.04 + x) (x) / (0.06 – x)
1.8 x 10-5 = 0.04 x / 0.06
(0.06 mol/L) (0.04 mol/L) 0 x = 2.7 x 10-5 M = [H+]
x x pH = 4.57
x
(pH(0.06 – x) from
changes (0.04 + x)to 4.57
4.74 x – only a 3.6 % change in pH)
 Problem: Calculate the pH of a 0.30 M NH3 / 0.36 M NH4Cl
buffer system. What is the pH after the addition of 20.0 mL of
0.050 M NaOH to 80.0 mL of the buffer? (Kb for NH3 = 1.8 x 10-5)

NH4+ = (0.36M)(0.080L)
Initally: pOH = pKb+ log [NH4+] / [NH3]
= 0.0288 mol
pOH = -log(1.8 x 10-5)+ log [0.36] / [0.30] After addition
= 0.0288 – 0.0010
pOH = 4.74 + 0.079 = 4.82 = 0.0278 mol
pH = 9.18 [NH4+] = 0.0278 / 0.100L
= 0.278 M
NH3 + H2O ↔ NH4+ + OH-
NH3 = (0.30M)(0.080L)
= 0.024 mol
After addition
Add (0.020 L)(0.050M) = 0.024 + 0.0010
= 0.0010 mol NaOH = 0.025 mol
NH3 increases by NH4+ decreases by [NH3] = 0.025 / 0.100L
0.0010 mol 0.0010 mol = 0.250 M

NOW: pH = pKa + log [NH3] / [NH4+]

pH = 9.26 + (-0.046) = 9.21
pH = 9.26+ log [0.250] / [0.278]
 A Review: Acid-Base Indicators

Acid-Base indicators are substances that have distinctly

different colours in solutions with different pHs.

Typically these substances tend to be weak acids or

weak bases that have different colours in their
molecular and ionic forms.

Recall: HA ↔ H+ + A- HIn ↔ H+ + In-

molecule ion

To generalize, use HIn as the formula for a weak

monoprotic acid. To be an effective indicator HIn and In-
must have very different colours.
 shift
In solution: HIn ↔ H+ + In-

in acid

According to Le Chatelier’s Principle, if this indicator is

added to an acidic solution, the [H+] increases and
reaction shifts left. [HIn] increases, and the colour of
HIn predominates. This is why the molecule will show
up at lower (acidic) pH’s.

[HIn] / [In-] > 10

colour of acid HIn
predominates.
 shift
[In-] / [HIn] > 10
HIn ↔ H+ + In- colour of base In-
predominates.
in base
If the indicator is added to a base, the [H +] decreases since it
combines with OH- to produce water. The reaction shifts forward
to produce more H+ . The ionic form, In-, will predominate at
higher pH’s.

If [HIn] = [In-] then the colour is a combination of the two

colours.
The choice of the indicator depends on the strength of the acid
and base being used. To get an accurate description of pH of a
solution using indicators, several indicators need to be used.
 Colour it turns if pH Colour it turns if pH
Indicator pH Range is below range is above range

methyl violet 0.0 – 1.6 yellow green blue

methyl yellow 2.9 – 4.0 red yellow

orange
bromophenol blue 3.0 – 4.6 yellow blue

methyl orange 3.2 – 4.4 red yellow

methyl red 4.8 – 6.0 red yellow

litmus 5.5 – 8.0 red blue

bromothymol blue 6.0 – 7.6 yellow blue

phenol red 6.6 – 8.0 yellow red

phenolphthalein 8.2 – 10.6 colourless red

thymolphthalein 9.4 – 10.6 colourless blue
alizarin yellow 10.0 – 12.0 yellow red
Universal Indicator colour chart

Many acid-base
indicators are plant
pigments
(red cabbage, tea
leaves, …)
17.3 Review: Acid-Base Titrations

This is a quantitative analysis of a neutralization

reaction, commonly used to determine the unknown
concentration of either an acid or a base. It can be
used for any reaction that is rapid and goes to
completion.

