lecture 3 acids and base - chemistry for all…. · lecture 3: acids and bases who theory: acid=...

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LECTURE 3: ACIDS AND BASES

Who Theory: Acid=

When

Arrhenius increases H+ 1880’s

Brønsted proton donor 1923

Lowry proton donor 1923

Lewis electron-pair acceptor 1923

DEFINITIONS OF AN ACID

Johannes Nicolaus Brønsted !(February 22, 1879-December 17, 1947)!Danish physical chemist!

Svante August Arrhenius !(February 19, 1859 – October 2, 1927)!Swedish chemist; Nobel Prize in Chemistry, 1903!* Arrhenius equation (activation energy)

* Greenhouse effect

Thomas Martin Lowry !(October 26, 1874–November 2, 1936)!English organic chemist!

Gilbert Newton Lewis !(October 23, 1875-March 23, 1946)!American physical chemist!

•  Arrhenius acids and bases – Acid: Substance that, when dissolved in water,

increases the concentration of hydrogen ions (protons, H+).

– Base: Substance that, when dissolved in water, increases the concentration of hydroxide ions.

ARRHENIUS DEFINITIONS

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•  Brønsted–Lowry: must have both 1. an Acid: Proton donor

and 2. a Base: Proton acceptor

BRØNSTED–LOWRY DEFINITION

The Brønsted-Lowry acid donates a proton,

while the Brønsted-Lowry base accepts it.

Brønsted-Lowry acids and bases are always paired.

Which is the acid and which is the base in each of these rxns?

A Brønsted–Lowry acid… …must have a removable (acidic) proton.

HCl, H2O, H2SO4 A Brønsted–Lowry base…

…must have a pair of nonbonding electrons. NH3, H2O

If it can be either…

...it is amphiprotic.

HCO3–

HSO4 –

H2O

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What Happens When an Acid Dissolves in Water?

•  Water acts as a Brønsted–Lowry base and abstracts a proton (H+) from the acid.

•  As a result, the conjugate base of the acid and a hydronium ion are formed.

•  From the Latin word conjugare, meaning “to join together.” •  Reactions between acids and bases always yield their

conjugate bases and acids.

CONJUGATE ACIDS AND BASES

•  Strong acids are completely dissociated in water. –  Their conjugate bases are quite

weak. •  Weak acids only dissociate

partially in water. –  Their conjugate bases are weak

bases.

ACID AND BASE STRENGTH

•  Substances with negligible acidity do not dissociate in water. –  Their conjugate bases are

exceedingly strong.

ACID BASE STRENGTH

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In any acid-base reaction, the equilibrium favors the reaction that moves the proton to the stronger base.

HCl(aq) + H2O(l) ⎯⎯→ H3O+(aq) + Cl–(aq)

H2O is a much stronger base than Cl–, so the equilibrium lies so far to the right K is not measured (K>>1).

ACID BASE STRENGTH

Acetate is a stronger base than H2O, so the equilibrium favors the left side (K<1). The stronger base “wins” the proton.

HC2H3O2(aq) + H2O H3O+(aq) + C2H3O2–(aq)

ACID BASE STRENGTH

As we have seen, water is amphoteric. •  In pure water, a few molecules act as bases and a few act

as acids.

This process is called autoionization.

AUTOIONIZATION OF WATER

•  The equilibrium expression for this process is Kc = [H3O+] [OH–]

•  This special equilibrium constant is referred to as the

ion-product constant for water, Kw. •  At 25°C, Kw = 1.0 × 10-14

EQUILIBRIUM CONSTANT FOR WATER

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pH is defined as the negative base-10 logarithm of the hydronium ion concentration.

pH = –log [H3O+]

pH pH

•  In pure water,

Kw = [H3O+] [OH–] = 1.0 × 10-14

•  Because in pure water [H3O+] = [OH-],

[H3O+] = (1.0 × 10-14)1/2 = 1.0 × 10-7

pH •  Therefore, in pure water,

pH = –log [H3O+] = –log (1.0 × 10-7) = 7.00

•  An acid has a higher [H3O+] than pure water, so its pH is <7 •  A base has a lower [H3O+] than pure water, so its pH is >7.

pH

These are the pH values for several common substances.

