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Aqueous Ionic Equilibrium: Buffers, Ksp, Kf

Homework: Read Ch 17

Work out sample/practice exercises in the sections as you read,

Bonus Ch 17: 27, 29, 41, 45, 53, 59, 65, 83, 95, 97, 99, 103, 109, 111

Check MasteringChemistry for due dates

The Danger of Antifreeze:

Each year, thousands of pets and wildlife species die from consuming antifreeze.

Most brands of antifreeze contain ethylene glycol which has a sweet taste and an

initial effect of drunkenness

Ethylene Glycol is metabolized in the liver to glycolic acid HOCH2COOH. If

enough is consumed the natural buffering system in blood is overwhelmed, causing

acidosis and leading to death. It interferes with the equilibrium needed to transport

oxygen in the bloodstream…

HbH+(aq) + O2(g) HbO2(aq) + H

+(aq)

One treatment is to give the patient ethyl alcohol, which has a higher affinity for the

enzyme that catalyzes the metabolism of ethylene glycol

The Common Ion Effect:

We have studied the acid-base equilibrium reactions in just water.

What happens when we add other substances?

Additions of common ions and even uncommon (spectator) ions affect activity

values according to the Debye-Huckel equation and therefore affect the

equilibrium. This can be quite complicated so we generally assume the activity

value is equal to the molarity of a solution or the atmospheric pressure of a gas. We

will continue with this simplified assumption in the lecture. You will have an

opportunity in lab to calculate adjusted activity values in the PbI2 lab.

When the same ion is produced by two different components the common ion

effect takes place. By adding a salt containing a weak acid’s conjugate base

(product) into the solution, the dissociation of the weak acid will decrease and the

pH will increase according to Le Chatellier’s Principle.

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Example 1:

a) Solve for the pH of a 0.40 M HF in pure water? Ka = 6.8 x 10-5

b) Solve for the pH of a 0.40 M HF in 0.20 M KF solution? Ka = 6.8 x 10-5

Example 2:

a) Solve for the pH of a 0.80 M CH3NH2 in pure water? Kb = 4.4 x 10-4

b) Solve for the pH of a 0.80 M CH3NH2 in 0.30 M CH3NH3Cl solution?

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Buffered Solutions:

Buffered solutions resist change in pH even after a strong acid or a strong base

is added to the solution. This is extremely important in many situations. Examples:

Human Blood is buffered around a pH of 7.4, too high or too low causes sickness

and death. Fish tanks, medicines, enzymes, biological applications all generally

require a specific pH range to function properly. Without buffers, the simple act of

consuming too much acid by drinking soda or lemonade can have terrible effects.

Buffers act by neutralizing acid or base that is added to the buffered solution.

But just like everything else, there is a limit to what they can do, and eventually

the pH changes. Buffers have a buffer capacity. Both conjugates of weak acid –

base pair are needed to maintain a buffer. If you run out of either while adding a

strong acid or base then you have reached the end of the buffer capacity.

Many buffers are made by mixing a solution of a weak acid with a solution of

soluble salt containing its conjugate base anion

blood has a mixture of H2CO3 and HCO3 –

It is possible to buffer a solution at any pH. Buffers do not mean neutralized to

a pH around 7. Choose an appropriate weak acid or weak base conjugate pair

which has a pKa or pKb close to the pH required.

HC2H3O2 (aq) H+1

(aq) + C2H3O2-1

(aq)

NH3(aq) +H2O(l) NH4+1

(aq) +OH-1

(aq)

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Composition of Effective Buffers:

a) Weak acid and weak base conjugate pairs: Buffers need to eliminate strong

acids or strong bases when they are added. They can do this because a reaction

will always favor creating a weaker acid or weaker base. Adding a strong acid

will cause the weak conjugate base in the buffer to accept the extra H+1

, forming

its weak conjugate acid. Adding a strong base will cause the weak conjugate

acid to donate its H+1

, creating its weak conjugate base. Buffers will never be

created using the conjugate acid/base pairs of something strong/negligible.

b) Ratio of Conjugate acid base pairs should be within 1 to 10 or 10 to 1:

Generally a buffer solution will have close to equivalent concentrations of each

weak acid/weak base conjugate pair. The pH range should be within +1 or – 1 of

the pKa

c) Relatively high concentrations of each: A buffer will generally have a minimum

concentration of 0.05 M of each conjugate pair. The greater the concentrations,

the better the buffer capacity (amount of acid or base it is able to take on before

losing its buffering capabilities).