A solution with an accurately known concentration

(standard solution) is gradually added to another
solution of unknown concentration, until the
chemical reaction between the two solutions is
complete.
 The acid-base indicator is
used to signal when the
solution is neutralized.
The point where the
indicator changes colour
is called the end point.

Since indicators change

colours over a pH range,
the colour change might
acid
not occur exactly at the
solutions’ “equivalence
A known amount of the standard
point” – where the
solution is placed in an Erlenmeyer flask
[H+] = [OH-].
and to it is added an appropriate
indicator (see later).
 Monoprotic Acids/ Monohydroxylic Bases:
Titration Calculations: quick review

acids give 1H+ / bases give 1 OH-

HBr → H+ + Br- / KOH → K+ +OH-
[acid] = [H+ ] / [base] = [OH-]

At the equivalence point: moles H+ = moles OH-

and moles = concentration x volume, therefore,

at the equivalence point: CaVa =CbVb

acid base
Polyprotic Acids/Polyhydroxylic Bases:
These substances can either donate or accept more
than one proton. Note – polyprotic acids require a
strong base to donate more than one hydrogen ion.
Therefore the ratio of H+ to OH- is not going to be 1:1.

For example: H2SO4 → 2H+ + SO42-

Ca 2Ca

H2SO4 + 2 NaOH → Na2SO4 + H2O

Mg(OH)2 → Mg2+ + 2OH-

Cb 2Cb
Example 1:
10.0 mL of an acid solution (HA) is needed to
neutralize 35.0 mL of a 0.050 M NaOH base solution.
What is the concentration of the monoprotic acid?

moles H+ = moles OH-

CaVa =CbVb
Ca(0.010L) = (0.050 M)(0.035L)

Ca = 0.175 M
Example 2: What volume of 0.50 M NaOH is needed
to neutralize 25.0 mL of 0.30 M H2SO4?

moles H+ = moles OH-

2CaVa = CbVb
2(0.30 M)(0.0250L) = (0.50 M)Vb

0.030 L = Vb
3. A 20.0 mL solution of strontium hydroxide is titrated
with 25.0 mL of 0.500 M HCl. What was the original
concentration of the Sr(OH)2?

CaVa = 2CbVb
(0.500M)(0.025L) = 2Cb(0.020L)

Cb = 0.313 M
 e.g. HCl + NaOH → NaCl + H2O
Titration Curves:
After the equivalence point—
Strong Acid – Strong Base: The pH of the solution is
determined by the excess [base]
At the equivalence point, equal moles of in the solution. All acid has been
neutralized.
the OH- and H+ have reacted, leaving only
their salt. pH is 7.00, because the salt ions
do not hydrolyze.

Between initial pH and the equivalence

point—As base is added, pH ↑ slowly at
first and then rapidly as it nears equivalence
point. pH is determined by the [acid]
not yet been neutralized

The initial pH—The pH before the

addition of any base. pH is low.
 Strong acid/Strong base: no salt hydrolysis to change final pH.
Problem: Use excess to determine pH.

Calculate the pH when the following quantities of 0.100 M

NaOH solution have been added to 50.0 mL of 0.100 M HCl
solution:
a) 49.0 mL b) 51.0 mL
• moles H+ = C x V = (0.100M) • moles H+ = C x V = (0.100M)
(0.050L) = .00500 (0.050L) = .00500
• moles OH- = C x V = (0.100M) • moles OH- = C x V = (0.100M)
(0.049L) = .00490 (0.051L) = .00510
• Excess (H+) moles • Excess (OH-) moles
= 0.0050 - 0.00490 = 1 x 10-4 = 0.0051 - 0.0050 = 1 x 10-4
[H ] = 1 x 10 / (0.050 + 0.049)
+ -4 [OH-] = 1 x 10-4 / (0.050 + 0.051)
= 1.01 x10-3 M = 9.90 x10-4 M
pH = -log(1.01 x10-3 ) = 2.996 pH =14 – (-log(9.90 x10-4 )) = 11.000
 Choosing an appropriate indicator for your
titration:
Optimally, an indicator would change colour at the
equivalence point.