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Other “p” Scales

•  The “p” in pH tells us to take the negative log of the quantity (in this case, hydronium ions).

•  Some similar examples are – pOH –log [OH-] – pKw –log Kw

Watch This!

Because [H3O+] [OH−] = Kw = 1.0 × 10-14,

we know that

–log [H3O+] + – log [OH−] = – log Kw = 14.00

or, in other words, pH + pOH = pKw = 14.00

If you know one, you know them all:

[H+] [OH-] pH

pOH

How Do We Measure pH?

–  Litmus paper •  “Red” paper turns blue

above ~pH = 8 •  “Blue” paper turns red below

~pH = 5 –  An indicator

•  Compound that changes color in solution.

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How Do We Measure pH? pH meters measure the voltage in the

solution

Strong Acids •  You will recall that the seven

strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4.

•  These are strong electrolytes and exist totally as ions in aqueous solution.

•  For the monoprotic strong acids, [H3O+] = [acid].

Strong Bases •  Strong bases are the soluble hydroxides, which are the alkali metal

(NaOH, KOH)and heavier alkaline earth metal hydroxides (Ca(OH)2, Sr(OH)2, and Ba(OH)2).

•  Again, these substances dissociate completely in aqueous solution. [OH-] = [hydroxide added].

Dissociation Constants •  For a generalized acid dissociation,

the equilibrium expression is •  This equilibrium constant is called the acid-

dissociation constant, Ka.

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DISSOCIATION CONSTANTS

The greater the value of Ka, the stronger the acid.

CALCULATING Ka FROM THE pH

•  The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature.

•  We know that

Calculating Ka from the pH

The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature.

To calculate Ka, we need all equilibrium concentrations. We can find [H3O+], which is the same as [HCOO−], from the

pH.

Calculating Ka from the pH

pH = –log [H3O+] – 2.38 = log [H3O+]

10-2.38 = 10log [H3O+] = [H3O+]

4.2 × 10-3 = [H3O+] = [HCOO–]

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Calculating Ka from pH

In table form:

[HCOOH], M [H3O+], M [HCOO−], M

Initially 0.10 0 0

Change –4.2 × 10-3 +4.2 × 10-3 +4.2 × 10-3

At Equilibrium 0.10 – 4.2 × 10-3 = 0.0958 = 0.10

4.2 × 10-3 4.2 × 10 - 3

Calculating Ka from pH

[4.2 × 10-3] [4.2 × 10-3] [0.10]

Ka =

= 1.8 × 10-4

CALCULATING PERCENT IONIZATION

In the example: [A-]eq = [H3O+]eq = 4.2 × 10-3 M [A-]eq + [HCOOH]eq = [HCOOH]initial = 0.10 M

Calculating Percent Ionization

Percent Ionization = × 100

4.2 × 10-3 0.10

= 4.2%

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Calculating pH from Ka

Calculate the pH of a 0.30 M solution of acetic acid, C2H3O2H, at 25°C.

Ka for acetic acid at 25°C is 1.8 × 10-5. Is acetic acid more or less ionized than formic acid (Ka=1.8

x 10-4)?


The equilibrium constant expression is:


Use the ICE table: [C2H3O2], M [H3O+], M [C2H3O2

−], M

Initial 0.30 0 0

Change –x +x +x

Equilibrium 0.30 – x x x



−], M

Initial 0.30 0 0

Change –x +x +x

Equilibrium 0.30 – x x x

Simplify: how big is x relative to 0.30?

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CALCULATING pH FROM Ka


−], M

Initial 0.30 0 0

Change –x +x +x

Equilibrium 0.30 – x ≈ 0.30 x x



Now,

(1.8 × 10-5) (0.30) = x2

5.4 × 10-6 = x2 2.3 × 10-3 = x

Check: is approximation ok?

CALCULATING Ka from pH

The pH of a 0.01M hypochlorous acid (HClO) is 4.76. Calculate its Ka.

