George Mason University General Chemistry 212 Chapter 19 Ionic Equilibria Acknowledgements

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George Mason UniversityGeneral Chemistry 212

Chapter 19Ionic Equilibria

Acknowledgements

Course Text: Chemistry: the Molecular Nature of Matter and Course Text: Chemistry: the Molecular Nature of Matter and

Change, 7 Change, 7thth edition, 2011, McGraw-Hill edition, 2011, McGraw-Hill Martin S. Silberberg & Patricia Amateis Martin S. Silberberg & Patricia Amateis

The Chemistry 211/212 General Chemistry courses taught at George Mason are intended for those students enrolled in a science /engineering oriented curricula, with particular emphasis on chemistry, biochemistry, and biology The material on these slides is taken primarily from the course text but the instructor has modified, condensed, or otherwise reorganized selected material.Additional material from other sources may also be included. Interpretation of course material to clarify concepts and solutions to problems is the sole responsibility of this instructor.

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Ionic Equilibria in Aqueous Solutions

Equilibria of Acid-Base Buffer Systems The Common Ion Effect The Henderson-Hasselbalch Equation Buffer Capacity and Range Preparing a buffer

Acid-Base Titration Curves Acid-Base Indicators Strong-Acid-Strong Base Titrations Weak Acid-Strong Base Titrations Weak Base-Strong Acid Titrations Polyprotic Acid Titrations Amino Acids as Polyprotic Acids

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Equilibria of Slightly Soluble Ionic Compounds

The Solubility-Product Constant Calculations involving Ksp

The Effect of a Common Ion

The Effect of pH Qsp vs. Ksp

Equilibria involving Complex Ions

Formation of Complex Ions

Complex Ions and Solubility

Amphoteric Hydroxides

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The simplest acid-base equilibria are those in which a single acid or base solute reacts with water

In this chapter, we will look at:

solutions of weak acids and bases through acid/base ionization

The reactions of salts with water

Titration curves

All of these processes involve equilibrium theory

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Equilibria of Acid-Base Buffer Systems

Buffer

An Acid-Base Buffer is a species added to a solution to minimize the impact on pH from the addition of [H3O+] or [OH-] ions

Small amounts of acid or base added to an unbuffered solution can change the pH by several units

Since pH is a logarithmic term, the change in [H3O+] or [OH-] can be several orders of magnitude

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The Common Ion Effect

Buffers work through a phenomenon known as the:

Common Ion Effect

The common ion effect occurs when a given ion is added to an “equilibrium mixture of a weak acid or weak base that already contains that ion

The additional “common ion” shifts the equilibrium away from its formation to more of the undissociated form; i.e., the acid or base dissociation decreases

A buffer must contain an “acidic” component that can react with the added OH- ion, and a “basic” component than can react with the added [H3O+]

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The buffer components cannot be just any acid or base

The components of a buffer are usually the conjugate acid-base pair of the weak acid (or base) being buffered

Ex: Acetic Acid & Sodium Acetate

Acetic acid is a weak acid, slightly dissociated

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)

Acid Base Conjugate Conjugate

Base Acid

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Ex: Acetic Acid & Sodium Acetate (Con’t)

Now add Sodium Acetate (CH3COONa), a strong electrolyte (Acetate ion is the conjugate base)

CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO-(aq) (added)

The added Acetate ions shift the equilibrium to the left forming more undissociated acetic acid

This lowers the extent of acid dissociation, which lowers the acidity by reducing the [H3O+] and increasing the pH

Similarly, if Acetic Acid is added to a solution of Sodium Acetate, the Acetate ions already present act to suppress the dissociation of the acid

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When a small amount of [H3O+] is add to acetic acid/acetate buffer, that same amount of acetate ion (CH3COO-) combines with it, increasing the concentration of acetic acid (CH3COOH).

The change in the [HA]/[A-] ratio is small; the added [H3O+] is

effectively tied up; thus, the change in pH is also small

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Essential Features of a Buffer

A buffer consists of high concentrations of the undissociated acidic component [HA] and the conjugate base [A-] component

When small amounts of [H3O+] or [OH-] are added to the buffer, they cause small amounts of one buffer component to convert to the other component

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Ex. When a small amount of H3O+ is added to an Acetate buffer, that same amount of CH3COO- combines with it, increasing the amount of undissociated CH3COOH tying up potential [H3O+]

Similarly, a small amount of OH- added combines with undissociated CH3COOH to form CH3COO- & H2O tying up potential [OH-]

In both cases, the relative changes in the amount of buffer components is small, but the added [H3O+] or [OH-] ions are tied up as undissociated components; thus little impact on pH

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The equilibrium perspective

The [H3O+] (pH) of the solution depends directly on the buffer-component concentration ratio

If the ratio [HA]/[A-] goes up, [H3O+] goes up (pH down)

If the ratio [HA]/[A-] goes down, [H3O+] goes down([OH-] goes up) and pH increases (less acidic)

When a small amount of a strong acid is added, the increased amount of [H3O+] reacts with a stoichiometric amount of acetate ion from the buffer to form more undissociated acetic acid

+ -+ -3 3

a3

[H O ][CH COO ][H ][A ]K = =

HA [CH COOH]

+ 33 a a --

3

HA [CH COOH][H O ] = K × = K ×

[CH COO ]A

+ -3 2 3 3CH COOH + H O = H O + CH COO

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Practice ProblemWhat is the pH of a buffer solution consisting of 0.50 M CH3COOH (HAc) and 0.50 M CH3COONa (NaAc)?

Ka (CH3COOH) = 1.8 x 10-5

+3 2[H O ] from autoionization of H O is neglible and is neglected

Concentration HAc(aq) + H2O(l) Ac-(aq) + H3O+

Initial Ci 0.50 - 0.50 0

- X - + X +X

Equilibrium Cf 0.50 - X - 0.50 + X X

If x is small, i.e., < 5% CH3COOHinit , then:

3[CH COOH] = 0.50 - x 0.50-

3[CH COO ] = 0.50 + x 0.50

+3 dissoc 3x = [CH COOH] = [H O ]

(Con’t)

NaAc is a sodium salt : it dissociates completely

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Practice Problem (Con’t)

+ 33 a -

3

[CH COOH]x = [H O ] = K ×

[CH COO ]

-+ -33 3

a3 3

x ×[CH COO ][H O ][CH COO ]K = =

[CH COOH] [CH COOH]

+ -5 -53

0.5x = [H O ] = 1.8 10 × = 1.8×10 M

0.5

+ -53pH = - log[H O ] = - log(1.8 10 )

pH = 4.74

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Practice Problem (Con’t)Calculate the pH of the previous buffer solution after adding 0.020 mol of solid NaOH to 1 L of the solution