However, due to
Large pH change for
the drastic change minimal volume
in pH close to the moles H+ = moles OH- change.
equivalence point,
any indicator that
changes colours
during the rapid-
rise portion would
be good.
 Weak-Acid and Strong Base Titrations:
Final pH is still high
(due to presence of
E.g. CH3COOH + NaOH ↔ CH3COONa + H2O strong base).
After the equivalence [OH–] from the
excess NaOH. determines the pH

At the equivalence point: The anion of the

salt is weak base, and pH > 7.

Between the initial pH and the equivalence

point
Two main steps:
• neutralization reaction between the weak
acid and OH– to determine [HA] and [A-]
• calculate the pH of this buffer pair.

The initial pH of the weak acid.

(is higher than strong acid)
The pH titration curves for weak acid-strong base
titrations differ from strong acid-strong base titrations in 3
ways:
1.The solution of the weak acid has
a higher initial pH than a solution
of a strong acid of the same
concentration.

2. The pH change at the rapid-rise

portion of the curve
near the equivalence point is
smaller for the weak acid than it
is for the strong acid.
3. The pH at the equivalence point is above 7.00 for the
weak acid-strong base titration. (due to hydrolysis of
conjugate base)
Influence of acid strength on the shape of the curve for titration
with NaOH.

Each curve represents

titration of 50.0 mL of 0.10
M acid with 0.10 M NaOH

Because the pH change near the equivalence point becomes

smaller as Ka decreases, the choice of indicator for a weak acid-
strong base titration is more critical than it is for a strong acid-
strong base titration.
Titration of a weak base with a strong acid:
 Calculation Example:
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 x10–5).

Stoichiometric calculation: CH3COOH + NaOH → CH3COONa + H2O

Moles acid: C X V = (0.100M)(0.050 L) = 0.005 mol acid

Moles base: C X V = (0.100M)(0.045) = 0.0045 mol base
CH3COOH + OH -→ CH3COO- + H2O
0.005 0.0045 0

0.0045 0.0045 0.0045

0.0005 0 0.0045
moles remain moles produced
 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 x10–5).

Concentration:
[CH3COOH] = 0.0005 mol/ (0.045 + 0.050) = 0.0053 M
[CH3COO-] = 0.0045 / 0.095 = 0.047 M
Equilibrium calculation:
Or: pH = pKa + log [A-] / [HA]
Ka = [CH3COO- ][H+]
pH = -log(1.8 x 10-5) + log [0.047]
[CH3COOH] [0.0053]
1.8 x 10-5 = [0.047] ][H+] pH = 5.69
[0.0053]
*Could use mole
[H+] = 2.0 x 10-6 M ratio in HH
pH = -log(2.0 x 10-6 M) = 5.69 equation too!
Example: Calculate the pH at the equivalence point in the
titration of 50.0 mL of 0.100 M HC2H3O2 with 0.100 M NaOH.
(Kb of acetate: 5.6 x 10-10) (what volume of base is added?)

CH3COOH + OH- → CH3COO- + H2O

neutralizes Determines the pH when it
undergoes hydrolysis
Moles acid: (0.100M)(0.050 L) = 0.005 mol = mol NaOH
Following stoichiometry: CH3COOH : CH3COO-
0.005 mol reacts: X = 0.005 mol produced.
Following Hydrolysis:
Acetate is a weak base: CH3COO- + H2O ↔ CH3COOH + OH -
Kb = [CH3COOH][OH-] 5.6 x 10-10 = [X][X]
[CH3COOH] [(0.005/.100) - X] (omit X)

X = [OH-] 5.3 10–6 M which gives a pOH = 5.28 and pH = 8.72.