Example 3:

Identify which of the following pairs can function as a buffer. For those pairs

that cannot function as a buffer, explain why not.

a) HNO3 ; KNO3 e) CH3NH2 ; CH3NH3Br

b) NH4Cl ; NH3 f) H2CO3 ; HCO3-1

c) NaC2H3O2 ; HC2H3O2 g) NaCl ; NaNO3

d) HF ; HNO2 h) KOH ; H2O

Effective Buffer pH Range (Buffering Range):

The effective pH range of a buffer over which it can act is generally pKa -1 to

pKa+1 of the weak acid. This is within the 1:10 and 10: 1 ratio between weak

acid-base conjugates.

For an acid: the pKa = –log Ka

For a base: the pKb = –log Kb

Conversion: pKw = 14.00 = pKa + pKb for conjugate acid-base pairs

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Example 4:

Determine the effective buffer range:

pH range (pKa +/- 1)

pOH range (pKb +/-1)

a) NH4Cl ; NH3 Kb of NH3 = 1.8 x 10-5

b) NaC2H3O2 ; HC2H3O2 Ka of HC2H3O2 = 1.8 x 10-5

c) H2CO3 ; HCO3-1

Ka1 of H2CO3 = 4.3 x 10-7

d) CH3NH2 ; CH3NH3Br Kb of CH3NH2 = 4.4 x 10-4

e) Which combination would be the best choice to make a buffer with

a pH 5.00? a pH 10.75?

f) What ratio of the chosen combination would be required to make a

buffer with a pH 5.00? a pH 10.75?

g) If you mix 0.24 M of the acid, what is the necessary concentration

for the base?

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How Buffers Work:

Buffers work by applying Le Châtelier’s Principle to weak acid equilibrium

Buffer solutions contain significant amounts of the weak acid molecules, HA and

its conjugate base A-1

.

These molecules react with added acid or base to neutralize it

HA(aq) + OH−(aq) → A

−(aq) + H2O(l)

Henderson-Hasselbalch Equation to Calculate pH of Buffers :

Buffer solutions have a special formula to calculate pH. This is derived by taking

the negative log of both sides of the equilibrium constant using the RICE equation , it

assumes that the amount of change is small and negligible compared to the original

concentrations of the weak acid/weak base conjugate pairs

For ACIDS:

pH = pKa + log [ conjugate base ion/weak acid]

For BASES:

pOH = pKb + log [conjugate acid ion/weak base]

Example 5: Solve for the pH of the 0.40 M HF in 0.20 M KF solution in Example 1 using

the Henderson-Hasselbalch equation. Ka = 6.8 x 10-5

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Adding Strong Acids and Strong Bases to Buffers:

When adding a strong acid or a strong base to a buffer solution the conjugate weak

acid or base will decrease as it consumes/neutralizes the strong and its conjugate

pair will increase.

H+ added will decrease the conjugate weak base and increase the weak acid

concentrations

OH-1

added will decrease the conjugate weak acid and increase the weak base

concentrations

This can only occur as long as the concentrations remain in equilibrium.

Once the buffer capacity is breached you will no longer have a buffered

solution.

Example 6: Solve for the pH of 100 ml of the 0.40 M HF/ 0.20 M KF solution after the

following is added

a) 0.010 mole HCl

b) 0.010 mole NaOH

c) What amount in moles of NaOH would exceed the buffer capacity?