Set up reaction table for the stoichiometry of NaOH (strong base) to Acetic acid (weak acid)

Concentration HAc(aq) + OH-(aq) Ac-(aq) + H2O(l)

Initial Ci 0.50 0.020 0.50 -

- 0.020 -0.020 + 0.020 -

Equilibrium Cf 0.48 0.0 0.52 -

NaOH0.020mol

Molarity (M) = = 0.020 mol / L1L

Set up reaction table for acid dissociation using newconcentrations

Concentration HAc(aq) + H2O Ac-(aq) + H3O+(aq)

Initial Ci 0.48 - 0.52 0

- X - + X + X

Equilibrium Cf 0.48 - X - 0.52 + X X(Con’t)

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Practice Problem (Con’t)Assuming x is small:

-3 3[CH COOH] = 0.48 M - x = 0.48 M and [CH COO ] = 0.52 M + x = 0.52 M

+ -533 a -

3

[CH COOH] 0.48x = [H O ] = K × = (1.8×10 ) ×

0.52[CH OO ]

+ -53[H O ] = 1.7 ×10 M

+ -53pH = -log[H O ] = -log(1.7 ×10 )

pH = 4.77

The addition of the strong base increased the

concentration of the basic buffer [CH3COO-] component to 0.52 M at the expense of the acidic buffer component [CH3COOH] (0.48 M)

The concentration of the [H3O+] (x) decreased, but

only slightly, thus, the pH increased only slightly from 4.74 to 4.77

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The Henderson-Hasselbalch Equation

For any weak acid, HA, the dissociation equation and Ka expression are:

The key variable that determines the [H3O+] is the ratio of acid species (HA) to base species (A-)

+ -2 3HA + H O H O + A

+ -3

a

[H O ][A ]K =

[HA]

+3 a -

[HA] (acid species)[H O ] = k ×

[A ] (base species)+

3 a -

[HA]-log[H O ] = - logK - log

[A ]-

a[A ]

pH = pK + log[HA]

a[base]

pH = pK + log[acid]

Note the change in sign of “log” term and inversion of acid-base terms

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Practice ProblemWhat is the pH of a buffer formed by adding 0.15 mol of acetic acid and 0.25 mol of sodium acetate to 1.5 L of water?

Ans: Both the acid and salt forms are added to the solution at somewhat equal concentrations

Neither form has much of a tendency to react

Dissociation of either form is opposed in bi-directional process

The final concentrations will approximately equal the initial concentrations[HA] = 0.15 mol / 1.5 L = 0.10 M

-[A ] = 0.25 mol / 1.5 L = 0.167 M-5

aK for acetic acid = 1.7 × 10 (from table "C")-5

a apK = - log(K ) = - log(1.7 10 ) = 4.77 (Con’t)

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Practice ProblemExample Problem (con’t)

From the Henderson-Hasselbalch equation-

a[A ]

pH = Pk + log[HA]

0.167MpH = 4.77 + log = 4.77 + log(1.67) = 4.77 + 0.22

0.10M

pH = 4.99

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Buffer Capacity A buffer resists a pH change as long as the

concentration of buffer components are large compared with the amount of strong acid or base added

Buffer Capacity is a measure of the ability to resist pH change

Buffer capacity depends on both the absolute and relative component concentrations Absolute - The more concentrated the components

of a buffer, the greater the buffer capacity Relative – For a given addition of acid or base, the

buffer-component concentration ratio changes less when the concentrations are similar than when they are different

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Ex. Consider 2 solutions

Solution 1 – Equal volumes of 1.0M HAc and 1.0 M Ac-

Solution 2 – Equal volumes of 0.1M HAc and 0.1M Ac-

Same pH (4.74) but 1.0 M buffer has much larger buffer capacity

(Con’t)

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Ex. Buffer #1 [HA] = [A-] = 1.000 M

Add 0.010 mol of OH- to 1.00 L of buffer

[HA] changes to 0.990 M [A-] changes to 1.010 M

-

init

init

[A ] 1.000 M = = 1.000

[HA] 1.000 M

-

final

final

[A ] 1.010 M = = 1.020

[HA] 0.990 M

1.020 -1.000Percent change = ×100 = 2%

1.000

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Buffer #2 [HA] = 0.250 M [A-] = 1.75 M

Add 0.010 mol of OH- to 1.00 L of buffer

[HA] changes to 0.240 M [A-] changes to 1.760 M

-

init

init

[A ] 1.750 M = = 7.000

[HA] 0.250 M

-

final

final

[A ] 1.760 M = = 7.330

[HA] 0.240 M

7.330 - 7.000Percent change = × 100 = 4.7%

7.000

Buffer-component concentration ratio is much larger when the initial concentrations of the components are very different

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A buffer has the highest capacity when the component acid / base concentrations are equal:

A buffer whose pH is equal to or near the pKa of its acid component has the highest buffer capacity

-

a a a a

[A ]pH = pK + log = pK + log1 = pK + 0 = pK

[HA]

-ABase = = 1

Acid HA

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Buffer Range pH range over which the buffer acts effectively is

related to the relative component concentrations The further the component concentration ratio is

from 1, the less effective the buffering action If the [A-]/[HA] ratio is greater than 10 or less

than 0.1 (or one component concentration is more than 10 times the other) the buffering action is poor

Buffers have a usable range within:

± 1 pH unit or pKa value of the acid components

a a

10pH = pK + log = pK + 1

1

a a

1pH = pK + log = pK - 1

10

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Equilibria of Acid-Base Systems

Preparing a Buffer

Choose the Conjugate Acid-Base pair

Driven by pH

Ratio of component concentrations close to 1 and pH ≈ pKa

Con’t

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Ex. Assume you need a biochemical buffer whose pH is 3.91. pKa of acid component should be close to 3.9

Ka = 10-3.9 = 1.3 x 10-4

2. From a table of pKa values select buffer possibilitiesLactic acid (pKa = 3.86)Glycolic acid (pKa = 3.83)Formic Acid (pKa = 3.74)

3. To avoid common biological species, select Formic Acid

Buffer components of Formic Acid

Formic Acid – HCOOH (Acid)

Formate Ion – HCOO- (Conjugate Base)

Obtain soluble Formate salt – HCOONa Con’t

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4. Calculate Ratio of Buffer Component Concentrations ([A-]/[HA]) that gives desired pH

1.4 mol of HCOONa is needed for every mole HCOOH

-[A ]pH = Pka + log

[HA]

