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More Acid/Base Equilibria !!! 17.1 – 17.3. 17.1 The Common Ion Effect Consider the ionization of a weak acid, acetic acid: HC 2 H 3 O 2 ( aq )  H + ( aq ) + C 2 H 3 O 2 – ( aq )

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slide2
17.1 The Common Ion Effect

Consider the ionization of a weak acid, acetic acid: HC2H3O2(aq)  H+(aq) + C2H3O2–(aq)

If we increase the [C2H3O2–] ions by adding NaC2H3O2, the equilibrium will shift to the left. (Le Chatelier)

This reduces the [H+] and raises the pH (less acidic)

This phenomenon is called the common-ion effect.

Common ion equilibrium problems are solved following the same pattern as other equilibrium problems (ICE charts) EXCEPT the initial concentration of the common ion must be considered (it is NOT zero).

slide3
Example 1: Does the pH increase, decrease, or stay the same on addition of each of the following?

(a) NaNO2 to a solution of HNO2

(b) (CH3NH3)Cl to a solution of CH3NH2

(c) sodium formate to a solution of formic acid

(d) potassium bromide to a solution of hydrobromic acid

(e) HCl to a solution of NaC2H3O2

  • HNO2 H+ + NO2-

increases

  • CH3NH2 + H2O  CH3NH3+ + OH-

decreases

  • HCHO2 H+ + CHO2-

increases

(d) HBr H+ + Br-

no change

(e) C2H3O2-1 + H2O  HC2H3O2 + OH-1

decreases

slide4
Example 2: Using equilibrium constants from Appendix D, calculate the pH of the solution containing 0.060 M KC3H5O2 and 0.085 M HC3H5O2

change: -x +x +x

Equilibrium:0.085-x x .060+ x

x2+.060013x-(1.105*10-6)=0

x=1.84*10-5

pH = -log(1.84*10-5) = 4.74

slide5
17. 2 Buffered Solutions

A buffered solution or buffer is a solution that resists a change in pH after addition of small amounts of strong acid or strong base.

A buffer consists of a mixture of a weak acid (HX) and its conjugate base (X– ) or weak base (B) and its conjugate acid (HB+)

Thus a buffer contains both:

an acidic species to neutralize added OH–

When a small amount of OH– is added to the buffer solution, the OH– reacts with the acid in the buffer solution.

a basic species to neutralize added H+

When a small amount of H+ is added to the buffer solution, the H+ reacts with the base in the buffer solution.

slide6
Composition of a Buffer - 4 ways to make a buffer solution:

1.) Weak acid + salt of the acid

HCN and NaCN

weak acid: HCN weak base: CN-1

2.) Weak base + salt of the base

NH3 and NH4Cl

weak acid: NH4+1 weak base: NH3

3.) EXCESSWeak acid + strong base

2 mol HCN + 1 molNaOH 1 molHCN + 1 molNaCN + H2O

weak acid: HCN weak base: CN-1

2 mol NH4Cl + 1 molNaOH 1 molNH4Cl + 1 molNH3+ NaCl

weak acid: NH4+1 weak base: NH3

4.) EXCESS Weak base + strong acid

2 mol NH3 + 1 molHCl 1 molNH3 and 1 molNH4Cl

weak acid: NH4+1 weak base: NH3

2 molNaF + 1 molHCl 1 molNaF + 1 mole HF + NaCl

weak acid: HF weak base: F-1

slide7
Example 3: Explain why a mixture of HCl and KCl does not function as a buffer, whereas a mixture of HC2H3O2 and NaC2H3O2 does.

HCl is a strong acid - Cl-1 is a negligible base and will NOT react with added H+ - added H+will significantly change the pH of the solution

HC2H3O2 and C2H3O2-1 are a weak conjugate acid/base pair which act as a buffer

HC2H3O2 reacts with added base

C2H3O2-1 reacts with added acid

leaving the [H+1] and pH relatively unchanged

slide8
Buffer Capacity and pH

Buffer capacityis the amount of acid or base that can be neutralized by the buffer before there is a significant change in pH.