Titration: In an acid-base titration, a solution of

unknown concentration (titrant) is

slowly added to a solution of known

concentration from a burette until the

reaction is complete. The endpoint of

the titration is generally determined by

adding an indicator that changes color

at an appropriate pH.

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Indicators: Many dyes change color depending on the pH of the solution

Indicators are weak acid-conjugate base dyes that have one color in its acid form and

another color in its conjugate base form.

HIn (aq)+ H2O (l) In

(aq)+ H3O+

(aq)

The color of the solution depends on the relative concentrations of In:HIn

when In:HIn ≈ 1, the color will be mix of the colors of In

and HIn

when In:HIn > 10, the color will be mix of the colors of In

when In:HIn < 0.1, the color will be mix of the colors of HIn

For most titrations, the titration curve shows a very large change in pH for very

small additions of titrant near the equivalence point

An indicator can be used to determine the endpoint of the titration if it changes

color within the same range as the equivalence point: pKa of HIn ≈ pH at eq.

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Titration Curves:

A plot of pH vs. amount of added titrant

The inflection point of the curve is the equivalence point of the titration

The pH of the equivalence point depends on the pH of the salt solution

equivalence point of neutral salt, pH = 7

equivalence point of acidic salt, pH < 7

equivalence point of basic salt, pH > 7

Acid-Base Titration Curves:

StrongAcid/Strong Base:

Example 7: An acid-base titration experiment begins with 20.0 ml of 0.300 M HCl

in the flask and 0.150 M NaOH in the buret. Both are strong.

a) Draw the expected titration graph where pH is the y axis and ml of NaOH added

is the x axis, Sketch and label the titration curve including SA or SB, salt,

equivalence pt, pH at equiv pt <, =, or > 7, buffer, where color change of

indicator should occur.

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b) What are the initial [H+] and pH, in the flask? Clearly identify answers.

c) What is the pH after 18.0 ml of 0.150 M NaOH has been added to the flask

containing 20.0 ml of 0.300 M HCl?

d) What volume of 0.150 M NaOH must be added to the original flask of 20.0 ml

of 0.300 M HCl to reach the equivalence point?

e) Write the balanced whole, complete ionic, and net ionic equations of hydrolysis

of the salt sodium chloride (NaCl) and indicate whether the salt is acidic, basic,

or neutral.

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f) What is the pH of the solution in the flask at the equivalence point?

g) Which is the best indicator to use for determining the endpoint of this titration

Bromocresol green (pH 4-6) or phenolphthalein (pH 8-10) or either?

h) What is the pH after 45.0 ml of 0.150 M NaOH has been added to the original

flask containing 20.0 ml of 0.300 M HCl?

i) Titration curve for a SB/SA titration (reversed):

Strong Base (NaOH in flask) and Strong Acid (HCl in buret).

Draw the expected titration graph where pH is the y axis and ml of HCl added is

the x axis, Sketch and label the titration curve including SA or SB, salt,

equivalence pt, pH at equiv pt <, =, or > 7, buffer, where color change of

indicator should occur.

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Weak Acid/Strong Base:

Example 8: An acid-base titration experiment begins with 20.0 ml of 0.300 M HF

in the flask and 0.150 M NaOH in the buret. Ka for HF = 6.6 x 10-4.

a) Draw the expected titration graph where pH is the y axis and ml of NaOH added

is the x axis, Sketch and label the titration curve including WA or WB, SA or

SB, salt, equivalence pt, pH at equiv pt <, =, or > 7, buffer, where color change

of indicator should occur.

b) What are the initial [H+], [OH-], pH, pOH, and % ionization for the 0.300 M HF

in the flask? Clearly identify answers. Ka for HF = 6.6 x 10-4

c) What is the pH after 18.0 ml of 0.150 M NaOH has been added to the flask

containing 20.0 ml of 0.300 M HF?