 And more:
Calculate the pH at the equivalence point when 25.0 mL of
0.100 M NH3 is titrated by a 0.100 M HCl solution?
(Ka of ammonium is 5.6x 10-10)
NH3 + HCl → NH4+ + Cl- Stoich: moles NH3 = (0.025L)(0.100M) = 0.0025
0.0025 : 0.0025 0 0

Neutralization reaction produces

NH4+ = 0.0025 moles
pH = 5.27

NH4+ + H2O ↔ NH3 + H3O+ 5.6 x 10-10 = [X][X]

[0.0025/.050 - X] (omit X)
Ka = [NH3][H3O+]
[NH4+] X = [H+] = 5.4x10-6 M

Sketch the titration curve for the titration between HF and NaOH.
Place a star where you would find the equivalence point.
Half-Equivalence Point:
When a reaction in a titration is half way to the equivalence
point, we are at the “half equivalence point”.

For example, the titration of 2M HA reaches the half-

equivalence point when 1 M of the acid has been neutralized,
and 1 M is still left.
HA + OH- ↔ H2O + A-
[I] 2 1 0
[C] 1 1 1
[E] 1 0 1

The half-equivalence point can be

used to determine the Ka and pKa
of the acid used in titration
At half equivalence point, the concentrations of
the weak acid and its conjugate base are equal.
pH = pKa + log [A-] / [HA]
= pKa + log(1)
pH = pKa
 Titrations of Polyprotic Acids

Weak acids that contain more than Titration curve for the
one ionizable H+, will react with reaction of 25.0 mL of 0.10 M
Na2CO3 with 0.10 M HCl.
OH– in a series of steps:

H2CO3(aq) + OH-(aq) → HCO3-(aq) + H2O(l)

HCO3-(aq) + OH-(aq) → CO32-(aq) + H2O(l)

We saw before that the relative concentrations

of the species involved are:

[H2CO3] >> [HCO3–] >> [CO32–]

*Note the multiple equivalence points.

 17.4 Solubility Equilibria

Recall: In a saturated solution, the solute is in equilibrium

with its ions. Therefore, for a saturated solution of copper
(II) chloride:
CuCl2(s)  Cu2+(aq) + 2Cl-(aq)
We can write out an expression for the equilibrium:
Kc = products = [Cu2+][Cl-]2
reactants
Remember, CuCl2 is a solid
(as will be all solutes), so we omit it
from the expression.
Therefore Kc = Ksp= [Cu2+][Cl-]2
In order to determine the Ksp expression for an ionic salt:

• write out the balanced dissociation equation for the

substance
E.g. Na2SO4 (s)→ 2 Na+ + SO42-
• write your expression with the products of the ions,
using the molar coefficient as the exponent
Ksp = [Na+]2[SO42-]
Ksp is known as the solubility product constant.

It represents the solubility of some salts at their equilibrium

at a specific temperature.

The larger the Ksp value, the more soluble the salt.
 Problem Solving

A) Finding the solubility of a substance (given Ksp)

If Ksp is known, it can be used to determine the actual

solubility of the ionic salt in water.
(either in moles/L (molar solubility), or g/L)

This is done by using an ICE chart and plugging in “x”

as the amount of solute that will dissolve (i.e. its
solubility).
 For example: The solubility product (Ksp) for Al(OH)3 is
5.0 x 10-33. Calculate the solubility of Al(OH)3 in water.

Step 1: Write your dissociation equation:

Step 2: Use an ICE chart, plugging in “x” as the amount dissolved.
We are assuming that “x” represents the solubility – all of the solute
should dissolve and there should be none remaining at equilibrium.

Equation: Al(OH) ↔ Al3+ + 3 OH-

3(s) (aq) (aq)
[I] Not in water, so
Before dissolving.
X 0 0 no ions present.
[C] Solute dissociates;
Amount dissolved. X X 3X Watch ratio!
We assume ALL
[E]
0 X 3X dissolves and forms
ions.
You do not need to show your ICE chart if you can determine the [E] amounts
without it. However – you still need to show your reasoning!

Step 3: Set up your mass action expression equal to Ksp and

solve for x. (remember molar coefficients as the exponents).