Buffer capacity depends on the concentrations of the components of the buffer - the greater the concentrations of the conjugate acid-base pair, the greater the buffer capacity.

The pH of the buffer is related to Kaand to the relative concentrations of the acid and base.

slide9
Henderson-Hasselbalch equation – used for buffer solutions

(on AP equation sheet!!)

These equations technically use the equilibrium concentrations of the acid (base) and the conjugate base (acid).

However, since the acid/base in the buffer is WEAK – the amount of the conjugate produced by dissociation is generally small compared to the amount of the conjugate added as a salt. IF this is true (it is for all AP buffer problems!) we do not need to do an equilibrium problem – just use the INITIAL concentrations.

slide10
Example 4 (Example 2 again!): Using equilibrium constants from Appendix D, calculate the pH of the solution containing 0.060 M KC3H5O2 and 0.085 M HC3H5O2
slide11
Example 5: Calculate the pH of a buffer that is 0.12 M in lactic acid and 0.11 M in sodium lactate
slide12
Example 6:A buffer is prepared by adding 20.0 g of acetic acid, HC2H3O2 and 20.0 g of sodium acetate to enough water to form 2.00 L of solution.

(a) Determine the pH of the buffer

(b) Write the complete ionic equation for the reaction that occurs when a few drops of hydrochloric acid are added to the buffer

(c) Write the complete ionic equation for the reaction that occurs when a few drops of sodium hydroxide are added to the buffer

(b) C2H3O2-1(aq) + H+1(aq) + Cl-1(aq) HC2H3O2 (aq) + Cl-1 (aq)

(c) HC2H3O2 (aq) + Na+1 (aq) + OH-1 (aq)  C2H3O2-1 (aq) + H2O (l) + Na+1 (aq)

slide13
17.3 Acid-Base Titrations – Titration Curves

In an acid-base titration:

A solution of base (or acid) of known concentration (called standard) is added to an acid (or base).

Acid-base indicators or a pH meter are used to signal the equivalence point (when moles acid = moles base).

The plot of pH versus volume during a titration is called a pH titration curve.

slide14
Strong acid added to strong base

Starts high

Ends low

Equivalence point = 7

equal moles of acid and base present

end

start

slide15
Strong base added to strong acid

Starts low

Ends high

Equivalence point = 7

end

start

slide16
Strong acid added to weak base

Starts med-high

Ends low

Equivalence point < 7

Buffer area – in this area there is weak base and some salt of the weak base

Actual pH depends on the salt formed

but it will be < 7

slide17
Weak base added to strong acid

Starts low

Ends med-high

Equivalence point < 7

start

end

slide18
Weak acid added to strong base

Starts high

Ends med-low

Equivalence point > 7

Actual pH depends on the salt formed

but it will be > 7

start

end

slide19
Weak acid added weak base

Starts med-high

Ends med-low

Equivalence point = 7

start

end

slide21
Example 7: Predict whether the equivalence point of each of the following titrations is below, above or at pH 7:

a) NaHCO3 titrated with NaOH b) NH3 titrated with HCl

c) KOH titrated with HBr

At the equivalence point, only products are present in solution, so determine the products of the reaction and then determine if the solution is acidic, basic or neutral

a) NaHCO3 + NaOH Na2CO3 + H2O

weak acid strong base pH > 7

CO3-2 is basic, Na+ is neutral, H2O is neutral

b) NH3 + HCl NH4Cl

weak base strong acid pH < 7

NH4+1 is acidic, Cl- is neutral

c) KOH + HBrKBr + H2O

strong base strong acid pH = 7

K+ and Br- are both neutral

slide22
Example 8: How many mL of 0.0850 M NaOH solution is required to titrate 40.0 mL of 0.0900 M HNO3?