SAMPLE PROBLEM

Calculate the pH of a 0.02M Hydroflouric acid solution. Ka (HF) = 6.8 x 10-4

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POLYPROTIC ACIDS

Have more than one acidic proton. If the difference between the Ka for the first dissociation and

subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation.

SAMPLE PROBLEM

Calculate the pH of a 0.1 M H3PO4. Ka1 = 7.5 x 10-3

Ka2 = 6.8 x 10-8 Ka3 = 4.2 x 10-13

WEAK BASES

Bases react with water to produce hydroxide ion.

WEAK BASES

The equilibrium constant expression for this reaction is

where Kb is the base-dissociation constant.

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WEAK BASES Kb can be used to find [OH–] and, through it, pH.

pH of Basic Solutions

What is the pH of a 0.15 M solution of NH3?

[NH4+] [OH−]

[NH3] Kb = = 1.8 × 10-5

pH OF BASIC SOLUTIONS

Tabulate the data.

[NH3], M [NH4+], M [OH−], M

Initial 0.15 0 0

Equilibrium 0.15 - x ≈ 0.15 x x


pH OF BASIC SOLUTIONS

(1.8 × 10-5) (0.15) = x2 2.7 × 10-6 = x2 1.6 × 10-3 = x2

(x)2 (0.15) 1.8 × 10-5 =

Check: is approximation ok?

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pH of Basic Solutions

Therefore, [OH–] = 1.6 × 10-3 M pOH = –log (1.6 × 10-3) pOH = 2.80 pH = 14.00 – 2.80 pH = 11.20

SAMPLE PROBLEM

A 0.01M solution of caffeine, a weak organic base, has a pH of 11.3. Calculate its dissociation constant.

Ka and dissociation constant of a conjugate base

HCN + H2O CN- + H3O+

CN- + H2O HCN + OH- KaKb = Kw

SALT HYDROLYSIS

Salts of strong acids and bases Salts of strong base and a weak acid Salt of strong acid and a weak base.

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Ka and Kb are linked:

Combined reaction = ?

Ka and Kb are linked:

Combined reaction = ?

Ka and Kb

Ka and Kb are related in this way: Ka × Kb = Kw

Therefore, if you know one of them, you can calculate the other.

SAMPLE PROBLEM

Calculate the pH of 0.10 M NH4Cl solution. Kb(NH3)=1.8 x 10-5

Calculate the % hydrolysis of a 0.36M CH3COONa. Ka=1.75 x 10-5

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Polyprotic acids

H3PO4 + H2O H2PO4- + H3O+

𝐾↓𝑎1 = 7.5 x 10 -3

H2PO4- + H2O HPO4

2- + H3O+ 𝐾↓𝑎2  = 6.2 x 10-8

HPO42- + H2O PO4

3- + H3O+

𝐾↓𝑎3 = 4.8 x 10-13

Polyprotic Acids Have more than one acidic proton. If the difference between the Ka for the first dissociation and

subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation.

SAMPLE PROBLEM What is the pH of 0.025 M H2S solution? K1= 5.7 x 10-8 K2 = 1.2 x 10-15

SAMPLE PROBLEM What is the pH of 0.012 M Na2CO3 solution? K1= 4.2 x 10-7 K2 = 4.8 x 10-11

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SAMPLE PROBLEM A 50 ml of 0.05M formic acid solution (Ka = 1.77 x 10-4) is titrated with 0.05 M NaOH solution. What is the pH at equivalence point?

A 0.020 M solution of niacin has a pH of 3.26. (a) What percentage of the acid is ionized in this solution? (b) What is the acid-dissociation constant, Ka, for niacin?

PRACTICE EXERCISES 1. Niacin, one of the B vitamins, has the following molecular structure:

2. What is the pH of (a) a 0.028 M solution of NaOH, (b) a 0.0011 M solution of Ca(OH)2? What percentage of the bases are ionized?

3. Calculate the percentage of HF molecules ionized in (a) a 0.10 M HF solution, (b) a 0.010 M HF solution. Ka for HF is 6.8 x10-4.