Al(OH)3(s) ↔ Al3+(aq) + 3OH-(aq)

Ksp = [Al3+][OH-]3
5.0 x 10-33 = [x][3x]3
5.0 x 10-33 = 27x4
1.85 x 10-34 = x4
3.69 x 10-9 M = x = molar solubility
 B) Looking for Ksp (solubility is given).
Example: It was found that 6.35 x 10-3 moles of Ag2CO3
would saturate 50.0 L of water. What is the Ksp for the
Ag2CO3?
Step 1: Write out the dissociation equation.
Step 2: Calculate the [solute] that will form a saturated solution.
Then determine what [ions] you will have once the solute C = moles
dissociates. You can use an ICE chart if preferred. volume
Equation:
Ag2CO3(s) ↔ 2Ag+(aq)+ CO32-(aq)
[I] = 6.35 x 10-3 = 1.27 x 10-4 M Not in water, so
0 0 no ions present.
50.0 L
[C]
1.27 x 10 M -4 2.54 x 10-4 1.27 x 10-4 Solute dissociates
Watch ratio!
[E]
0 2.54 x 10-4 1.27 x 10-4
Step 3: Write out your Ksp expression (using your
dissociation equation).
Ag2CO3(s) ↔ 2Ag+(aq)+ CO32-(aq)

Ksp = [Ag+ ]2[CO32-]

Step 4: Plug in your amounts at equilibrium ([E]) into

your expression and solve for Ksp.

Ksp = [2.54x10-4 ]2[1.27x10-4] = 8.19x10-12

17.5: Factors that Affect Solubility

1) Common Ion Solubility

Simply put, if you add an ion, found already in an

equilibrium mixture, the system will shift to compensate
for the presence of the ion.

For example, if we place calcium fluoride in water it

will reach equilibrium as follows: CaF2

CaF2(s)  Ca2+(aq) + 2F-(aq) Ca2+ F-

 NaF
CaF2(s)  Ca2+(aq) + 2F-(aq)
CaF2

The addition of sodium fluoride (NaF) will provide the

common ion (fluoride).

This additional stress causes the reaction to shift to the

left, producing more CaF2 (as a precipitate).
The presence of a common ion
will decrease the solubility of
that ion in solution. (Less will
dissolve).
 Look at dissociation ratio:
This can be quantified: NaF ↔ Na+ + F-
0.20 → 0.20 + 0.20
1) CaF2 has a Ksp of 3.9 x 10-11. Find the molar solubility of this
substance in:
a) pure water b) 0.20 M NaF
CaF2(s)  Ca2+(aq) + 2F-(aq) CaF2(s)  Ca2+(aq) + 2F-(aq)
X 0 0 X 0 0.20
X X 2X X X 2X
0 X - 2 2X 0 2+ - X2 0.20 + 2X
Ksp = [Ca ][F ]
2+
Ksp = [Ca ][F ]
3.9 x10-11 = (X)(2X)2 3.9 x10-11 = (X)(0.20 + 2X)2
= 4X3
X = 2.1x10-4 M = (X)(0.20)2 = 0.04X OMIT
(solubility) X = 9.8x10-10 M
(solubility ion NaF)
2) AgCl has a Ksp = 1.6 x 10-10. Calculate the solubility of AgCl in

a) pure water and b) 6.5 x 10-3 M AgNO3.

AgCl(s) ↔Ag+(aq) + Cl-(aq) AgCl(s) ↔ Ag+(aq) + Cl-(aq)

X X 6.5 x 10-3 + X X

1.6 x 10-10 = x2
1.6 x 10-10 = (6.5 x 10-3+ x)(x)
x = solubility =
x = solubility =

X = 1.26 x 10-5 M X = 2.5 x 10-8 M

2. Solubility and pH
The solubility of any substance whose anion is basic will be
affected to some extent by the pH of the solution.

For example: Mg(OH)2 (s)↔ Mg2+(aq) + 2OH-(aq)

Decreasing pH (↓OH-) causes the Increasing pH (↑OH-) causes the reaction

reaction to shift forward– more to shift backwards – less soluble.
soluble.