-[HCOO ]log = 3.9 - 3.74 = 0.16

[HCOOH]

-[HCOO ]3.9 = 3.74 + log

[HCOOH]

Con’t

-0.16[HCOO ]

Thus : = 10 = 1.4[HCOOH]

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Preparing a Buffer (Con’t)

Determine the Buffer Concentrations

The higher the concentration of the components, the higher the buffer capacity

Assume 1 Liter of buffer is required and you have a stock of 0.40 M Formic Acid (HCOOH)

Compute moles and then grams of Sodium Formate (CHOONa) needed to produce 1.4/1.0 ratio

0.40 mol HCOOHMoles of HCOOH = 1.0 L × = 0.40 mol HCOOH

1.0 L soln

1.40 mol HCOONaMoles of HCOONa = 0.40 mol HCOOH × = 0.56 mol HCOONa

1.0 mol HCOOH

68.01 g HCOONaMass of HCOONa = 0.56 mol HCOONa × = 38 g HCOONa

1 mol HCOONa Con’t

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Preparing a Buffer (Con’t)

Mix the solution and adjust the pH

The prepared solution may not be an ideal solution (see Chapter 13, Section 6)

The desired pH (3.9) may not exactly match the actual value of the buffer solution

The pH of the buffer solution can be adjusted by a few tenths of a pH unit by adding strong acid or strong base.

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Practice ProblemExample of preparing a buffer solution without using the Henderson-Hasselbalch equationHow many grams of Sodium Carbonate (Na2CO3) must be added to 1.5 L of a 0.20 M Sodium Bicarbonate (NaHCO3) solution to make a buffer with pH = 10.0?

Ka (HCO3-) = 4.7 x 10-11 (From Appendix

C)

Con’t

Conjugate Pair - HCO3- (acid) and CO3

2- (base)

Convert pH to [H3O+]

- + 2-

3 2 3 3HCO (aq) + H O(l) H O (aq) + CO (aq)

-10+ -pH -10

3

[H O ] = 10 = 10 = 1×10 M

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Practice Problem (Con’t)3. Use Ka to compute [CO3

2-]

2-2- 2-3

3 30.094 mol CO

Moles CO = 1.5 L soln × = 0.14 mol CO1 L soln

5. Compute the mass (g) of Na2CO3 needed

2 32 3 2 3

2 3

105.99 g Na COMass (g) of Na CO = 0.14 mol Na CO × = 14.8 g

1mol Na CO

Con’t

4. Compute the amount of CO32- needed for 1.5 L of solution

2-- -113

3 a + -103

[HCO ] 0.20[CO ] = K = (4.7 10 ) = 0.094 M

[H O ] 1.0 10

+ 2-3 3

a -3

[H O ][CO ]K =

[HCO ]

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Practice Problem (Con’t)6. Dissolve the 14.8 g Na2CO3 in slightly less than 1.5 L of the

NaHCO3, say 1.3 L

7. Add additional NaHCO3 to make 1.5 L; adjust pH to 10.0with a strong acid

8. Check for a useful buffer range:

The concentration of the acidic component [HCO3-]

must be within a factor of “10” of the concentration of the basic component [CO3

2-]

- - -3 3 3HCO = 1.5 L × 0.20 M HCO = 0.30 mol HCO

- 2-3 30.30 mol HCO vs. 0.14 mol CO

0.30 / 0.14 = 2.1 << 10

Therefore, the buffer solution is reasonable

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Acid-Base Titration Curves Acid-Base Indicators

Weak organic acid (HIn) that has a different color than its conjugate base (In-)

The color change occurs over a relatively narrow pH range

Only small amounts of the indicator are needed; too little to affect the pH of the solution

The color range of typical indicators reflects a100 - fold range in the [HIn]/In-] ratio

This corresponds to a pH range of 2 pH units

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Mixing Acids & Bases Acids and bases react through neutralization reactions The change in pH of an acid mixed with a base is

tracked with an Acid-Base Titration Curve Titration:

Titration Curve: Plot of pH vs. the volume of the “Strong” acid or “Strong” base being added via buret

Solution in buret is called the “titrant” Equivalence Point: Point in a titration curve where

stoichiometric amounts of acid and base have been mixed (point of complete reaction)

3 Important Cases:

Strong Acid + Strong Base (& vice versa)

Weak Acid + Strong Base

Weak Base + Strong Acid

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Titration Curves Titration Curve: Plot of measured pH versus Volume

of acid or base added during a “Neutralization” experiment

All Titration Curves have a characteristic

“Sigmoid (S-shaped) profile

Beginning of Curve: pH changes slowly

Middle of Curve: pH changes very rapidly

End of Curve: pH changes very slowly again

pH changes very rapidly in the titration as the equivalence point (point of complete reaction) is reached and right after the equivalence point

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Neutralization Reactions Acids and Bases react with each other to form salts

(not always) and water (not always) through

Neutralization Reactions Neutralization reaction between a strong acid and

strong baseHNO3(aq) + KOH(aq) KNO3(aq) + H2O(l)

acid base salt water Neutralization of a strong acid and strong base

reaction lies very far to the right (Kn is very large); reaction goes to completion; net ionic equation is

H3O+(aq) + OH-(aq) 2 H2O(l)

H3O+ and OH- efficiently react with each other to form water

2142

3 3

n - -w

+ +[H O] 1 1

K = = = = 1×10K[H O ][OH ] [H O ][OH ]

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Titration Curves Curve for Titration of a Strong Acid by a Strong Base

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Practice Problem

Calculate pH after the addition of 10, 20, 30 mL NaOH during the titration of 20.0mL of 0.0010 M HCL(aq) with 0.0010 M NaOH(aq)

Initial pH HCl soln: [HCl] = [H3O+] = 0.0010 M

+ -3 2

Mixture

Strong Acid + Strong Base

H O (aq) + OH (aq) 2H O(l)€

+3pH = - log[H O ] = - log(0.0010) = 3.0

Con’t

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Practice ProblemAdd 10 mL 0.0010 M NaOH

1 mol H3O+ disappears for each mole OH- added

moles = Molarity(mol / L) × Volume(L)

+ +3 3moles of H O remaining = 0.00001 mol H O

+ +3 3Initial moles of H O = 0.0010 mol / L x 0.020L = 0.00002 mol H O

- -- moles OH added = 0.0010 mol / L x 0.010L = 0.00001 mol OH

Con’t

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Practice Problem (Con’t) Calculate [H3O+] & pH, taking “Total Volume” into

account

+

+ 33

0.000010 mol H O[H O ] = = 0.000033M

0.020L + 0.010L

+3pH = - log[H O ] = - log(0.0000333) = 3.48

Con’t

++ 3

3

amount(mol) of H O remaining[H O ] =

original volume of acid + volume of added base

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Practice Problem (Con’t) pH at “Equivalence Point”