? mL 40.0 mL

0.0850 M 0.0900 M

NaOH + HNO3 H2O + NaNO3

1 mole 1 mole

slide23
Example 9: A 20.0 mL sample of 0.200 M HBr solution is titrated with 0.200 M NaOH solution. Calculate the pH of the solution after the following volumes of base solution have been added:

(a) 15.0 mL (b) 19.9 mL (c) 20.0 mL (d) 20.1 mL (e) 35.0 mL

mL mL mL mol H+1mol OH-1 M of pH

HBrNaOH Total (M) (V) (M) (V) excess ion

(mol / tot vol)

(a) 20.0 15.0 35.0 0.00400 0.00300 0.0286 M H+1 1.544

(b) 20.0 19.9 39.9 0.00400 0.00398 0.0005 M H+1 3.3

(c) 20.0 20.0 40.0 0.00400 0.004001 x 10-7 M H+1* 7.0

(d) 20.0 20.1 40.1 0.00400 0.00402 0.0005 M OH-1 10.7

  • 20.0 35.0 55.0 0.00400 0.00700 0.0545 M OH-1 12.736

When molarity of H+ (or OH-) is less than 10-6 we must consider the autoionization of water!

(H+ = 1.0*10-7)

slide24
Example 10: Calculate the pH at the equivalence point for titrating 0.200 M solutions of each of the following bases with 0.200 M HBr: (a) NaOH (b) NH2OH

(a) strong acid/strong base titration so pH = 7

(b) HBr + NH2OH  Br- + NH2OH2+

strong acid weak base

.200M .200M

all product at equivalence point – no excess & Br- is neutral and will have no affect on pH

Volume doubles (equal molarity and 1:1 stoich ratio) so molarity halves

[NH2OH2+] = 0.200mol / 2 = 0.100 M

NH2OH2+ H+1 + NH2OH Kb = 1.1 x 10-8(appendix)

I .100 0 0

C -x +x +x

E 0.100 – x xx

Ka = Kw / Kb = 1  10-14 / 1.1  10-8 = 9.1  10-7

(x2) / (0.100-x) = 9.1 x 10-7

x = 3.0 x 10-4 M = [H+1]

pH = - log(3.0 x 10-4) = 3.52

acid base indicators
The equivalence point of an acid-base titration can be determined by measuring pH, but it can also be determined by using an acid-base indicator which marks the end point of a titration by changing color.

Although the equivalence point (defined by the stoichiometry) is not necessarily the same as the end point (where the indicator changes color), careful selection of the indicator can ensure that the difference between them is negligible.

Acid-base indicators are complex molecules that are themselves, weak acids (represented by HIn).

They exhibit one color when the proton is attached and a different color when the proton is absent.

Acid – Base Indicators
acid base indicators1
Acid – Base Indicators

Bromthymol Blue Indicator

In Acid In Base

acid base indicators2
Acid – Base Indicators

Methyl Orange Indicator

In Acid In Base

acid base indicators3
Acid – Base Indicators

Phenolphthalein color at different pH values

pH values 5 6 7 8 9

acid base indicators4
Consider a hypothetical indicator, HIn, a weak acid with Ka=1.0x10-8. It has a red color in acid and a blue color in base.

HIn(aq)  H+1(aq) + In-1(aq)

red blue

Acid – Base Indicators
acid base indicators5
HIn(aq)  H+1(aq) + In-1(aq)

red blue

This ratio shows that the predominant form of the indicator is HIn, resulting in a red solution. As OH-1 is added (like in a titration) [H+1] decreases and the equilibrium shifts to the right, changing HIn to In-. At some point in the titration, enough of the In- form will be present so we start to notice a color change.

Acid – Base Indicators
acid base indicators6
It can be shown (using the Henderson-Hasselbalch equation) that for a typical acid-base indicator with dissociation constant, Ka, the color transition occurs over a range of pH values given by pKa ± 1.

For example, bromthymol blue with Ka = 1.0 x 10-7 (pKa = 7), would have a useful pH range of 7 ± 1 or from 6 to 8.