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d) What volume of 0.150 M NaOH must be added to the original flask of 20.0 ml

of 0.300 M HF to reach the equivalence point?

e) Write the balanced whole, complete ionic, and net ionic equations of hydrolysis

of the salt sodium fluoride (NaF) and indicate whether the salt is acidic, basic,

or neutral.

f) Solve for the numerical Kb value for the net ionic equation in the previous step.

g) What is the pH of the solution in the flask at the equivalence point?

h) Which is the best indicator to use for determining the endpoint of this titration

Bromocresol green (pH 4-6) or phenolphthalein (pH 8-10)?

i) What is the pH after 45.0 ml of 0.150 M NaOH has been added to the original

flask containing 20.0 ml of 0.300 M HF?

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Strong Acid/Weak Base:

Example 9: An acid-base titration experiment begins with 20.0 ml of 0.300 M NH3

in the flask and 0.150 M HCl in the buret. Kb for NH3 = 1.8 x 10-5.

a) Draw the expected titration graph where pH is the y axis and ml of HCl added is

the x axis, Sketch and label the titration curve including WA or WB, SA or SB,

salt, equivalence pt, pH at equiv pt <, =, or > 7, buffer, where color change of

indicator should occur.

b) What are the initial [H+], [OH-], pH, pOH, and % ionization for the 0.300 M

NH3 in the flask? Clearly identify answers. Kb for NH3 = 1.8 x 10-5

c) What is the pH after 18.0 ml of 0.150 M HCl has been added to the flask

containing 20.0 ml of 0.300 M NH3?

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d) What volume of 0.150 M HCl must be added to the original flask of 20.0 ml of

0.300 M NH3 to reach the equivalence point?

e) Write the balanced whole, complete ionic, and net ionic equations of hydrolysis

of the salt ammonium chloride (NH4Cl) and indicate whether the salt is acidic,

basic, or neutral.

f) Solve for the numerical Ka value for the net ionic equation in the previous step.

g) What is the pH of the solution in the flask at the equivalence point?

h) Which is the best indicator to use for determining the endpoint of this titration

Bromocresol green (pH 4-6) or phenolphthalein (pH 8-10)?

i) What is the pH after 45.0 ml of 0.150 M HCl has been added to the original

flask containing 20.0 ml of 0.300 M NH3?

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Polyprotic Weak Acid/Strong Base:

Polyprotic acids have two or more acidic hydrogens that may be lost in a titration

with the base. Some special “extra” equations associated with a diprotic acid (H2A)

follows…

At the first half equivalence point: pH = pKa1

At the first equivalence point (salt, NaHA): pH = (pKa1 + pKa2)/2

At the second half equivalence point: pH = pKa2

At the second equivalence point (salt, Na2A): Kb = Kw/Ka2

Example 10: An acid-base titration experiment begins with 20.0 ml of 0.300 M

H2CO3 in the flask and 0.400 M NaOH in the buret.

a) Sketch the expected titration curve where pH is the y axis and ml of NaOH

added is the x axis.

b) Solve for the initial pH of the 20.0 ml of 0.300 M H2CO3

b) Solve for the pH at first half equivalence point

c) Solve for the pH at the first equivalence point

d) Solve for the pH at second half equivalence point

e) Solve for the pH at the second equivalence point

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Solubility Equilibria, Ksp:

The Solubility Product equilibrium constant (Ksp) involves slightly soluble solids,

(ionic compounds that we earlier labeled as solid when looking at the solubility

rules). These compounds dissolve less than 0.020 M at 25°C in water. All ionic

compounds dissolve in water to some degree however; many compounds have such

low solubility in water that we classify them as insoluble. A small amount of the

“insoluble” ionic compound dissolves into its individual ions creating equilibrium

with the ionic solid as a reactant and its ions in the product. Ksp values are

generally small (much less than 1) so few ions (products) exist at equilibrium.

Ksp reactions affect the world around us, examples include…dissolving tooth

enamel in acid (soda) solutions, development of kidney stones, and cave

formations.