Reactions of Anions with Water

•  Anions are bases. •  As such, they can react with water in a hydrolysis

reaction to form OH– and the conjugate acid:

X–(aq) + H2O(l) HX(aq) + OH–(aq)

Reactions of Cations with Water •  Cations with acidic protons (like

NH4+) lower the pH of a solution by

releasing H+.

•  Most metal cations (like Al3+) that are hydrated in solution also lower the pH of the solution; they act by associating with H2O and making it release H+.

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Reactions of Cations with Water •  Attraction between nonbonding electrons

on oxygen and the metal causes a shift of the electron density in water.

•  This makes the O-H bond more polar and the water more acidic.

•  Greater charge and smaller size make a cation more acidic.

Effect of Cations and Anions

1.  An anion that is the conjugate base of a strong acid will not affect the pH.

2.  An anion that is the conjugate base of a weak acid will increase the pH.

3.  A cation that is the conjugate acid of a weak base will decrease the pH.

Effect of Cations and Anions

4.  Cations of the strong Arrhenius bases will not affect the pH.

5.  Other metal ions will cause a decrease in pH.

6.  When a solution contains both the conjugate base of a weak acid and the conjugate acid of a weak base, the affect on pH depends on the Ka and Kb values.

What effect on pH? Why?

An anion that is the conjugate base of a strong acid does not affect pH. = very weak base

An anion that is the conjugate base of a weak acid increases pH. = strong base A cation that is the conjugate acid of a weak base decreases pH. = strong acid

Cations of the strong Arrhenius bases (Na+, Ca2+) do not affect pH.

= very weak acid (not really acidic at all)

Other metal ions cause a decrease in pH. = moderate bases (cations)

Weak acid + weak base Depends on Ka and Kb

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Factors Affecting Acid Strength

•  The more polar the H-X bond and/or the weaker the H-X bond, the more acidic the compound.

•  Acidity increases from left to right across a row and from top to bottom down a group.


In oxyacids, in which an OH is bonded to another atom, Y,

the more electronegative Y is, the more acidic the acid.


For a series of oxyacids, acidity increases with the number of oxygens.

Factors Affecting Acid Strength Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the conjugate acid more acidic.

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Lewis Acids

•  Lewis acids are defined as electron-pair acceptors. •  Atoms with an empty valence orbital can be Lewis acids. •  A compound with no H’s can be a Lewis acid.

Lewis Bases

•  Lewis bases are defined as electron-pair donors. •  Anything that is a Brønsted–Lowry base is also a Lewis base. (B-

L bases also have a lone pair.) •  Lewis bases can interact with things other than protons.

The Common-Ion Effect

•  Consider a solution of acetic acid:

•  If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left.

HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq)


“The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.”

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Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.

Ka for HF is 6.8 × 10−4.

[H3O+] [F−] [HF] Ka = = 6.8 × 10-4


Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M.

[HF], M [H3O+], M [F−], M

Initially 0.20 0.10 0

Change −x +x +x At Equilibrium 0.20 − x ≈ 0.20 0.10 + x ≈ 0.10 x

HF(aq) + H2O(l) H3O+(aq) + F−(aq)


= x

1.4 × 10−3 = x

(0.10) (x) (0.20) 6.8 × 10−4 =

(0.20) (6.8 × 10−4) (0.10)


•  Therefore, [F−] = x = 1.4 × 10−3

[H3O+] = 0.10 + x = 0.10 + 1.4 × 10−3 = 0.10 M •  So, pH = −log (0.10)

pH = 1.00

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Buffers:

•  Solutions of a weak conjugate acid-base pair.

•  They are particularly resistant to pH changes, even when strong acid or base is added.

Buffers

If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.

Buffers

If acid is added, the F− reacts to form HF and water.

Buffer Calculations

Consider the equilibrium constant expression for the dissociation of a generic acid, HA:

[H3O+] [A−] [HA] Ka =

HA + H2O H3O+ + A−

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Buffer Calculations

Rearranging slightly, this becomes

[A−] [HA] Ka = [H3O+]

Taking the negative log of both side, we get

[A−] [HA] −log Ka = −log [H3O+] + −log

pKa pH acid

base

Buffer Calculations

•  So pKa = pH − log [base]

[acid] •  Rearranging, this becomes

pH = pKa + log [base] [acid]

•  This is the Henderson–Hasselbalch equation.