You want to select an indicator whose pKa value is close to the pH you want to detect (usually the pH at the equivalence point)

Acid – Base Indicators
slide33
The pH curve for the titration of 100.0 mL of 0.10 MHCl with 0.10 MNaOH. Neither of theindicators shown wouldbe useful for atitration. Bromthymolblue (pKa=7) would beuseful.
slide34
The pH curve for the titration of 50 mL of 0.1 M HC2H3O2 with 0.1 M NaOH. Here, phenolphthalein is the indicator of choice. It has a pKa value of about 9.
slide35
Example 11: Use the following table to determine which of the following would be the best indicator to use to indicate the equivalence point of the titrations described in Example 10.
  • pH at equivalence point was 7.0

Bromthymol Blue

b. pH at equivalence point was 3.52

Methyl Yellow

slide37
Know your solubility rules:

SOLUBILITY GUIDELINES

Soluble CompoundsExceptions

NOT precipitates PRECIPITATES

Nitrates None

Acetates None

Chlorates None

Chlorides Ag+1, Hg2+2, Pb+2

Bromides Ag+1, Hg2+2, Pb+2

Iodides Ag+1, Hg2+2, Pb+2

Sulfates Ca+2, Sr+2, Ba+2, Hg2+2, Pb+2

Insoluble CompoundsExceptions

PRECIPITATES NOT Precipitates

Sulfides NH4+1, Li+1, Na+1, K+1, Ca+2, Sr+2, Ba+2

Carbonates NH4+1, Li+1, Na+1, K+1

Phosphates NH4+1, Li+1, Na+1, K+1

Hydroxides Li+1, Na+1, K+1, Ca+2, Sr+2, Ba+2

Chromates NH4+1, Li+1, Na+1, K+1, Ca+2, Mg+2

We classify these based on the

Solubility - maximum

amount of solute that dissolves

in water.

slide38
17.4 Solubility Equilibria

The Solubility-Product Constant, Ksp

Consider a saturated solution of BaSO4 in contact with solid BaSO4.

We can write an equilibrium expression for the dissolving of the solid.

BaSO4(s)  Ba2+(aq) + SO42–(aq)

Since BaSO4(s) is a pure solid, the equilibrium expression depends only on the concentration of the ions.

Ksp= [Ba2+][ SO42–]

Ksp is the equilibrium constant for the equilibrium between an ionic solid solute and its saturated aqueous solution.

Ksp is called the solubility-product constant

slide39
In general: the solubility product is equal to the product of the molar concentration of ions raised to powers corresponding to their stoichiometric coefficients.

Al2(CO3)3 2 Al+3 + 3 CO3-2

Ksp = [Al+3]2 [CO3-2]3

slide40
Solubility and Ksp

Solubility is the amount of substance that dissolves to form a saturated solution.

This can be expressed as grams of solid that will dissolve per liter of solution.

Molar solubility- the number of moles of solute that dissolve to form a liter of saturated solution.

Solubility can be used to find Ksp and Ksp can be used to find solubility (see problems)

slide41
Example 1:

a. If the molar solubility of CaF2 at 35oC is 1.24  10-3mol/L, what is Ksp at this temperature?

CaF2 Ca+2+ 2 F-1

E .00124M actually dissolves

Ksp= [Ca+2] [F-1]2 = (.00124 M)(.00248 M)2 = 7.63 x 10-9

b. It is found that 1.1  10-2 g of SrF2 dissolves per 100 mL of aqueous solution at 25oC. Calculate the solubility product of SrF2.

[SrF2] = (.011 g / 125.6 g/mole) / .100 L = .00088 M

SrF2 Sr+2 + 2 F-1

E .00088 M .00088 M 2(.00088) = .00176 M

Ksp = [Sr+2] [F-1]2 = (.00088 M)(.00176)2 = 2.7 x 10-9

.00124 M 2(.00124) = .00248 M

slide42
c. The Ksp of Ba(IO3)2 at 25oC is 6.0  10-10. What is the molar solubility of Ba(IO3)2?

Ba(IO3)2 Ba+2 + 2 IO3-1

E x

(x) (2x)2 = 4 x3= 6.0 x 10-10

x = 5.3 x 10-4M

x 2 x

slide43
17.5 Factors That Affect Solubility

Factors that have a significant impact on solubility are:

- The presence of a common ion

- The pH of the solution

Common-Ion Effect

Solubility is decreased when a common ion is added.