In pure water, Mg(OH)2 has a Ksp = 1.8 x 10-11

1.8 x 10-11 = [Mg2+][OH-]2 [Mg2+]= X = 1.65X10-4 M
= 4x3 [OH-] = 2X = 3.3x10-4 M
x = 1.65 x 10-4 M
A saturated solution has a [OH-] of 3.3x10-4 M
And a pH = 10.52
If solid Mg(OH)2 is dissolved in a solution with a pH of 9.0
(more acidic):
The solution now has a pOH of 5.0 and [OH–] = 1.0 x 10–5M.

Looking at the solubility equilibrium of Mg(OH)2:

1.8 x 10-11 = [Mg2+][OH-]2 To maintain equilibrium,

= [Mg2+](1 x 10-5)2 Mg(OH)2 dissolves in the
[Mg2+] =0.18 M solution until [Mg2+] = 0.18M
(more dissolves).
If the concentration of OH– were reduced further by making the
solution more acidic, the Mg2+ concentration would have to
increase to maintain the equilibrium condition.

Generally, if a compound contains a basic anion, solubility

will increase as the solution becomes more acidic.
 Another example: the solubility of CaF2 increases as the
solution becomes more acidic, (F– ion is a weak base).

Which of the following compounds

will be more soluble in acidic
solutions than in water?
a) CuS Basic S2- more
soluble.

b) AgCl No change

Basic SO42-
c) PbSO4 more soluble.
Basic OH-
d) Ca(OH)2 more soluble.

No change.
e) PbBr2
Advanced Problem:
Calculate the concentration of aqueous ammonia necessary to
initiate the precipitation of iron (II) hydroxide from a 0.0030 M
solution of FeCl2. (Ksp for Fe(OH)2 = 1.6 x10-14 Kb for NH3 = 1.8 x 10-5)
FeCl2 → Fe2+ + 2Cl- and NH3 + H2O(l)↔ NH4+ + OH-
1st – find [ ] at which ppt of Fe(OH)2 will form.
Fe(OH)2 ↔ Fe2+ + 2OH-
1.6 x10-14 = [Fe2+][OH-]2
1.6 x10-14 = [0.0030][OH-]2
[OH-] =2.3 x10-6 M
Now find [NH3 ] that will produce this much OH-
NH3 + H2O(l)↔ NH4+ + OH- 1.8x10-5 = [2.3 x10-6 ][2.3 x10-6 ]
x 0 0 (x - 2.3 x10-6)
- 2.3 x10-6 +2.3 x10-6 +2.3 x10-6 X = [NH3] = 2.6x10-6 M
X – 2.3x10-6 2.3 x10-6 2.3 x10-6 [NH3] needs to be >2.6x10-6M
 Al3+, Cr3+, Fe3+, Bi3+,
3. Formation of Complex Ions
Be2+
Recall that some metal ions can act as Lewis acids toward water
molecules, which act as Lewis bases.
Lewis bases other than water (NH3, CN-, OH- ) can also interact
with metal ions, particularly with transition-metal ions. Such
interactions can have an effect on the solubility of a metal salt.
For example, AgCl, whose Ksp = 1.8 x 10–10, will dissolve in the
presence of aqueous ammonia because of the interaction
between Ag+ and the Lewis base NH3
AgCl(s) ↔ Ag+(aq) + Cl-(aq)
and Ag+(aq) + 2NH3(aq) ↔ Ag(NH3)2+(aq)
Overall: AgCl(s) + 2NH3(aq) ↔ Ag(NH3)2+(aq) + Cl-(aq)
→The presence of NH3 drives the top reaction, the solubility equilibrium of
AgCl, to the right as Ag+(aq) is removed to form Ag(NH3)2+.
A measure of the tendency of a metal ion to form a complex ion is
given by the formation constant Kf (the equilibrium constant for
the complex ion formation).
e.g. Cu2+(aq) + 4 NH3(aq) ↔ Cu(NH3)42+(aq) Ag+(aq) + 2NH3(aq) ↔ Ag(NH3)2+(aq)