Henderson–Hasselbalch Equation

What is the pH of a buffer that is 0.12 M in lactic acid, HC3H5O3, and 0.10 M in sodium lactate? Ka for lactic acid is 1.4 × 10−4.

Henderson–Hasselbalch Equation

pH = pKa + log [base] [acid]

pH = −log (1.4 × 10−4) + log (0.10) (0.12)

pH = 3.85 + (−0.08) pH = 3.77

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pH Range

•  The pH range is the range of pH values over which a buffer system works effectively.

•  It is best to choose an acid with a pKa close to the desired pH.

When Strong Acids or Bases Are Added to a Buffer…

…it is safe to assume that all of the strong acid or base is consumed in the reaction.

Addition of Strong Acid or Base to a Buffer

1.  Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution.

2.  Use the Henderson–Hasselbalch equation to determine the new pH of the solution.

Calculating pH Changes in Buffers A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. a) Calculate the pH of this solution after 0.020 mol of NaOH is added. Ka = 1.8 x 10-5

b) calculate the pH after 0.020 mole HCl is added.

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Calculating pH Changes in Buffers

Before the reaction, since mol HC2H3O2 = mol C2H3O2

−

pH = pKa = −log (1.8 × 10−5) = 4.74

Calculating pH Changes in Buffers The 0.020 mol NaOH will react with 0.020 mol of the acetic acid:

HC2H3O2(aq) + OH−(aq) ⎯⎯→ C2H3O2−(aq) + H2O(l)

HC2H3O2 C2H3O2− OH−

Before reaction 0.300 mol 0.300 mol 0.020 mol

After reaction 0.280 mol 0.320 mol 0.000 mol

Calculating pH Changes in Buffers Now use the Henderson–Hasselbalch equation to calculate the new pH:

pH = 4.74 + log (0.320) (0. 200)

pH = 4.74 + 0.06 pH = 4.80

Titration

A known concentration of base (or acid) is slowly added to a solution of acid (or base).

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Titration

A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

Titration of a Strong Acid with a Strong Base

From the start of the titration to near the equivalence point, the pH goes up slowly.


Just before and after the equivalence point, the pH increases rapidly.


At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.

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As more base is added, the increase in pH again levels off.

Titration of a Weak Acid with a Strong Base

•  Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed.

•  The pH at the equivalence point will be >7.

•  Phenolphthalein is commonly used as an indicator in these titrations.


At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time.


With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

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Titration of a Weak Base with a Strong Acid

•  The pH at the equivalence point in these titrations is < 7.

•  Methyl red is the indicator of choice.

Titrations of Polyprotic Acids

In these cases there is an equivalence point for each dissociation.

Solubility Equilibria

Solubility Rules •  Salts are generally more soluble in HOT water (Gases are more

soluble in COLD water) •  Alkali Metal salts are very soluble in water.

NaCl, KOH, Li3PO4, Na2SO4 etc... •  Ammonium salts are very soluble in water.

NH4Br, (NH4)2CO3 etc… •  Salts containing the nitrate ion, NO3

-, are very soluble in water. •  Most salts of Cl-, Br- and I- are very soluble in water - exceptions are

salts containing Ag+ and Pb2+. soluble salts: FeCl2, AlBr3, MgI2 etc... “insoluble” salts: AgCl, PbBr2 etc...

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Dissolving a salt... •  A salt is an ionic compound - usually a

metal cation bonded to a non-metal anion. •  The dissolving of a salt is an example of

equilibrium. •  The cations and anions are attracted to

each other in the salt. •  They are also attracted to the water

molecules. •  The water molecules will start to pull out

some of the ions from the salt crystal.

•  At first, the only process occurring is the dissolving of the salt - the dissociation of the salt into its ions.

•  However, soon the ions floating in the water begin to collide with the salt crystal and are “pulled back in” to the salt. (precipitation)

•  Eventually the rate of dissociation is equal to the rate of precipitation.