Equal molar amounts of HCL & NaOH have been mixed

20 mL x 0.0010 M HCl reacts with20 mL x 0.0010 M NaOH

At the equivalence point all H3O+ has been neutralized by the addition of all the NaOH, except for [H3O+] from the autoionization of water, i.e., negligible

[H3O+] = pure water = 1 x 10-7 M

Con’t

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Practice Problem (Con’t) After the equivalence point

After the equivalence point, the pH calculations are based on the moles of excess OH- present

A total of 30 mL of NaOH has been added

- -moles of excess OH = 0.00001 mol OH

-[OH ] = 0.00020M

-pOH = - log[OH ] = - log(0.00020) = 3.70

wpH = pK - pOH = 14.0 - 3.7 = 10.3

- -- amount excess OH 0.000010 moles OH

[OH ] = =original volume acid + volume base added 0.020 L + 0.030 L

- -Total moles of OH added = 0.0010 mol / L x 0.030L = 0.00003 mol OH

+ +-3 3- moles H O consumed = 0.0010 mol / L x 0.020L = 0.00002 mol H O

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Neutralization Reactions Neutralization reactions between weak acids and

strong bases (net ionic equation shown below) C6H5COOH(aq) + NaOH C6H5COONa(aq) + H2O(l)

benzoic acid base benzoate “salt” water

The above equilibrium lies very far to the right The two equilibria below can be added together to

provide the overall neutralization reaction shown above

(1) C6H5COOH(aq) + H2O(l) C6H5COO-(aq) + H3O+

(aq)

Ka = 1.7 x 10-5 (from Appendix C)

(2) H3O+(aq) + OH-(aq) 2 H2O(l)

+ - -14w 3

14

w

Recall : K = [H O ][OH ] = 1.0 × 10

1 = 1.0×10K

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Neutralization Reactions The overall neutralization equilibrium

constant (Kn) is the product of the two intermediate equilibrium constants

(Ka & 1/Kw)

KN = Ka x 1/Kw = (1.7 x 10-5)(1 x 1014) = 1.9 x 109

(very large!)

Weak acids react completely with strong bases

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Titration CurvesCurve for Titration of a Weak Acid by a Strong Base

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Weak Acid-Strong Base Titration Curve

Consider the reaction between Propanoic Acid (weak acid) and Sodium Hydroxide (NaOH) (strong base)

Ka for CH3CH2COOH (HPr) - 1.3 x 10-5

The Titration Curve (see previous slide)

The bottom dotted curve corresponds to the strong acid-strong base titration

The Curve consists of 4 regions, the first 3 of which differ from the strong acid case

The initial pH is “Higher”

The weak acid dissociates only slightly producing less [H3O+] than with a strong acid

The gradually arising portion of the curve before the steep rise to the equivalence point is called the “buffer region”

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As HPr reacts with the strong base, more and more of the conjugate base (Pr-) forms creating an (HPr/Pr-) buffer

At the midpoint of the “Buffer” region, half the original HPr has reacted

pH at midpoint of “Buffer” region is common method of estimating pKa of an “unknown” acid

-

- [Pr ]HPr = [Pr ] or = 1

[HPr]

-

a a a[Pr ]

pH = Pk + log = pK + log(1) = pK[HPr]

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The pH at the “equivalence point” is greater than 7.00 The weak-acid anion (Pr-) acts as a

weak base to accept a proton from H2O forming OH-

The additional [OH-] raises the pH Beyond the equivalence point, the ph

increases slowly as excess OH- is added

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Practice ProblemCalculate the pH during the titration of 40 ml of 0.1000 M of the weak acid, Propanoic acid (HPr); Ka = 1.3 x 10-5 after the addition of the following volumes of 0.1000 M NaOH

0.00 mL 30.0 mL 40.0 mL 50.0 mL

a. 0.0 mL NaOH added

+

-53a

init

-[H O ][Pr ] (X)(X)K = = 1.3 10

[HPr] [HPr]

+ -2 3HPr(aq) + H O H O (aq) + Pr (Aq)

init dissoc init[HPr] = [HPr] - [HPr] [HPr]+ -

3[H O ] = [Pr ] = x

2 + 2 -5

3 a initX = [H O ] = K × HPr = 1.3 × 10 × 0.1000 M

+ -3

3[H O ] = 1.14 10 M Con’t3pH = - log[H O ] = 2.94

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Practice Problem (Con’t)b. 30.00 mL 0.1000 M NaOH added

For each mol of NaOH that reacts, 1 mol Pr- forms

Last line shows new initial amounts of HPr & Pr- that react to attain a “new” equilibrium

At equilibrium:

Original moles of HPr = 0.0400L x 0.1000 M = 0.004000 mol HPr-moles of NaOH added = 0.03000 L x 0.1000 M = 0.003000 mol OH

Amount (mol) HPr(aq) + OH-(aq) → Pr- + H2O(l)

Initial 0.004000 0.003000 0 -

Change - 0.003000 -0.003000 + 0.003000 -

Final 0.001000 0 0.003000 -

-

[HPr] 0.001000 mol = = 0.3333

[Pr ] 0.003000 mol

+3

a

-[H O ][Pr ]K =

[HPr]+ -5 -6

3 a -

[HPr][H O ] = K × = (1.3 x 10 ) (0.3333) = 4.3 x 10 M

[Pr ]

+ -6

3pH = - log[H O ] = - log(4.3 10 ) = 5.37 Con’t

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Practice Problem (Con’t)c. 40.00 mL NaOH (0.00400 mol) added

Reaction is at “equivalence point” All the HPr has reacted, leaving 0.004000 mol Pr-

Calculate concentration of Pr-

Calculate Kb

[HPr] = [OH-]

- 0.004000 mol[Pr ] = = 0.05000 M

0.04000 L + 0.004000 L-14

-10wb -5

a

K 1.0 10K = = = 7.7 10

K 1.3 10

- - 2

b - --

x x[HPr][OH ] [OH ]K = =

[Pr ] [Pr ]Pr

- -

b[OH ] = K x [Pr ]+ -

w 3K = [H O ][OH ]

-14

+ -9w3 - -10

b

K 1.0 10[H O ] = = 1.6 10 M

K [Pr ] (7.7 10 ) (0.05000)

