TOPIC 18 ACIDS AND BASES

TOPIC 18

ACIDS AND BASES

18.3

pH Curves

By: Merinda Sautel Alameda Int’l Jr/Sr High School Lakewood, CO [email protected]

ESSENTIAL IDEA

pH curves can be investigated experimentally but are mathematically determined by the dissociation constants of the acid and base. An indicator with an

appropriate end point can be used to determine the equivalence point of the reaction.

NATURE OF SCIENCE (3.7)

Increased power of instrumentation and advances in available techniques – development in pH meter technology has allowed for more reliable and ready

measurement of pH.

THEORY OF KNOWLEDGE

Is a pH curve an accurate description of reality or an artificial

representation? Does science offer a

representation of reality?

UNDERSTANDING/KEY IDEA 18.3.A

The composition and action of a

buffer solution.

What is a Buffer?



A buffer solution is resistant to changes in pH on the addition of small amounts of acid or alkali.



The most practical example is our blood

which can absorb the acids and bases

produced in biologic reactions without

changing its pH.

Composition of buffer solutions



A buffer solution may contain a weak acid and its salt or a weak base and its salt.



Ex: HF and NaF



Ex: NH

and NH

Cl



Acidic buffers maintain a pH below 7 and

basic buffers maintain a pH above 7.



Buffer solutions are a mixture containing both an acid and a base of a weak

conjugate pair.



The buffer’s acid neutralizes added alkali or OH

and the buffer’s base neutralizes

added acid or H

.

Example 1 (Basic weak acid)



Calculate the [H

] and percent dissociation of a 1.0 mol/dm

solution of HF (K

= 7.2x10

).

 What are the major species?

 HF and H₂O

 Set up your ICE table.



Assume 1.0-x = 1.0 because K_a is so small.

 Set up your K_a expression and solve for x.

K_a = [H+][F-] = 7.2x10^-4 = x2 x = [H⁺]=.027

[HF] 1.0

 Calculate the % dissociation.

% dissoc = x = .027 x 100% = 2.7%

[conc]_ini 1.0

Example 2 (Addition of common ion)



Calculate the [H

] and percent dissociation of a 1.0

mol/dm

solution of HF (K

= 7.2x10

) and 1.0 mol/dm

NaF.

 What are the major species?

 HF , H₂O, Na⁺ and F^-

 Set up your ICE table.



Assume 1.0-x = 1.0 and 1.0+x = 1.0 because K_a is so small.

 Set up your K_a expression and solve for x.

K_a = [H+][F-] = 7.2x10^-4 = x (1.0) x = [H⁺]= 7.2x10^-4

[HF] 1.0

 Calculate the % dissociation.

% dissoc = x = 7.2x10^-4 x 100% = .072%

[conc]_ini 1.0

 Notice that with the addition of the common ion, equilibrium was pushed back to the reactant side due to LeChatelier’s principle and less of the acid dissociated.

Example 3 (Buffered solution)

 A buffered solution contains 0.50 mol/dm³ HC₂H₃O₂

(K_a = 1.8x10^-5) and 0.50 mol/dm³ NaC₂H₃O₂. Calculate the pH of this solution.

 What are the major species?

 HC₂H₃O₂, H₂O, Na⁺ and C₂H₃O_2-

 The Na⁺ has no effect on the equilibrium.

 HC₂H₃O₂ ↔ H⁺ + C₂H₃O_2- I .50 0 .50 C -x +x +x E .50- x x .50+x

Assume .50-x = .50 and .50+x = .50 because K_a is so small.

 Set up your K_a expression and solve for x.

K_a = [H+][C₂H₃O_2-] = 1.8x10^-5 = x (.50) x = [H⁺]= 1.8x10^-5

[HC₂H₃O₂] .50

 Solve for pH pH = -log[H+] = -log 1.8x10^-5 = 4.74

Example 4 (Adding OH ^- to buffer)



Calculate the pH when 0.010 mol NaOH is added to 1 dm

of the solution in Ex 3.



The major species in solution before any reaction occurs are: HC

H

O

, H

O, Na

, OH

and C

H

O



The strong base (OH

) will react with the best source of protons until there are no OH

ions left in solution.