•  The solution is now “saturated”. It has reached equilibrium.

Solubility Equilibrium: Dissociation = Precipitation

In a saturated solution, there is no change in amount of solid precipitate at the bottom of the beaker. Concentration of the solution is constant. The rate at which the salt is dissolving into solution equals the rate of precipitation.

Dissolving NaCl in water

Na+ and Cl - ions surrounded by

water molecules

NaCl Crystal

Dissolving silver sulfate, Ag2SO4, in water •  When silver sulfate dissolves it dissociates into ions. When the

solution is saturated, the following equilibrium exists:

Ag2SO4 (s) D 2 Ag+ (aq) + SO42- (aq)

•  Since this is an equilibrium, we can write an equilibrium expression

for the reaction:

Ksp = [Ag+]2[SO42-]

Notice that the Ag2SO4 is left out of the expression! Why? Since K is always calculated by just multiplying concentrations, it is called a “solubility

product” constant - Ksp.

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Writing solubility product expressions...

•  For each salt below, write a balanced equation showing its dissociation in water.

•  Then write the Ksp expression for the salt.

Iron (III) hydroxide, Fe(OH)3

Nickel sulfide, NiS

Silver chromate, Ag2CrO4

Zinc carbonate, ZnCO3

Calcium fluoride, CaF2

Some Ksp Values

Note: These are experimentally determined, and may be slightly different on a different Ksp table.

Calculating Ksp of Silver Chromate •  A saturated solution of silver chromate, Ag2CrO4, has [Ag+] = 1.3 x

10-4 M. What is the Ksp for Ag2CrO4?

Ag2CrO4 (s) D 2 Ag+ (aq) + CrO42- (aq)

---- ----

1.3 x 10-4 M

Calculating the Ksp of silver sulfate •  The solubility of silver sulfate is 0.014 mol/L. This means that 0.0144

mol of Ag2SO4 will dissolve to make 1.0 L of saturated solution. Calculate the value of the equilibrium constant, Ksp for this salt.

Ag2SO4 (s) D 2 Ag+ (aq) + SO42- (aq)

--- ---

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Calculating solubility, given Ksp •  The Ksp of NiCO3 is 1.4 x 10-7 at 25°C. Calculate its molar solubility.

NiCO3 (s) D Ni2+ (aq) + CO32- (aq)

--- ---

Other ways to express solubility... •  We just saw that the solubility of nickel (II) carbonate is 3.7 x 10-4 mol/

L. What mass of NiCO3 is needed to prepare 500 mL of saturated solution?

0.022 g of NiCO3 will dissolve to make 500 mL solution.

g 0.022 NiCO mol 1

g 118.72 x L 0.500 x L1

NiCO mol10x3.73

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=−

Calculate the solubility of MgF2 in water. What mass will dissolve in 2.0 L of water? Ksp = 7.4 x 10-11

MgF2 (s) D Mg2+ (aq) + 2 F- (aq)

Solubility and pH •  Calculate the pH of a saturated solution of silver hydroxide, AgOH.

Ksp = 2.0 x 10-8.

AgOH (s) D Ag+ (aq) + OH- (aq)

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The solubility of MgF2 in pure water is 2.6 x 10-4 mol/L. What happens to the solubility if we dissolve the MgF2 in a solution of NaF, instead of pure water?

The Common Ion Effect on Solubility Calculate the solubility of MgF2 in a solution of 0.080 M NaF.

MgF2 (s) D Mg2+ (aq) + 2 F- (aq)

Explaining the Common Ion Effect The presence of a common ion in a solution will lower the

solubility of a salt.

•  LeChatelier’s Principle: The addition of the common ion will shift the solubility equilibrium backwards. This means that there is more solid salt in the solution and therefore the solubility is lower!

Ksp and Solubility •  Generally, it is fair to say that salts with very small solubility product

constants (Ksp) are only sparingly soluble in water.

•  When comparing the solubilities of two salts, however, you can sometimes simply compare the relative sizes of their Ksp values.

•  This works if the salts have the same number of ions!