• For an ionic solid MnXm, the dissociation reaction is:

MnXm(s) nMm+

(aq) + mXn−

(aq)

• The solubility product would be

Ksp = [Mm+

]n[X

n−]

m

Example 11:

Write Ksp reactions and Ksp expressions.

Example: PbI2 (s) Pb+2

(aq) + 2 I-1

(aq) Ksp = [Pb+2

][I-1

]2

a) CuCO3 (s)

b) Fe(OH)3 (s)

c) Ca3(PO4)2 (s)

Solubility:

The solubility (g/L) or molar solubility (mol/L) is the quantity of the solid that

dissolves to form a saturated solution. When comparing Ksp values to determine

relative solubility strengths , the number of ions the solid breaks into is important.

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Example 12:

Comparing the Ksp, estimate which solid you think will have the greatest molar

solubility before any calculations are performed.

Second calculate the molar solubilities and equilibrium ion concentrations for the

solids in just water.

Given:

AgCl Ksp = 1.8 x 10-10

Ag2CrO4 Ksp = 1.2 x 10-12

Ca3(PO4)2 Ksp = 2.0 x 10-29

Example 13:

Saturated Mg3(AsO4)2 contains 16 mg/L of Mg3(AsO4)2 at 25°C. Calculate the

molar solubility and Ksp. (MW = 350.7)

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Reaction Quotient, Qsp: When two ionic solutions are mixed, they may form a solid in a double

displacement reaction. To determine if a precipitate will form, compare Qsp to Ksp

Qsp > Ksp precipitate will form: The human eye is not very sensitive, If

Qsp/Ksp > 1000 the solid will become visible (cloudy) to the

naked eye.

Qsp = Ksp solution is just saturated, but no solid

Qsp < Ksp Unsaturated solution, no solid

Note: Some solutions with Q > Ksp will not precipitate unless disturbed. These

are supersaturated solutions

Example 14:

Will a CaSO4 solid form, will it be visible to a human eye if the following

separate solutions are mixed in equal volumes (total volume doubles) in

each part.

Ksp for CaSO4 = 2.4 x 10-5

a) 0.0040 M Ca(NO3)2 added to 0.0040 M Na2SO4

b) 0.040 M Ca(NO3)2 added to 0.040 M Na2SO4

c) 0.40 M Ca(NO3)2 added to 0.40 M Na2SO4

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Factors that Affect Solubility:

Temperature:

The Ksp value will change as the temperature changes. For example, PbCl2

has a greater solubility as the temperature increases. This is useful in our

separation of Group 1 ions lab.

Ksp at 25°C Molar solubility

at 25°C

Ksp at 100°C Molar solubility

at 100°C

PbCl21.7 x 10

-5 0.016 M 6.9 x 10

-3 0.12 M

Common ion:

Solubility decreases in the presence of a common ion.

Solubility increases if a basic compound is added to an acidic solution or

an acidic compound is added to a basic solution.

Effect of pH:

For insoluble ionic hydroxides, the higher the pH, the lower the solubility

of the ionic hydroxide

For insoluble ionic hydroxides, the lower the pH, the higher the solubility

M(OH)n(s) Mn+

(aq) + nOH−(aq)

For insoluble ionic compounds that contain anions of weak acids, the

lower the pH, the higher the solubility

M2(CO3)n(s) 2 Mn+

(aq) + nCO32−

(aq)

H3O+(aq) + CO3

2− (aq) HCO3

− (aq) + H2O(l)

Example 15: Ca(OH)2 Ksp = 6.5 x 10-6

Solve for the molar solubility, ion concentrations, and pH for…

(a) solid Ca(OH)2 in water

(b) solid Ca(OH)2 in a solution of 0.10 M CaCl2 (aq)

(c) solid Ca(OH)2 in a buffered solution of pH = 11.20

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Simultaneous Equilibria:

Many chemical reactions involve simultaneous Equilibria, more than one

competing at the same time. An example to consider is a slightly soluble salt

(Ksp) that ionizes to make a weak acid or weak base (Ka or Kb).