The acetic acid is a better source for H

than water so the reaction is as follows:

 OH^- + HC2H3O2 ↔ C2H3O2- + H2O



Notice that adding OH

will use up the acid and produce

more of the conj base.



There are always 2 steps when adding acids and bases to buffer solutions.



A stoichiometry problem where you deal with the addition of the OH

or the H

added.



Followed by an equilibrium problem with the new

concentrations of the acid and conj base.

Back to Example 4



Stoichiometry problem (Before and After)

Always use moles in this type of problem, not concentration.

^1.0dm3 x .50mol/dm³ 1.0dm³ x .50mol/dm³

OH^- + HC₂H₃O₂ ↔ C₂H₃O_2- + H₂O Before .010 .50 .50 -- -.010 -.010 +.010

After 0 .49 .51

Notice that all the OH- has been used up and no longer exists.

Before we proceed to the equilibrium problem, we must convert back to concentration.

[HC₂H₃O₂] = .49mol/1dm³ and [C₂H₃O_2-] = .51mol/1dm³



Equilibrium problem (ICE table with new values)

 HC₂H₃O₂ ↔ H⁺ + C₂H₃O_2- I .49 0 .51 C -x +x +x E .49- x x .51+x

Assume .49-x = .49 and .51+x = .51 because K_a is so small.

Ka = [H+][C₂H₃O_2-] = 1.8x10^-5 = x (.51) x = [H⁺]= 1.7x10^-5

[HC₂H₃O₂] .49 The new pH = -log1.7x10^-5 = 4.76

Henderson Hasselbalch Equation



K

= [H+][A-]

[HA]

Rearranging to solve for [H+] gives: [H+] = K

[HA]

[A

] Take the negative logarithms of both sides to give:

pH = pKa + log [salt] or pOH = pKb + log [salt]

[acid] [base]

More on buffers



When the concentrations of the salt and the acid or the salt and the base are equal,

then pH = pK

and pOH = pK

.



The pH of a buffer solution depends upon the pK

or pK

of the acid or base and the ratio of the initial concentrations of the

salt/acid or salt/base pairs.

Factors influencing buffers



Dilution – does not change K

or K

and it does not change the ratio of acid or base to salt concentration. Dilution does not

change its pH. It does lower the buffering capacity because dilution affects how much strong acid or base can be absorbed.



Temperature – changes K

or K

so pH is

affected. Temp needs to be held constant.

APPLICATION/SKILLS

Know that the nature of acid-base buffers remains the same, but they

can be prepared in two different ways:

1. Half neutralization with strong acid or base

2. Equal parts weak acid/conj salt

Making Buffer Solutions



Start with an acid or a base with a pK

or pK

as close as possible to the required pH of the buffer needed.



Then react the acid or base with enough OH- or H+ to exactly half neutralize it.



Then you can use pH of buffer = pK

or pK

.

Example 5



How would you prepare a buffer solution of pH 3.75 starting with methanoic acid, HCOOH?

 From the IB data booklet we have pK_a of HCOOH = 3.75, so a buffer with equal amounts of the acid and its salt NaCOOH will have a pH of 3.75.

 The solution is prepared by reacting the acid with enough NaOH so that half of it is converted into its salt so

[HCOOH]=[HCOO^-].



When acids and bases react, they form salt and water.



These salts can be neutral, basic or acidic.



A strong acid with a strong base produces a neutral salt.



A strong acid with a weak base produces an acidic salt. (strong acid overrules)



A strong base with a weak acid produces a

basic salt. (strong base overrules)



Salt is another name for ionic compound.



When they dissolve in water, the ions can behave as acids or bases.



Remember that strong acids and bases

dissolve completely in water so their ions

have no affinity for the H

or OH

ions.



The anions of strong acids such as Cl

, NO

, I

, ClO

will not combine with H

so they have no affect on pH and are therefore true spectator ions.



The cations of strong bases such as Na

, Li

, Ba

will not combine with OH

so they are also true spectator ions and will have no affect on pH.