•  For example… CuI has Ksp = 5.0 x 10-12 and CaSO4 has Ksp = 6.1 x 10-5. Since the Ksp for calcium sulfate is larger than that for the copper (I) iodide, we can say that calcium sulfate is more soluble.

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But be careful... Salt Ksp Solubility

(mol/L)

CuS 8.5 x 10-45 9.2 x 10-23

Ag2S 1.6 x 10-49 3.4 x 10-17

Bi2S3 1.1 x 10-73 1.0 x 10-15

Mixing Solutions - Will a Precipitate Form? If 15 mL of 0.024-M lead nitrate is mixed with 30 mL of 0.030-M potassium chromate - will a precipitate form?

Pb(NO3)2 (aq) + K2CrO4 (aq) D PbCrO4 (s) + 2 KNO3 (aq)

Pb(NO3)2 (aq) + K2CrO4 (aq) D PbCrO4 (s) + 2 KNO3 (aq) Step 1: Is a sparingly soluble salt formed?

We can see that a double replacement reaction can occur and produce PbCrO4. Since this salt has a very small Ksp, it may precipitate from the mixture. The solubility equilibrium is:

PbCrO4 (s) D Pb2+ (aq) + CrO42- (aq)

Ksp = 2 x 10-16 = [Pb2+][CrO4

2-]

If a precipitate forms, it means the solubility equilibrium has shifted BACKWARDS. This will happen only if Qsp > Ksp in our mixture.

Step 2: Find the concentrations of the ions that form the sparingly soluble salt. Since we are mixing two solutions in this example, the concentrations of the Pb2+ and CrO4

2- will be diluted. We have to do a dilution calculation!

Dilution: C1V1 = C2V2

[Pb2+] =

[CrO42-] =

+== 2

211 Pb M 0.0080

mL) (45mL) M)(15 (0.024

VVC

-24

211 CrO M 0.020

mL) (45mL) M)(20 (0.030

VVC

==

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Step 3: Calculate Qsp for the mixture.

Qsp = [Pb2+][CrO42-] = (0.0080 M)(0.020 M)

Qsp = 1.6 x 10-4

Step 4: Compare Qsp to Ksp.

Since Qsp >> Ksp, a precipitate will form when the two solutions are mixed!

Note: If Qsp = Ksp, the mixture is saturated If Qsp < Ksp, the solution is unsaturated

Either way, no ppte will form!

A common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when 0.100 L of 0.30 M Ca(NO3)2 is mixed with 0.200 L of 0.060 M NaF?

Solubility Products

Consider the equilibrium that exists in a saturated solution of BaSO4 in water:

BaSO4(s) Ba2+(aq) + SO42−(aq)

Solubility Products

The equilibrium constant expression for this equilibrium is

Ksp = [Ba2+] [SO42−]

where the equilibrium constant, Ksp, is called the solubility product.

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Solubility Products •  Ksp is not the same as solubility. •  Solubility is generally expressed as the mass of solute

dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M).

Factors Affecting Solubility

•  The Common-Ion Effect –  If one of the ions in a solution equilibrium is already

dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease.

BaSO4(s) Ba2+(aq) + SO42−(aq)

Factors Affecting Solubility •  pH

–  If a substance has a basic anion, it will be more soluble in an acidic solution.

–  Substances with acidic cations are more soluble in basic solutions.


•  Complex Ions –  Metal ions can act as Lewis acids and form complex ions

with Lewis bases in the solvent.

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•  Complex Ions –  The formation of

these complex ions increases the solubility of these salts.


•  Amphoterism –  Amphoteric metal oxides and

hydroxides are soluble in strong acid or base, because they can act either as acids or bases.

–  Examples of such cations are Al3+, Zn2+, and Sn2+.

Will a Precipitate Form?

•  In a solution, –  If Q = Ksp, the system is at equilibrium and the

solution is saturated. –  If Q < Ksp, more solid will dissolve until Q = Ksp. –  If Q > Ksp, the salt will precipitate until Q = Ksp.

Selective Precipitation of Ions

One can use differences in solubilities of salts to separate ions in a mixture.

lecture 3 acids and base - chemistry for all…. · lecture 3: acids and bases who theory: acid=...

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