This is an application of Le Châtelier’s principle:

Consider the solubility of CaF2:

CaF2(s)  Ca2+(aq) + 2F–(aq)

If more F– is added (say by the addition of NaF), the equilibrium shifts left to offset the increase.

Therefore, more CaF2(s) is formed (precipitation occurs).

slide44
Example 2: Using Appendix D, calculate the molar solubility of AgBr in (a) pure water (b) 3.0  10-2 M AgNO3 solution (c) 0.50 M NaBr solution

(a) AgBr Ag+1 + Br-1Ksp = 5.0  10-13

E x x x

5.0  10-13 = x2

x = 7.1  10-7 M

(b) AgBr Ag+1 + Br-1Ksp = 5.0  10-13

E x .030 + x x

5.0  10-13 = (.030+x) x

x = 1.7  10-11 M

(c) AgBr Ag+1 + Br-1Ksp = 5.0  10-13

x x .50 + x

5.0  10-13 = x (.50+x)

x = 1.0  10-12 M

notice the DECREASED solubility with the common ion in (b) and (c)

slide45
pH effects

Consider: Mg(OH)2(s)  Mg2+(aq) + 2 OH–(aq)

If OH– is removed, then the equilibrium shifts right and Mg(OH)2 dissolves.

OH– can be removed by adding a strong acid (lowering the pH):

OH–(aq) + H+(aq)  H2O(aq)

Another example:

CaF2(s)  Ca2+(aq) + 2 F–(aq)

If the F– is removed, then the equilibrium shifts right and CaF2 dissolves.

F– can be removed by adding a strong acid (or lowering pH):

F–(aq) + H+(aq)  HF(aq)

slide46
Example 3: Calculate the molar solubility of Mn(OH)2 at (a) pH 7.0 (b) pH 9.5 (c) pH 11.8

the [OH-1] is set by the pH (or pOH)

(a) pH = 7.0 so pOH = 7.0 so [OH-1] = 1.00  10-7

Mn(OH)2 Mn+2 + 2 OH-1Ksp = 1.6  10-13

x x 1.00 x 10-7

Ksp = [Mn+2][OH-1]2

1.6  10-13 = (x) (1.00  10-7)2

x = 16 M

(b) pH = 9.5 so pOH = 4.5 so [OH-1] = 3.16  10-5

Mn(OH)2 Mn+2 + 2 OH-1Ksp = 1.6  10-13

x x 3.16 x 10-5

Ksp = [Mn+2][OH-1]2

1.6  10-13 = (x) (3.16  10-5)2

x = 1.7  10-4 M

(c) pH = 11.8 so pOH = 2.2 so [OH-1] = 6.31  10-3

Mn(OH)2 Mn+2 + 2 OH-1Ksp = 1.6  10-13

x x 6.31 x 10-3

Ksp = [Mn+2][OH-1]2

1.6  10-13 = (x) (6.31  10-3)2

x = 4.0  10-9 M

Common ion effect – increasing [OH-] decreases solubility

slide47
Example 4: Which of the following salts will be substantially more soluble in acidic solution than in pure water:

(a) ZnCO3 (b) ZnS (c) BiI3 (d) AgCN (e) Ba3(PO4)2

If the anion of the salt is the conjugate base of a weak acid, it will combine with H+1, reducing the concentration of the anion and making the salt more soluble

ZnCO3 Zn+2 + CO3-2

the CO3-2 ion will react with the added H+

CO3-2 + H+ HCO3-1

Le Chatelier effect of removing CO3-2

more soluble in acid: ZnCO3, ZnS, AgCN, Ba3(PO4)2

slide48
17.6 Precipitation and Separation of Ions

Consider the following:

BaSO4(s)  Ba2+(aq) + SO42–(aq)

At any instant in time, Q = [Ba2+][ SO42– ]

If Q > Ksp, (too many ions) precipitation occurs until Q = Ksp. If Q = Ksp equilibrium exists (saturated solution)

If Q < Ksp, (not enough ions) solid dissolves until Q = Ksp.

Selective Precipitation of Ions

Removal of one metal ion from a solution of two or more metal ions is called selective precipitation.

Ions can be separated from each other based on the solubilities of their salt compounds.