+ -9

3pH = - log[H O ] = - log(1.6 10 ) = 8.80 Con’t

1/23/2015 53

Practice Problem (Con’t)d. 50 mL of 0.1000 M NaOH added

Beyond equivalence point, just excess OH- is being added

Calculation of pH is same as for strong acid-strong base titration

Calculate the moles of excess OH- from the initial concentration (0.1000 M) and the remaining volume (50.0 mL - 40.0 mL)

+ -w3 -

K moles of excess OH -[H O ] = where [OH ] =

[OH ] total volume

- moles of excess OH - 0.001000 mol[OH ] = = = 0.01111 M

total volume 0.04L + 0.05L

-Moles excess OH = (0.1000 M)(0.05000 L - 0.04000 L) = 0.001000 mol

-14+ -13w

3 -

K 1.0 10[H O ] = = = 9.0 10

[OH ] 0.01111

+ -13

3pH = - log[H O ] = - log(9.0 10 ) = 12.05

1/23/2015 54

Weak Base-Strong Acid Titration Neutralization reactions between weak bases and strong

acids (net ionic equation shown below)

NH3(aq) + H3O+(aq) NH4+(aq) + H2O(l) KN=?

base acid salt water

Above equilibrium also lies very far to the right; the two equilibria below can be added together to provide the overall neutralization reaction shown above

NH3(aq) + H2O(aq) NH4+ + OH-(aq) Kb = 1.8 x 10-5

H3O+(aq) + OH-(aq) 2 H2O(l) 1/Kw = 1.0 x 1014

KN = Kb X 1/Kw = (1.8 x 10-5)(1 x 1014) = 1.8 x 109 (very large!)

Weak bases react completely with strong acids

1/23/2015 55

Weak Base-Strong Acid Titration

Curve for Titration of a Weak Base by a Strong Acid

1/23/2015 56

Polyprotic Acid Titrations Polyprotic Acids have more than one ionizable

proton Except for Sulfuric Acid (H2SO4), the common

polyprotic acids are all weak acids Successive Ka values for a polyprotic acid differ by

several orders of magnitude The first H+ is lost much more easily than

subsequent ones All “1st” protons (H+) are removed before any of the

“2nd” protons - +

2 3 2 3 3H SO (aq) + H O(l) HSO (aq) + H O (aq)

- 2- +

3 2 3 3HSO (aq) + H O(l) SO (aq) + H O (aq)

-2

a1 a1K = 1.4 x 10 and pK = 1.85

-8

a2 a2K = 6.5 x 10 and pK = 7.19

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Polyprotic Acid Titrations Each mole of H+ is titrated separately In a diprotic acid two OH- ions are required to

completely remove both H+ ions All H2SO3 molecules lose one H+ before any HSO3

- ions lose a H+ to form SO3

-

- -- 2-1 mol OH 1 mol OH

2 3 3 3H SO HSO SO

Titration curves looks like two weak acid-strong base curves joined end-to-end

HSO3- is the conjugate

base of H2SO3 (Kb = 7.1x10-13)

HSO3- is also an acid

(Ka=6.5x10-8) dissociating to form its conjugate base SO3

2-

1/23/2015 58

Slightly Soluble Compounds Most solutes, even those called “soluble,”

have a limited solubility in a particular solvent

In a saturated solution, at a particular temperature, equilibrium exists between the dissolved and undissolved solute

Slightly soluble ionic compounds, normally called “insoluble,” reach equilibrium with very little of the solute dissolved

Slightly soluble compounds can produce complex mixtures of species

Discussion here is to assume that the small amount of a slightly soluble solute that does dissolve, dissociates completely

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Equilibria - Slightly Soluble Ionic Compounds

Solubility Rules for Ionic Compounds in Water

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Equilibrium exists between solid solute and the aqueous ions

PbSO4(s) Pb2+(aq) + SO42-(aq)

Set up “Reaction Quotient”

Note: concentration of a solid - [PbSO4(s)] = 1

Define Solubility Product

When solid PbSO4 reaches equilibrium with Pb2+ and SO4

2- at saturation, the numerical value of Qsp attains a constant value called the Solubility-Product constant (Ksp)

2+ 2-

4

4

c[Pb ][SO ]

Q = [PbSO (s)]

sp spQ at saturation = K = Solubility - Product constant

2+ 2-

4 4cQ PbSO (s) = Pb SO

2+ 2-

4 4sp cQ = Q [PbSO (s)] = [Pb ][SO ]

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The Solubility Product Constant (Ksp)

Value of Ksp depends only on temperature, not individual ion concentrations

Saturated solution of a slightly soluble ionic compound, MpXq, composed of ions Mn+ and Xz-, the equilibrium condition is:

n+ p z- q

sp spQ = [M ] [X ] = K

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Single-Step Process

Multi-Step Process

2+ -

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

2+ - 2

spK = [Cu ][OH ]

2+ 2-MnS(s) Mn (aq) + S (aq)

2- - -

2S (aq) + H O(l) HS (aq) + OH (aq)

2+ - -

2MnS(s) + H O(l) Mn (aq) + HS (aq) + OH (aq)

2+ - -

spK = [Mn ][HS ][OH ]

1/23/2015 63


1/23/2015 64

Practice ProblemDetermine Ksp from solubility

The solubility of PbSO4 at 25oC is 4.25 x10-3 g /100 mL solution. Calculate the Ksp

Convert solubility to molar solubility

Some PbSO4 dissolves into Pb2+ & SO42-

2+ 2-

4 4PbSO (s) = Pb (aq) + SO (aq)

4 4

4

0.00425 g PbSO 1 mol PbSO1000 mlMolar Solubility = x x

100 mL 1L 303.3 g PbSO

2+ 2- -4 2 -8

sp 4K = [Pb ][SO ] = (1.40 10 ) = 1.96 10

2+ 2-

4[Pb ] = SO

2+ 2- -4

4[Pb ] = SO = 1.40 10 M

-4

4= 1.40 10 M PbSO

1/23/2015 65

Practice ProblemDetermine Solubility from Ksp

Determine the solubility of Calcium Hydroxide, Ca(OH)2

Ksp for Ca(OH)2 = 6.5 x 10-6

Set up reaction table (S = molar solubility)

2+ -

2Ca(OH) (s) Ca (aq) + 2OH (aq)

Concentration (M) Ca(OH)2(s) ⇄ Ca2+(aq) + 2OH-(aq)