Therefore, mixtures of strong acids and strong

bases produce neutral salt solutions in water.



When given the salt, you can easily

determine which acid and base reacted to form it.

NaCl

NaCl breaks down into Na+ and Cl- ions.

The base has to be NaOH and the acid has

to be HCl. Do you know why?

Hydrolysis



When ions react with water, they can

hydrolyze the water by either releasing H

or OH

ions.

Anion Hydrolysis



An anion can be the conjugate base of the parent acid.



When the acid is weak, the conj base is strong enough to hydrolyze water.



A

+ H

O

↔ HA + OH



The release of OH- ions causes the pH of the

solution to increase.

Strong base with weak acid

NaOH + HC

H

O

↔ NaC

H

O

+ H

O



The NaC

H

O

is soluble so it breaks into Na

and C

H

O

ions. The strong cong base C

H

O

reacts with water in the solution as follows to produce OH

.

C

H

O

+ H

O ↔ HC

H

O

+ OH



This is the real reason that a strong base

with a weak acid produces a basic salt.

Example 1



Calculate the pH of a 0.30M solution of NaF. The K

for HF is 7.2x10

.



Step 1: Write the equation.

NaF ↔ Na

+ F



Step 2: Which species reacts with water?

F

+ H

O ↔ HF + OH



Step 3: This is actually a K

problem, but we only have K

. We can solve for K

by K

K

=K

.

K

= (1.00x10

)/(7.2x10

)= 1.4x10

F

+ H

O ↔ HF + OH

I .30 - 0 0 C -x - +x +x E .30 - x x

K

= [HF][OH

] = 1.4x10

[F

] so x

= 1.4x10

.30

x = [OH

] = 2.0x10

pOH = 5.69, pH = 14-5.69 = 8.31

Cation Hydrolysis



A cation can be the conjugate acid of the parent base.



When the base is weak and the conj acid a non-metal (NH

), it is able to hydrolyze

water.



M

+ H

O

↔ MOH + H



The release of H

into solution causes the

pH to decrease.



If the cation is a metal, the situation is more complex and depends upon the charge

density.



Metal ions from Group 1 and Group 2

(except Be

) do not have enough charge

density to hydrolyze water to generate H

so they do not show acidic behavior in

water.

Strong acid with weak base

HCl + NH

↔ NH

+ Cl



The NH

then reacts with water in the solution as follows to produce H+:

NH

↔ NH

+ H



This is the real reason that a strong acid

with a weak base produces an acidic salt.

Example 2



Calculate the pH of a 0.10 M NH

Cl

solution. The K

value for NH

is 1.8x10

.



Step 1: Write the equation.

NH

Cl ↔ NH

+ Cl



Step 2: Which species reacts with water?

NH

↔ NH

+ H



Step 3: This is actually a K

problem, but we only have K

. We can solve for K

by K

K

=K

.

K

= (1.00x10

)/(1.8x10

)= 5.6x10

NH

↔ NH

+ H

I .10 0 0 C -x +x +x E .10 x x K

= [H

][NH

] = 5.6x10

[NH

] so x

= 5.6x10

.10

x = [H

] = 7.5x10

pH = 5.12



Metal ions that are small and have 2 or 3 positive charges such as Be

, Al

and transition metals most notably Fe

can

hydrolyze water by pulling enough electron

density away from the O-H bond to release

the H

to cause a decrease the pH.

UNDERSTANDING/KEY IDEA 18.3.B

The characteristics of the pH

curves produced by the different

combinations of strong and weak

acids and bases.

APPLICATION/SKILLS

Know the general shapes of graphs of pH against volume for titrations

involving strong and weak acids and bases with an explanation of their

important features.

What is titration?



Technique in which one solution is used to analyze another.



The neutralization reactions between acids and bases can be carried out in a controlled way using titration.



Standardization is the calculation of the

exact concentration of one solution when

the other is known.



The equivalence point is the point when the acid and base exactly neutralize each

other. At this point the major species present are the salt and water.



The titrant is the solution of known

concentration delivered from a burette.



The titration or pH curve is the plot of pH vs

volume of titrant delivered.