Example 16: Mg(OH)2 Ksp = 1.6 x 10-12

NH3 Kb = 1.8 x 10-5

Consider that 0.10 mole NH3 and 0.10 mole Mg(NO3)2 are added to enough

water to dilute to 1.00 L.

a) Will the Mg(OH)2 precipitate and will it be visible to the eye? For this

problem the [Mg+2

] = 0.10M, the [OH-1

] must be solved using the Kb

reaction for NH3

b) How many moles of NH4Cl must be added to prevent the precipitation of

the Mg(OH)2 in the 1.00L solutions containing 0.10 mole NH3 and 0.10

mole Mg(NO3)2? By buffering the solution the [OH-1

] concentration can

remain low enough to avoid precipitation. Use your known Ksp to solve

for the max [OH-1

]: Ksp = [0.10M Mg+2

][OH-1

]2. Then convert to the

pOH and use the buffer equation to solve for the minimum molarity of

NH4Cl required to avoid precipitation.

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Formation of Complex Ions:

• Transition metals tend to be good Lewis acids

• They often bond to one or more H2O molecules or other Lewis base ligand to form

a complex ion…

Ag+(aq) + 2 H2O(l) Ag(H2O)2

+(aq)

• Ions that form by combining a cation with several anions or neutral molecules are

called complex ions

• The attached Lewis base ions or molecules are called ligands

• If a ligand is added to a solution that forms a stronger bond than the current ligand,

it will replace the current ligand

Ag(H2O)2+

(aq) + 2 NH3(aq) Ag(NH3)2+

(aq) + 2 H2O(l)

Ag+

(aq) + 2 NH3(aq) Ag(NH3)2+

(aq)

(water is often left out of the equilibrium )

• The equilibrium constant for the formation of a complex ion is called the formation

constant, Kf

• The reverse reaction is the dissociation constant (Kd) equilibrium; Kd = Kf-1

.

• Many metal ions have an ability to form a complex ion, which is an assembly of a

metal ion (a Lewis acid) and some simple multiple of the Lewis bases (those able to

donate a share of an electron pair) bonded to it. The Lewis bases can include H2O,

OH-1

, CN-1

, NH3, and others. Appendix II lists many Kf values (Kf >>1)

Complex ion constants: Example 17:

Fill in the chart:

Complex Ion Kf Kd

[Cu(NH3)4+2

1.7 x 1013

[HgI42−] 2 x 1030

[Pb(OH)3-1] 8 x 1013

[Fe(CN)6-3] 5 x 10-44

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Equilibrium of Complex Ions:

Example 18:

50.0 mL of 1.5 x 10−3

M Cu(NO3)2 is mixed with 100.0 mL of 0.24 M NH3

What is the [Cu+2

] at equilibrium?

b) Use the appropriate values of Ksp and Kf to find the equilibrium constant for

the reaction… PbCl2 (s) + 3 OH-1 (aq) Pb(OH)3-1 (aq) + 2 Cl-1 (aq)

Are reactants or products preferred?

c) Use the appropriate values of Ksp and Kf to find the equilibrium constant for

the reaction… Cu(OH)2 (s) + 4 NH3 (aq) Cu(NH3)4+2

(aq) + 2 OH-1

(aq)

Are reactants or products preferred?

a)

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Dissolving precipitates:

Precipitates dissolve when Qsp < Ksp. The following three reaction types

remove ions from the solution to shift the reaction forward and dissolve

the solid.

1) Convert ions to weak electrolytes:

Insoluble metal hydroxides dissolve in acids…

Zn(OH)2 (s) + 2H+1

2 H2O (l) + Zn+2

(aq)

In this case the OH-1

ions are removed by reacting with the acid and

creating water.

2) Convert ions to another species by oxidation reduction reactions:

Insoluble metal sulfides dissolve in hot HNO3…

3PbS (s) + 8H+1

+ 2 NO3-1 3 S (s) + 2NO (g) +4 H2O (l) +3 Pb

+2

In this case the S-2

ions are removed by reduction reaction and

converting the ions to solid sulfur.