Initial 0 0

Change +S +2S

Equilibrium S 2S2+ - 2 2 2 3 -6

spK = [Ca ][OH ] = (S)(2S) = (S)(4S ) = 4S = 6.5 x 10

-62+ -23

6.5x10[Ca ] = S = = 1.2 x 10 M = 0.012 M

4

- -2 -2[OH ] = 2S = 2 1.2 10 = 4.2 10 M = 0.042 M

1/23/2015 66


The Effect of the Common Ion

The presence of a “Common Ion” decreases the solubility of a slightly soluble ionic compound

Add some Na2CrO4, a soluble salt, to the saturated PbCrO4 solution

Concentration of CrO42- increases

Some of excess CrO42- combines with Pb2+ to

form PbCrO4(s)

Equilibrium shifts to the “Left” This shift “reduces” the solubility of PbCrO4

2+ 2-

4 4PbCrO (s) Pb (aq) + CrO (aq)€

2+ 2- -13

sp 4K = [Pb ][CrO ] = 2.3 x 10

1/23/2015 67

Practice ProblemWhat is the solubility of Ca(OH)2 in 0.10 M Ca(NO3)2? Ksp Ca(OH)2 = 6.5 x 10-6

The addition of the common ion, Ca2+, should lower the solubility of Ca(OH)2 Note: (Compare to result from slide 65)

Since Ksp is small, S << 0.10 then, 0.10 M + S 0.1 M

2+ -

2Ca(OH) (s) Ca (aq) + 2OH (aq)

Concentration (M) Ca(OH)2(s) ⇄ Ca2+(aq) + 2OH-(aq)

Initial 0.10 0

Change +S +2S

Equilibrium 0.10 + S = 0.1 2S

2+ - 2 -6 2

spK = [Ca ][OH ] = 6.5 x 10 (0.1)(2S)

-62 -56.5 10

4S = = 6.5 100.10

-52+ -36.5 x 10

S = [Ca ] = = 4.03 10 M4

-34.0 x 10 M 100 = 4.0% < 5%

0.10 M Assumption : S << 0.10 is valid

2+ 2+

this prob prev prob[Ca ] vs. [Ca ] ( 0.00403 M vs. 0.012 M)

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The Effect of pH on Solubility The Hydronium Ion (H3O+) of a strong acid

increases the solubility of a solution containing the anion of a weak acid (HA-)

Adding some strong acid to a saturated solution of Calcium Carbonate (CaCO3) introduces large amount of H3O+ ion, which reacts immediately with the CaCO3 to form the weak acid HCO3

-

Additional H3O+ reacts with the HCO3- to form

carbonic acid, H2CO3, which immediately decomposes to H2O and CO2

2+ 2-

3 3CaCO (s) Ca (aq) + CO (aq)€2- + -

3 3 3 2CO (aq) + H O HCO (aq) + H O(l)€

- +

3 3 2 3 2 2 2HCO (aq) + H O H CO (aq) + H O(l) CO (g) + 2H O(l)

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The equilibrium shifts to the “Right” and more CaCO3 dissolves – increased solubility

The overall reaction is:

Adding H3O+ to a saturated solution of a compound with a strong acid anion – AgCl

Chloride ion, Cl-, is the conjugate base of a strong acid (HCl)

It coexists with water, i.e. does not react with water

There is “No” effect on the equilibrium

3 3+ +

2+ 2- -

3 3 3

2+

2 3 2 2

H O H OCaCO (s) Ca + CO HCO

H CO CO (g) + H O + Ca

€

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Practice ProblemWrite balanced equations to explain whether addition of H3O+ from a strong acid affects the solubility of the following ionic compounds

Bromide ion does not react with water

Lead Bromide (PbBr2)

2+ -

2PbBr Pb + 2Br€

Addition of an Acid has No effect

-Bromide (Br ) is the anion of the strong acid HBr

1/23/2015 71

Practice Problem Cu(II) Hydroxide (Cu(OH)2

-

2more Cu(OH) dissolves to replace OH

2+ -

2Cu(OH) (s) Cu (aq) + 2OH (aq)€

Increases Solubility

-

2OH is anion of H O, a very weak acid

+

3 2It reacts with added H O to form water (H O)

- +

3 2OH (aq) + H O (aq) 2H O

1/23/2015 72

Practice ProblemIron(II) Sulfide (FeS)

2+ - -

2FeS(s) + H O(l) Fe (aq) + HS (aq) + OH (aq)€

+ - -

3Additional H O reacts with both HS and OH (both weak acid anions)

- +

3 2 2HS (aq) + H O (aq) H S(aq) + H O(l)

- +

3 2OH (aq) + H O 2H O(l)

+

3Additional H O Increases Solubility

) - -

2Some of the FeS dissolves in water (H O to form HS and OH

- -More FeS dissolves to replace the HS & OH

1/23/2015 73

Practice Problem Calcium Fluoride (CaF2)

2+ -

2CaF (s) Ca (aq) + 2F (aq)€

-F is the conjugate base of a weak acid (HF)

- +

3F reacts with added H O to form the acid and water

2The solubility of CaF increases

- +

3 2F (aq) + H O (aq) HF(aq) + H O(l)

2-More CaF dissolves to replace the F

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Predicting Formation of a Precipitate

Qsp = Ksp when solution is “saturated”

Qsp > Ksp solution momentarily “supersaturated”

Additional solid precipitates until Qsp = Ksp again

Qsp < Ksp solution is “unsaturated”

No precipitate forms at that temperature

More solid dissolves until Qsp = Ksp

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Practice ProblemDoes CaF2 precipitate when 0.100 L of 0.30 M Ca(NO3)2 is mixed with 0.200 L of 0.060 M NaF?

Ksp (CaF2) = 3.2 x 10-11

Ions present: Ca2+ Na+ NO3- F-

Note: All sodium & nitrate salts are soluble2+ 2+ 2+moles Ca = 0.30 M Ca x 0.100L = 0.030 mol Ca

2+2+ 2+

init

0.030 mol Ca[Ca ] = = 0.10 M Ca

0.100 L + 0.200 L moles F = 0.060 M F x 0.200L = 0.012 mol F

-- -

init

0.012 mol F[F ] = = 0.040 M F

0.100 L + 0.200 L2+ - 2 2 -4

sp init initQ = [Ca ] [F ] = (0.10)(0.040) = 1.6 10-4 -11

sp spQ > K (1.6 10 > 3.2 10 )

2 sp sp CaF will precipitate until Q = K

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Selective Precipitation of Ions Separation of one ion in a solution from another Exploit differences in the solubility of their

compounds The Ksp of the less soluble compound is much smaller

than the Ksp of the more soluble compound

Add solution of precipitating ion until the Qsp value of the more soluble compound is almost equal to its Ksp value

The less soluble compound continues to precipitate while the more soluble compound remains dissolved, i.e., the Ksp of the less soluble compound is always being exceeded, i.e. precipitation is occurring

At equilibrium, most of the ions of the less soluble compound have been removed as the precipitate

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Practice ProblemA solution consists of 0.20 M MgCl2 and 0.10 M CuCl2

Calculate the [OH-] that would separate the metal ions as their hydroxides

Ksp of Mg(OH)2 = 6.3 x 10-10 Ksp of Cu(OH)2 = 2.2 x 10-

20

Mg(OH)2 is 1010 more soluble than Cu(OH)2; thus Cu(OH)2 precipitates first

Solve for [OH-] that will just give a saturated solution of Mg(OH)2

This concentration of [OH-] is more than enough for the less soluble CuCl2 to reach its Ksp and precipitate completely.