NaOH + HCl → NaCl + H ₂ O



Initially, there is only the strong acid present, so to get the initial pH, you just take the negative log of the initial concentration.



You have 50.0 cm

of 0.10 mol/dm

HCl



pH = -log(0.1) = 1 Major species: H

and Cl



The next step is to add some base, so we are going to add 25.0cm

of 0.10mol/dm

NaOH.

 Major species before reaction: H⁺, Cl^-, Na⁺, OH^-, H2O

 All the OH^- will react with available H⁺ until all OH^- ions are consumed.

Adding 25.0 cm ³ base (con’t)

Remember to convert cm³ to dm³ by dividing by 1000:

50cm³ x 1dm³/1000cm³=0.050dm³ & 25cm³ x 1dm³/1000cm³ = 0.025dm³

We now need a before and after table using moles, not concentration.

H

+ OH

→ H

O

Before: 0.050dm³x.10mol/dm³= .025 x .10=

.005mol H⁺ .0025mol OH^- -.0025mol -.0025mol

After: .0025mol H⁺ 0.0mol OH^-

To find pH, we need to convert back to concentration of H⁺ using the new volume of 0.050dm³ + 0.025dm³ = 0.075dm³

So [H⁺] = .0025mol/0.075dm³ = 0.0333 pH= -log(.0333) = 1.48

Next add 49.0 cm ³ of the base

Follow the same procedure as before:

H

+ OH

→ H

O

Before: 0.050dm³x.10mol/dm³= .049 x .10=

.005mol H⁺ .0049mol OH^- -.0049mol -.0049mol

After: .0001mol H⁺ 0.0mol OH^-

To find pH, we need to convert back to concentration of H⁺ using the new volume of 0.050dm³ + 0.049dm³ = 0.0990dm³

So [H⁺] = .0001mol/0.0990dm³ = 0.00101 pH= -log(.00101) = 2.99

Add 50.0 cm ³ of strong base

Follow the same procedure as before:

H

+ OH

→ H

O

Before: 0.050dm³x.10mol/dm³= .050 x .10=

.005mol H⁺ .0050mol OH^- -.0050mol -.0050mol

After: 0.0 mol H⁺ 0.0 mol OH^-

At this point, all of the OH- ions have consumed all the H+ ions so we are at the equivalence point where only salt and water exist.

The pH = 7 at the equivalence point of a strong acid with a strong base.

Add 51.0 cm ³ of the strong base

Follow the same procedure as before:

H

+ OH

→ H

O

Before: 0.050dm³x.10mol/dm³= .051 x .10=

.005mol H⁺ .0051mol OH^- -.005mol -.005mol

After: 0.0 mol H⁺ 0.0001mol OH^- in excess

We are now past the equivalence point so we use up all the H⁺ and then have extra OH^- left in solution. We can calculate pH from pOH So [OH^-] = .0001mol/0.101dm³ = 0.00099 pOH= -log(.00099) = 3

pH = 14 – 3 = 11

Add 75.0 cm ³ of the strong base

Follow the same procedure as before:

H

+ OH

→ H

O

Before: 0.050dm³x.10mol/dm³= .075 x .10=

.005mol H⁺ .0075mol OH^- -.005mol -.005mol

After: 0.0 mol H⁺ 0.0025mol OH^- in excess

We are now past the equivalence point so we use up all the H⁺ and then have extra OH^- left in solution. We can calculate pH from pOH So [OH^-] = .0025mol/0.125dm³ = 0.02 pOH= -log(.02) = 1.7

pH = 14 – 1.7 = 12.3

Titration curve for strong acid and strong base.



Be sure to be able to draw and label a

titration curve for a strong acid with a strong

base and the opposite conditions.

 www.dlt.ncssm.edu

Points to deduce from graph



1. The initial pH = 1 (pH of strong acid)



2. pH changes gradually until equivalence.



3. There is a very sharp jump in pH at equivalence from pH 3 to pH 11.



Remember these were the points right before and right after the equivalence pt.