Example: If HCl is added to a solution containing Ag+ and Cu2+, the silver precipitates (as AgCl) while the Cu2+ remains in solution

Generally, the less soluble ion is removed first!

slide49
Example 5: Will Ca(OH)2 precipitate if the pH of a 0.050 M solution of CaCl2 is adjusted to 8.0?

if Q > than Ksp then precipitation will occur

pH = 8.0 so pOH = 6.0 so [OH-1] = 1.0  10-6 M

Ca(OH)2 Ca+2 + 2 OH-1Ksp = 6.5  10-6

.050 1.00 x 10-6

Q = [Ca+2] [OH-1]2

Q = (.050)(1.0  10-6)2 = 5.0  10-14

Q < K so no precipitation occurs

slide50
Example 6: A solution contains 0.00020 M Ag+1 and 0.0015 M Pb+2 . If NaI is added, will AgI or PbI2 precipitate first? Specify the [I-1] needed to begin precipitation for each cation.

the cation needing the lower [I-1] will precipitate first

AgI Ag+1 + I-1 Ksp = 8.3 x 10-17

.000200 x

Ksp = [Ag+1][x]

8.3  10-17 = (.00020)[x]

4.2  10-13 = x = [I-1]

PbI2 Pb+2 + 2 I-1 Ksp = 1.4 x 10-8

0.0015 x

Ksp = [Pb+2][x]2

1.4  10-8 = (.0015)[x]2

3.1  10-3 = x = [I-1]

AgI will precipitate first at an [I-1] = 4.2  10-13

complex ions
Complex ion – a metal ion bonded to one or more Lewis bases. (We saw this with water in chapter 16)Complex Ions
slide52
It can happen with other Lewis bases (things that have lone pairs of electrons)
  • Rule of thumb: The number of Lewis bases (ligands) that a metal ion attracts is equal to double its charge. (Works about 75% of the time!)
acid base indicators8
How much In- must be present for the human eye to detect that the color is different? For most indicators, about 1/10 of the initial form must be converted to the other form before a color change is apparent. We can assume that in the titration of an acid with a base, the color change will occur at a pH whereAcid – Base Indicators
acid base indicators9
Bromthymol blue, an indicator with a Ka = 1.0 x 10-7, is yellow in its HIn form and blue in its In- form. Suppose we put some strong acid in a flask, add a few drops of bromthymol blue and titrate with NaOH. At what pH will the indicator color change first be visible?

HIn(aq)  H+1(aq) + In-1(aq)

yellow blue

Acid – Base Indicators
slide57
Selective Precipitation of Ions (continued)

Sulfide ion is often used to separate metal ions.

Example: Consider a mixture of Zn2+(aq) and Cu2+(aq).

CuS (Ksp= 6 x 10–37)is less soluble than ZnS (Ksp= 2 x 10–25).

Because CuS is LESS SOLUBLEthan ZnS, CuS will be removed from solution before ZnS.

As H2S is bubbled through the acidified green solution, black CuS forms.

When the precipitate is removed, a colorless solution containing Zn2+(aq) remains.

When more H2S is added to the solution, a second precipitate of white ZnS forms.

slide58
Formula Type of Formula of Hydrolysis equation Hydrolysis equation

acid or conjugate of the acid of the base

base acid or base

HCl

HOCl

NH3

Ba(OH)2

KI

NaC2H3O2

Cl-1

HCl + H2O  H3O+1 + Cl-1

Strong acid

Cl-1 + H2O  X

Weak acid

OCl-1

HOCl + H2O  H3O+1 + OCl-1

OCl-1 + H2O  HOCl + OH-1

Weak base

NH4+1

NH3 + H2O  OH-1 + NH4+1

NH4+1 + H2O  H3O+1 + NH3

Strong base

OH-1 + H2O  X

H2O

H2O + H2O  H3O+1 + OH-1

K+1 or I-1

neutral

I-1 + H2O  X

K+1 + H2O  X

weak base

HC2H3O2

HC2H3O2 + H2O  H3O+1 + C2H3O2-1

C2H3O2-1 + H2O  HC2H3O2 + OH-1

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