Con’t

1/23/2015 78

Practice ProblemWrite the equations and ion-product expressions

Calculate [OH-] of saturated Mg(OH)2 solution

This is the maximum amount of [OH-] that will not precipitate Mg2+ ion

Calculate [Cu2+] remaining in solution

Since initial [Cu2+] was 0.10M, virtually all Cu2+ is precipitated

2+ - 2+ - 2

2 spMg(OH) (s) Mg (aq) + 2OH (aq) K = [Mg ][OH ]€2+ - 2+ - 2

2 spCu(OH) (s) Cu (aq) + 2OH (aq) K = [Cu ][OH ]€

-10sp- -5

2+

K 6.3 10[OH ] = = = 5.6 10 M

[Mg ] 0.20

-20sp2+ -12

- 2 -5 2

K 2.2 x 10[Cu ] = = = 7.0 x 10 M

[OH ] (5.6 x 10 )

1/23/2015 79


Equilibria Involving Complex Ions The product of any Lewis acid-base reaction is called

an “Adduct”, a single species that contains a new covalent bond

A complex ion consists of a central metal ion covalently bonded to two or more anions or molecules, called ligands

Ionic ligands – OH- Cl- CN-

Molecular ligands – H2O CO NH3

Ex Cr(NH3)63+

Cr3+ is the central metal ion

NH3 molecules are molecular ligands Metal acts as Lewis Acid by accepting electron pair;

ligands acts as Lewis base by donating electron pair All complex ions are Lewis adducts

A + : B A - B (Adduct)

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Complex Ions Acidic Hydrated metal ions are complex ions with

water molecules as ligands When the hydrated cation is treated with a

solution of another ligand, the bound water molecules exchange for the other ligand

At equilibrium

Water is constant in aqueous reactions Incorporate in Kc to define Kf (formation constant)

2+ 2+

2 4 3 3 4 2M(H O) (aq) + 4NH (aq) M(NH ) (aq) + 4H O(l)

2+ 4

3 4 2c 2+ 4

2 4 3

[M(NH ) ][H O]K =

[M(H O) ][NH ]

2+

c 3 4f 4 2+ 4

2 2 4 3

K [M(NH ) ]K = =

[H O] [M(H O) ][NH ]

1/23/2015 81


Complex Ions (Con’t) The actual process is stepwise, with ammonia

molecules replacing water molecules one at a time to give a series of intermediate species, each with its own formation constant (Kf)

The sum of the 4 equations gives the overall equation

The product of the individual formation constants gives the overall formation constant

The Kf for each step is much larger than “1” because “ammonia” is a stronger Lewis base than H2O

Adding excess NH3 replaces all H2O and all M2+ exists as M(NH3)4

2+

2+ 2+

2 4 3 2 3 3 2M(H O) (aq) + NH (aq) M(H O) (NH ) (aq) + H O(l)2+

2 3 3f1 2+

2 4 3

[M(H O) (NH )K =

[M(H O) ][NH ]

f f1 f2 f3 f4K = K x K x K x K

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Practice ProblemWhat is the final [Zn(H2O)4

2+] after mixing 50.0 L of0.002 M Zn(H2O)4

2+ and 25 L of 0.15 M NH3

Determine Limiting Reagent

2+ 2+

2 4 3 3 4 2Zn(H O) + 4NH (aq) Zn(NH ) + 4H O(l)

Con’t

2+ 8f 3 4K of Zn(NH ) = 7.8 10

2

2

3 4

f 4

2 4 3

[Zn(NH )K =

[Zn(H O) ][NH ]

3Moles NH = 25 L×0.15 mol / L = 3.75 mol

2+2 4thus, Zn(H O) is limiting

2+2 4Moles Zn(H O) = 50.0 L×0.002 mol / L = 0.100 mol

2+3 2 4Moles NH (3.75) >> 4 Moles Zn(H O) (4 0.1)

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Practice Problem (Con’t) Find the initial reactant concentrations:

2+ -3

2 4 init

50.0 L 0.0020 M[Zn(H O) ] = = 1.3 10 M

50.0 L + 25.0 L

-2

3 init

25.0 L 0.15 M[NH ] = = 5.0 10 M

50.0 L + 25.0 L

Con’t

1/23/2015 84

Practice Problem (Con’t)Assume most Zn(H2O)4

2+ is converted to Zn(NH3)42+

Solve for x, the [Zn(H2O]42+ remaining at equilibrium

Concentration (M) Zn(H2O)42+(aq) + 4NH3(aq) ⇄ Zn(NH3)4

2+(aq) + 4H2O(l)

Initial 1.3 x 10-3 5.0 x 10-2 0

Change (-1.3 x 10-3)4 x -1.3 x 10-3

(-5.2 x 10-3)(+1.3 x 10-3)

Equilibrium X 4.5 x 10-2 1.3 x 10-3

2+ -383 4

f 2+ 4 -2 4

2 4 3

[Zn(NH ) ] 1.3 10K = = 7.8 10

[Zn(H O) ][NH ] (x) (4.5 10 )

2+ -7

2 4x = [Zn(H O) ] 4.1 10 M

Kf is very large; [Zn(H2O)42+ should be “low”

-3 -33 reacted[NH ] 4(1.3 10 M) 5.2 10 M

2+ -33 4[Zn(NH ) ] 1.3 10 M

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Complex Ions – Solubility of Precipitates

Increasing [H3O+] increases solubility of slightly soluble ionic compounds if the anion of the compound is that of a weak acid

A Ligand increases the solubility of a slightly soluble ionic compound if it forms a complex ion with the cation

When 1.0 M NaCN is added to the above solution, the CN- ions act as ligands and react with the small amount of Zn2+(aq) to form a complex ion

2+ - -

2ZnS(s) + H O(l) Zn (aq) + HS (aq) + OH (aq)