4. After equivalence, the curve flattens out at a high value (pH of a strong base)



5. pH at equivalence = 7

Note



All example problems start with 50 cm

of

0.10 mol/dm

acid. The titrant also has the

concentration of 0.10 mol/dm

.

Weak acid (Ac = acetic acid) with strong base

HAc + NaOH ↔ NaAc + H

O



Initially, there is only the weak acid present, so to get the initial pH, you must use an ICE table.

 What are the major species?

 HAc and H2O (K_a = 1.8x10^-5)

 Set up your ICE table.

 HAc ↔ H⁺ + Ac- I 0.10 0 0

C -x +x +x E .10-x x x

Ka = [H+][Ac-] = 1.8x10^-5= x2 x = [H⁺]=.0013 pH = 3

[HAc] .10

Add 25.0cm ³ of strong base



Major species before reaction:

HAc, Na

, OH

and H

O



All the OH

will react with available H

until all OH

ions are consumed.



Each addition of OH

in the titration requires a “Before and After Table” using moles, followed by a new ICE Table using concentration.



***This is the halfway to the equivalence point where HAc and Ac

will be equal.

So pH = pK

.***



HAc + OH

→ Ac

+ H

O

Before: 0.050dm³x.10mol/dm³= .025 x .10=

.005mol H⁺ .0025mol OH- 0.0

-.0025mol -.0025mol +.0025mol

After: .0025mol H⁺ 0.0mol OH- .0025mol Ac-

 Next, change back to concentrations to use the ICE table.

.0025mol/.075dm³ = .033mol/dm³

 HAc ↔ H⁺ + Ac- I 0.033 0 0.033 C -x +x +x E .033-x x .033+x

K_a = [H+][Ac-] = 1.8x10^-5= x(.033) x = [H⁺]= 1.8x10^-5 pH = 4.74

[HAc] .033



Add 50cm

of strong base (Equiv pt)



HAc + OH

→ Ac

+ H

O

Before: 0.050dm³x.10mol/dm³= .050 x .10=

.0050mol H⁺ .0050mol OH- 0.0

-.0050mol -.0050mol +.0050mol

After: 0.0mol H⁺ 0.0mol OH- .0050mol Ac-

 You now have different major species: Ac-, Na+, H₂O

 Ac- + H₂O ↔ HAc + OH-

I 0.05 0 0.0 Kb = Kw/Ka

C -x +x +x =1.0x10^-14 E .05-x x x 1.8x10^-5 K_b = [HAc][OH-] = 5.6x10^-10= x² x = [OH^-]= 5.3x10^-6 pOH = 5.28

[Ac-] .05 pH = 8.72



Add 75cm

of strong base



HAc + OH

→ Ac

+ H

O

Before: 0.050dm³x.10mol/dm³= .075 x .10=

.0050mol H⁺ .0075mol OH- 0.0

-.0050mol -.0050mol +.0050mol

After: 0.0mol H⁺ 0.0025mol OH- .0050mol Ac-



The pH is now determined by the excess OH

in solution because it overrules the Ac

.

[OH-] = 0.0025mol/.125dm

= .02mol/dm

pOH = -log(.02) = 1.70

pH = 14-1.70 = 12.3

UNDERSTANDING/KEY IDEA 18.3.C

The buffer region on the pH curve

represents the region where small

additions of acids or base result in

little or no change in pH.

 www.people.bu.edu

Important points about graph for weak acid with a strong base



1. The initial pH is higher than 1. (pH of a weak acid)



2. pH stays fairly consistent until equivalence (buffer region).



3. There is a smaller jump in pH at the equivalence from about 7.0 – 11.0. (This is not as large a jump as with strong with strong.)



4. After equivalence, the pH curve flattens out at a high value (pH of a strong base).



5. pH at equivalence is greater than 7 (anion

hydrolysis releases OH

)

Titration curve for weak acid with a strong base



Be able to draw this titration curve and that

of a weak acid with a strong base.

Important to note



All of the above graphs can be reversed so that you are adding the acid to the base.



You are expected to be able to draw and

label both types of graphs.

Important points about graph for a strong acid with a weak base



1. The initial pH is 1. (pH of a strong acid)



2. pH stays fairly consistent until equivalence (buffer region).