Kf = [Cu(NH3)42+] Kf = [Ag(NH3)2+]
[Cu2+][NH3]4 [Ag+][NH3]2
The larger the Kf, the more stable the complex ion.
Problem: A 0.20 mole quantity of CuSO4 is added to a liter of
1.20 M ammonia solution. What is the concentration of Cu2+
ions at equilibrium? We know: CuSO4(s) ↔ Cu2+(aq) + SO42-(aq)
0.20 mole
But Cu2+(aq) will react with NH3

From the table: Cu2+ + 4 NH3 ↔ Cu(NH3)42+ Kf = 5x1012

A really large Kf 0.20 M 1.2 M 0
tells us that most ~0.20 4(0.2) 0.2
of the Cu2+ will react.
x 1.2 – 0.8 0.2
Answer is really
Kf = [Cu(NH3)4 ]
2+ 5x10 = [0.2]
12
small, so
[Cu2+][NH3]4 [x][0.4]4 assumption is
allowed.

X = [Cu2+] = 1.6 x 10-12 M

The formation of a complex ion generally increases the
solubility of a substance.

Another advanced problem:

Calculate the molar solubility of AgCl in a 1.0 M NH3 solution.
AgCl(s) ↔ Ag+(aq) + Cl-(aq) Ksp = 1.6 x 10-10
Ag+(aq) + 2NH3(aq) ↔Ag(NH3)2+(aq) Kf = 1.7x107

AgCl(s) + 2NH3(aq) ↔ Ag(NH3)2+ + Cl- K = (Ksp)(Kf) = 2.7 x 10-3

x 1.0 0 0 [Ag(NH3)2+ ] [Cl-]
x 2x x x [NH3]2

0 1-2x x x 2.7 x 10-3 = [x ] [x]

[1-2x]2
Square root both sides….. molar solubility = X = 0.047 M
Amphoteric metal oxides and hydroxides that are relatively
insoluble in water will dissolve in strongly acidic and basic
solutions.
those of: Al3+, Cr3+, Zn2+, Sn2+
Al2O3, ZnO….
Al(OH)3, Cr(OH)3,Zn(OH)2….

For example:
Al(OH)3(s) + 3H+ → Al3+(aq) + 3H2O(l)

Al(OH)3(s) + OH- → Al(OH)4-(aq)

 17.6: The Formation of a Precipitate
When you mix aqueous solutions together there is always the
possibility that a precipitate will be produced. Since we know
that a Ksp value represents the solubility of a saturated solution,
we can use this information to determine whether or not a given
mixture exceeds this amount.

There are three possibilities for ion products in solution. An ion

product is simply a calculation of a “trial Ksp” using the actual
concentrations of the ions found in the mixture:
ion product (Q) < Ksp
No precipitate will form in
ion product (Q) = Ksp either of these

ion product (Q) > Ksp  this will result in the formation
of a precipitate.
Here’s how it is done:
Will a precipitate of calcium sulfate form if 10.0 mL of 0.050 M
CaCl2 solution is mixed with 30.0 mL of a 0.020M MgSO4 solution?
(calcium sulfate has a Ksp of 9.1 x 10-6)
CaCl2 + MgSO4→ CaSO4 + MgCl2
1. If the substance that might precipitate is not provided, identify
the possible products of the reaction and determine which
substance is more likely to precipitate (has a lower Ksp value).