2+ - 2-

4Zn (aq) + 4CN (aq) Zn(CN) (aq)

-22

spK = 2.0 10

19

fK = 4.2 10

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Complex Ions – Solubility of Precipitates

Add the two reactions and compute the overall K2+ - -

2ZnS(s) + H O(l) Zn (aq) + HS (aq) + OH (aq)€2+ - 2-

4Zn (aq) + 4CN (aq) Zn(CN) (aq)€

- 2- - -

2 4ZnS(s) + 4CN (aq) + H O(l) Zn(CN) + HS (aq) + OH (aq)

-22 19 -3

overall sp fK = K K = (2.0 10 ) (4.2 10 ) = 8.4 10

The overall equilibrium constant, Koverall , is more than a factor of 1019 larger than the original Ksp (2.0 x 10-22)

This reflects the increased amount of ZnS in solution as Zn(CN4

2-)

1/23/2015 87

Practice Problem

Calculate the solubility of Silver Bromide (AgBr) in water (H2O) and 1.0 M of photographic “Hypo” – Sodium Thiosulfate (Na2S2O3), which forms the “complex” ion Ag(S2O3)2

3-

a. Solubility AgBr in Water

Con’t

-13 3- 13

sp f 2 3 2K (AgBr) = 5 10 K (Ag(S O ) ) = 4.7 10

+ -AgBr(s) Ag (aq) + Br (aq)

-13 + -

spK = 5.0 10 = [Ag ][Br ]

+ -

dissolvedS = [AgBr] = [Ag ] = [Br ]

-13 +

spK = 5.0 10 = [Ag ][Br-] = S x S

-7S = 7.1 10 M

-13

spK = 5 10

1/23/2015 88

Practice Problem (con’t)Solubility AgBr in 1.0 M Hypo - Sodium Thiosulfate (Na2S2O3)

Write the overall equation:

Calculate Koverall (product of Ksp & Kf):

+ -AgBr(s) Ag (aq) + Br (aq)€+ 2- 3-

2 3 2 3 2Ag (aq) + 2S O (aq) Ag(S O ) (aq)€2- 3- -

2 3 2 3 2AgBr(s) + 2S O (aq) Ag(S O ) (aq) + Br (aq)

]

3- --13 132 3 2

overall sp f2-

2 3

[Ag(S O ) ][Br ]K = = K K = (5.0 × 10 )(4.7 × 10 ) = 24

[(S O )

1/23/2015 89

Practice Problem (Con’t) Set Up the Reaction Table

3-

dissolved 2 3 2S = [AgBr] = [Ag(S O ) ]

Concentration (M) AgBr(s) + 2S2O32-(aq) ⇄ Ag(S2O3)2

3-(aq) + Br-(l)

Initial 1.0 0 0

Change -2S +S +S

Equilibrium 1.0 - 2S S S

243- 2

2 3 2overall 2- 2

2 3

[Ag(S O ) ][Br-] SK = =

[(S O ) ] (1.0 M - 2S)

S = 24 = 4.9

1.0 - 2SS = 4.9 - 2S(4.9) S + 2S(4.9) = 4.9

3-

2 3 2S = Solubility Ag(S O ) or AgBr = 0.45 M

S(1 + 2× 4.9) = 4.94.9

S = = 0.45 M1 + 9.8)

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Equilibria Involving Complex Ions Amphoteric Oxides & Hydroxides (Recall Chapter 8)

Some metals and many metalloids form oxides or hydroxides that are amphoteric; they can act as acids or bases in water

These compounds generally have very little solubility in water, but they do dissolve more readily in acids or bases Ex. Aluminum Hydroxide

Al(OH)3(s) ⇆ Al3+(aq) + 3OH-(ag)

Ksp = 3 x 10-34 (very insoluble in water)

In acid solution, the OH- reacts with H3O+ to form water

3H3O+(ag) + 3OH-(aq) 6H2O(l)

Al(OH)3(s) + H3O+(aq) Al3+(aq) + 6H2O(l)

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Equilibria Involving Complex Ions

Aluminum Hydroxide in basic solution

Al(OH)3(s) + OH-(aq) Al(OH)4-(aq)

The above reaction is actually a much more complex situation, involving multiple species

When dissolving an aluminum salt, such as Al(NO3), in a strong base (NaOH), a precipitate forms initially and then dissolves as more base is added

The formula for hydrated Al3+ is Al(H2O)63+

Al(H2O)63+ acts as a “weak polyprotic acid and

reacts with added OH- in a stepwise removal of the H2O ligands attached to the hydrated Al

1/23/2015 92


Amphoteric Aluminum Hydroxide in basic solution

Al(H2O)63+(aq) + OH-(aq) ⇆ Al(H2O)5OH2+(aq) + H2O(l)

Al(H2O)52+(aq) + OH-(aq) ⇆ Al(H2O)4(OH)2

+(aq) + H2O(l)

Al(H2O)4+(aq) + OH-(aq) ⇆ Al(H2O)3(OH)3(s) + H2O(l)

Al(H2O)3(OH)3(s) is more simply written Al(OH)3(s)

As more base is added, a 4th H+ is removed from a H2O ligand and the soluble ion Al(H2O)2(OH)4

-(aq) forms

Al(H2O)3(OH)3(s) + OH-(aq) ⇆ Al(H2O)2(OH)4-(aq)

The ion is normally written as Al(OH)4-(aq)

1/23/2015 93

Equation Summary- +

3[HA ][H O ]Ka =

[HA]

Concentration- A(aq) + B(aq) A-(aq) + H2O(l)

Initial Ci - - - 0

- - +X +X

Equilibrium Cf - - X X

+ -2 3HA + H O H O + A€

+ -3

a

[H O ][A ]K =

[HA]

a[base]

pH = pK + log[acid]

-

a a a a

[A ]pH = pK + log = pK + log1 = pK + 0 = pK

[HA]

-[A ]When : = 1

[HA]

1/23/2015 94

Equation Summary

n+ p z- q

sp spQ = [M ] [X ] = K

-14

w a bK = K K = 1.0 10

wpK = pH + pOH = 14

+ -

b[BH ][OH ]

K = [B]

2142

nw3 3

- -+ +[H O] 1 1

K = = = = 1×10K[H O ][OH ] [H O ][OH ]

2 3

+ -2H O H O + OH

3 2

+ -H O + OH 2H O

-143w 32

2

+ -+ -[H O ][OH ]

K = = [H O ][OH ] = 1×10 [H O]

Documents

George Mason University General Chemistry 212 Chapter 19 Ionic Equilibria Acknowledgements