3. There is a smaller jump in pH at the

equivalence from about 3.0 – 7.0. (This is not as large a jump as with strong with strong.)



4. After equivalence, the pH curve flattens out at a lower value (pH of a weak base).



5. pH at equivalence is less than 7 (cation

hydrolysis releases H

)

Important points about graph for weak acid with a weak base



1. The initial pH is fairly high. (pH of a weak acid)



2. Addition of a base causes pH to rise steadily.



3. The change in pH at equivalence is much less sharp than the other titrations.



4. After equivalence, the pH curve flattens out at a lower value (pH of a weak base).



5. This is not a good technique to determine the

equivalence point.



Indicators are substances that change color reversibly according to the pH of the

solution.



An indicator can be a weak acid or a weak base in which the undissociated and

dissociated forms have different colors.

HIn ↔ H ⁺ + In ^-



The HIn is one color and the In

ion is another color.



Remember that the H

is the factor that affects the pH.



Using Le Chatelier’s principle, you can predict how the equilibrium will respond to a change in the pH by the addition or deletion of H

ions.



Increasing [H

]: equilibrium shifts to HIn



Decreasing [H

]: equilibrium shifts to In-

UNDERSTANDING/KEY IDEA 18.3.D

An acid-base indicator is a weak acid or a weak base where the

components of the conjugate acid-

base pair have different colors.

Equivalence point vs End point



Do not confuse the equivalence point with the end point.



The equivalence point is where

stoichiometrically equal amounts of acids and bases have neutralized each other.



The end point is the pH at which the

indicator changes color. This occurs at the

pH equal to its pK

.

HIn ↔ H ⁺ + In ^-

K

= [H

][In

] [HIn]

When [In

] = [HIn], the indicator is exactly in the middle of its color change and K

=[H

] so pK

= pH. This is the end point.

Different indicators change colors at different

pH values.

BOH ↔ B ⁺ + OH ^-

color a color b

K

= [B

][OH

] [BOH]

When [B

] = [BOH], the indicator is exactly in the middle of its color change and K

=[OH

] so pK

= pH. This is the end point.

Different indicators change colors at different

pH values.

UNDERSTANDING/KEY IDEA 18.3.E

The relationship between the pH

range of an acid-base indicator,

which is a weak acid, and its pK

value.

GUIDANCE

Be able to select an appropriate indicator for a titration, given the

equivalence point of the titration and

the end point of the indicator.

Steps for choosing an indicator



1. Determine what type of titration you have.

(ex: strong acid with weak base, etc.)



2. Deduce the pH of the salt solution at the equivalence.



3. Choose an indicator with an end point in the range of the equivalence point from a table.



4. The end point range should be within +/- 1

pH unit on either side of the pK

.



Strong acid with strong base



pH range at equivalence is 3-11



Indicators: phenolphtalein

 pKa = 9.50

 End point range of pth = 8.2-10 Colorless to pink

methyl orange

End point range of methyl orange = 3.2-4.4 Red to yellow



Weak acid with strong base



pH range at equivalence is 7-11



Indicators: phenolphtalein

 pKa = 9.50

 End point range of pth = 8.2-10 Colorless to pink

phenol red

End point range of phenol red = 6.6-8.0 Yellow to red



Strong acid with weak base



pH range at equivalence is 3-7



Indicators: methyl orange

 pKa = 3.46

 End point range of methyl orange = 3.2-4.4 Red to yellow

bromophenol blue

End point range of bromophenol blue = 3.0-4.6 Yellow to blue

Citations

International Baccalaureate Organization. Chemistry Guide, First assessment 2016. Updated 2015.

Brown, Catrin, and Mike Ford. Higher Level Chemistry. 2nd ed.

N.p.: Pearson Baccalaureate, 2014. Print.

ISBN 978 1 447 95975 5 eBook 978 1 447 95976 2

Most of the information found in this power point comes directly from this textbook.

The power point has been made to directly complement the Higher Level Chemistry textbook by Brown and Ford and is used for direct instructional purposes only.