3) Convert ions into another form a complex ion:

The solubility of an ionic compound that contains a metal cation that

forms a complex ion increases in the presence of its ligands

AgCl(s) Ag+

(aq) + Cl−

(aq) Ksp = 1.77 x 10−10

Ag+

(aq) + 2 NH3(aq) Ag(NH3)2+

(aq) Kf = 1.7 x 107

Adding NH3 to a solution in equilibrium with AgCl(s) increases the

solubility of AgCl by consuming free Ag+ ions.

Another example… +2

(aq) + 2 OH-1

(aq)Cu(OH)2 (s) + 4 NH3 (aq) Cu(NH3)4

As seen in example 18c).

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Amphoteric Oxides and Hydroxides:

Substances that behave as both an acid and base are amphoteric

Many insoluble oxides and hydroxides can be made to dissolve in either acid or

base solutions.

• All metal hydroxides become more soluble in acidic solution by

removing OH−

• Some metal hydroxides become more soluble in basic solution, acting as

a Lewis base forming a complex ion

• Some cations that form amphoteric hydroxides include Al3+

, Cr3+

, Zn2+

,

Pb2+

, and Sb2+

Metal hydroxides can…

Dissolve in acid to create the metal ions…

Al(OH)3 + 3 H+1

3 H2O (l) + Al+3

(aq)

Dissolve in strong base solutions to create complex ions…

Al(OH)3 + OH-1

Al(OH)4-1

(aq)

• Small metal cations (example Al+3

) are hydrated in water to form an

acidic solution.

Al(H2O)63+

(aq) + H2O(l) Al(H2O)5(OH)2+

(aq) + H3O+

(aq)

• Additional OH− drives the equilibrium to the right

Al(H2O)5(OH)2+

(aq) + OH−

(aq) Al(H2O)4(OH)2+

(aq) + H2O (l)

Al(H2O)4(OH)2+

(aq) + OH−

(aq) Al(H2O)3(OH)3(s) + H2O (l)

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Precipitation and Separation of Ions: • A solution containing several different cations can often be separated by addition of

a reagent that will form an insoluble salt with one of the ions, but not the others

• A successful reagent can precipitate with more than one of the cations, as long as

their Ksp values are significantly different

• The end of this chapter goes into details of qualitative analysis to determine the

unknown ions in a solution. You should understand the general concepts and be

able to work out a problem dealing with fractional precipitation, but the test will not

question specific directions on how to separate specific ions. Labs will give you

experience with separations of cations and anions.

Qualitative Analysis is an analytical scheme that utilizes selective precipitation to

identify the ions present in a solution is called a qualitative analysis scheme

• A sample containing several ions is subjected to the addition of several precipitating

agents

• Addition of each reagent causes one of the ions present to precipitate out

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Fractional Precipitation (A Separation Process):

It is possible to separate ion mixtures if one ion will precipitate before

another due to the difference in molar solubilities.

Another influence is temperature: Ksp values alter as temperature changes.

Solid Ksp at 25°C Molar solubility

at 25°C

Ksp at 100°C Molar solubility

at 100°C

AgCl 1.8 x 10-10

1.3 x 10-5

M 2.3 x 10-8

1.5 x 10-4

M

PbCl21.7 x 10

-5 1.6 x 10

-2 M 6.9 x 10

-3 0.12 M

We use this difference in lab as we separate the Pb+2

ions by heating the

chloride solids in hot water.

Example 19:

A solution contains 0.10 M Ag+1

and 0.10 M Pb+2

.

HCl is added to form precipitates (ignore volume change due to HCl added).

Use the table above to solve for the concentration of chloride ion that is

required to initiate precipitation of each cation at 25°C?

What percent of the first ion to form a precipitate remains when the second

just begins to precipitate at 25°C