2. Write out the Ksp expression for this substance (that might
ppt). CaSO might precipitate
4

CaSO4 → Ca2+ + SO42- Trial Ksp =Q= [Ca2+ ][SO42- ]

Will a precipitate of calcium sulfate form if 10.0 mL of 0.050 M CaCl2
solution is mixed with 30.0 mL of a 0.020M MgSO4 solution? (calcium
sulfate has a Ksp of 9.1 x 10-6)

3. Calculate the concentrations of the ions found in the Ksp expression

using the info. given in the questionThis is called your “trial ion product”
(trial Ksp) or “reaction quotient” (Q).
Trial Ksp = Q = [Ca ][SO4 ]
2+ 2-

CaCl2 + MgSO4→ CaSO4 + MgCl2

from the CaCl2: from the MgSO4:
10.0 mL of 0.050 M 30.0 mL of 0.020 M
[ ] = moles = C x V
volume total volume Q =(0.050M)(0.010L) (0.020M)(0.030L)
(0.010 + 0.030) (0.010 + 0.030)

Q = [ 0.0125] [0.015]
Q = 1.9 x10-4
4. Compare your reaction quotient (Q) to the actual Ksp value
for this substance. Only if Q > Ksp will a precipitate form.

calcium sulfate has a Ksp of 9.1 x 10-6

Q = 1.9 x10-4 > Ksp

Precipitate will form.

(The concentration of ions is too large).
Another Example:
1) Will a precipitate form if 120.0 mL of 0.00020 M Fe(NO3)3 solution is
mixed with 25.0 mL of 0.00050 M NaOH solution?
Fe(NO3)3+ 3NaOH → Fe(OH)3 + 3NaNO3
More likely to precipitate. (compare Ksp)
(Ksp of Fe(OH)3 = 4 x 10-38)
Write Ksp expression for substance more likely to ppt.
Fe(OH)3→ Fe3+ + 3OH-
Trial Ksp = Q = [Fe3+ ][OH- ]3 Find concentrations using source.

from the Fe(NO3)3 from the NaOH

Q =(0.00020M)(0.120L) (0.00050M)(0.025L) 3

(0.120 + 0.025) (0.120 + 0.025)

Q =(1.66x10-4)(8.62 x 10-5)3
Q = 1.1 x 10-16 > Ksp = 4 x10-38 PPT forms!
Selective Precipitation of Ions:

It is sometimes desirable to remove one type of ion from

solution by precipitation (while leaving the other ions in
solution).

Even when both products are insoluble, we can obtain

some degree of precipitation.

For example: A solution containing Cl-, Br-, and I- could

be separated by adding AgNO3

Compound Ksp
AgCl 1.6x10-10
AgBr 7.7x10-13
AgI 8.3 x10-
17
Example:
Silver nitrate is slowly added to a solution that is 0.020 M in
Cl-, and 0.020 M in Br-. Calculate the concentration of Ag+
required to initiate the

a) precipitation of AgBr, and b) the precipitation of AgCl

Ksp values: AgBr = 7.7x10-13 , AgCl = 1.6 x 10-10

7.7x10-13 =[Ag+][ Br-] 1.6x10-10= [Ag+][ Cl-]

7.7x10-13 =[Ag+][ 0.020] 1.6x10-10= [Ag+][ 0.020]

[Ag+] = 3.9x10-11 M [Ag+]= 8.0x10-9 M

To precipitate AgBr without precipitating the Cl- ions [Ag+]

must be greater than 3.9 x 10-11, but less than 8.0 x 10-9M.
Example:
If a solution contains 4.3 x 10-4 M lead ions and 5.1 x 10-5M
silver ions, which would precipitate first, PbSO4 or Ag2SO4,
when a solution of K2SO4 is added one drop at a time? The Ksp
for PbSO4 is 1.62 x 10-8 and the Ksp for Ag2SO4 if 1.45 x 10-5.
PbSO4↔ Pb2+ + SO42- Ag2SO4↔ 2Ag+ + SO42-

1.62x10-8 =[Pb2+][ SO42-] 1.45x10-5= [Ag+]2[ SO42-]

1.62x10-8 =[4.3x10-4][ SO42-] 1.45x10-5= [5.1x10-5]2[ SO42-]

[ SO42-] = 3.9x10-5 M [ SO42-] = 5577 M

PbSO4 requires a much smaller concentration

of SO42- ions, so it will precipitate first.
That’s it for Ch. 17.

Review for Test