Acids & Bases and Titrations

Properties of Acids & Bases• Acids

– taste sour

– feel like water

– good electrolytes

– turn blue litmus paper “pink”

– pH less than 7 at 25°C

• Bases– taste bitter

– feel slippery

– good electrolytes

– turn pink litmus paper “blue”

– pH greater than 7 at 25°C

3 Definitions of Acids

• Arrhenius: acids are substances which produce H+ ions in water.

• Brønsted-Lowry: acids are substances which donate H+ ions to other substances.

• Lewis: acids are substances which accept electron pairs from other substances.

3 Definitions of Bases

• Arrhenius: Bases are substances that produce OH- ions in water.

• Brønsted-Lowry: Bases are substances that accept H+ ions from other substances.

• Lewis: Bases are substances that donate a pair of electrons to another substance.

Conjugate Acids-Base Pairs

HF (aq) + H2O (l) H3O+(aq) + F- (aq)

acid base acidbase

SO42-(aq) + HCO3

-(aq) HSO4-(aq) + CO3

2-

(aq) base acid acid base

Acid Strength

When an acid is placed in water, it dissociates as it donates its H+ ions to water molecules:

HCl(aq) + H2O (l) H3O+(aq) + Cl-(aq)

A strong acid is one which completely dissociates.

This means it has a LARGE ionization constant, Ka. (the equilibrium constant for this reaction)

It is a very good electrolyte.

Relative Acid Strength

• Hydrochloric acid, HCl

• Nitric acid, HNO3

• Hydrobromic acid, HBr

• Hydroiodic acid, HI

• Sulfuric acid, H2SO4

(only the first H+ dissociates strongly)

Common Strong Acids

Common Weak Acids

• Hydrofluoric acid, HF Ka = 6.8 x 10-4

• Acetic acid, CH3COOH Ka = 1.8 x 10-5

• Nitrous acid, HNO2 Ka = 4.5 x 10-4

• Benzoic acid, C6H5COOH Ka = 6.5 x 10-5

Rank these acids in order of increasing strength.

Write a dissociation reaction for each acid.

Write the Ka expression for each acid.

Which of these "molecular" pictures

best represents a concentrated

solution of the weak acid HA,

with Ka = 1 x 10-5?

Explain your choice.

Weak Acids

Polyprotic Acids

• A polyprotic acid is one that has more than one H+ to donate:

Monoprotic acids:

HF, HCl, HNO3

Diprotic acids:

H2SO4, H2CO3

Triprotic acid:

H3PO4

•Polyprotic acids dissociate “stepwise”:

e.g. Sulfuric Acid:

H2SO4 + H2O H3O+ + HSO4-

HSO4- + H2O H3O+ + SO4

2-

Write the 3 steps in the dissociation of phosphoric acid! Write the Ka expression for each step!

Dissociation Constants - Polyprotic Acids

Amphoteric Substances• An amphoteric substance is a substance

which can behave as either an acid or a base.

eg. HNO3 (aq) + H2O (l) H3O+(aq) + NO3- (aq)

NH3 (aq) + H2O (aq) NH4+ (aq) + OH- (aq)

To be amphoteric, a substance must have a hydrogen ion to donate. Most amphoteric substances are negatively charged.

(e.g. HCO3- , H2PO4

- etc…)

The pH Scale• Acids produce H+ ions in water that immediately react

with water to form hydronium ions, H3O+.

• The concentration of H+ (or H3O+) in a solution is a measure of the solution’s acidity.

• pH is one way to express the hydronium concentration.

pH = - log [H3O+]

When [H3O+] = 0.010 M in a solution, the pH = 2.00.

When the pH of a solution is 8.00, [H3O+] = 1.0 x 10-8 M.

What is [H3O+] when the pH = 3.25?

The pH Scale (continued)

Acidic, Basic, or Neutral?

Auto-Ionization of Water• We’ve seen that water is amphoteric. In pure water, a very

small fraction of the water molecules undergo an acid-base reaction with each other:

H2O + H2O H3O+ + OH-

• At 25°C, the equilibrium constant (Kw) for this reaction is very small.

Kw = 1.0 x 10-14 =[H3O+][OH-]

• A simple equilibrium calculation shows that in pure water at 25°C, [H3O+] = [OH-] = 1.0 x 10-7 M.

• Thus, at 25°C, the pH of pure water is 7.00 (and pOH = 7.00).

Ion product constant @ 25oC: Kw=[H3O+][OH-]= 1.0 X 10-14

A Logarithm Lesson

Kw = 1.0 x 10-14 =[H3O+][OH-]

- log (1.0 x 10-14) = - log ([H3O+][OH-])

14.00 = -log [H3O+] + -log[OH-]

14.00 = pH + pOH

pH Practice

[H3O+] [OH-] pH pOH

0.015 M

8.41

2.16

3.7 x 10-8 M

Calculate the pH of 0.10-M HNO3 solution.

HNO3 + H2O H3O+ + NO3-

0.10 M ---- ----

- 0.10 + 0.10 + 0.10

---- 0.10 M 0.10 MThus, [H3O+] = 0.10 M

pH = - log (0.010) = 2.00

HNO3 is a strong acid - Ka is large!

It dissociates completely.

Why do we let [H3O+] = 0 at the start? Isn’t there some H3O+ in pure water??Are there any HNO3 molecules in the solution??Would this solution be a good electrolyte??

Calculate the pH of 0.15-M HF solution.

HF + H2O H3O+ + F-

0.15 M ---- ----

- x + x + x

0.15 - x x x

][]][[

107.6 34

HFFOH

xKa

xx

x

15.0

107.62

4

Since Ka is small let’s assumethat x << 0.15 M

15.0107.6

24 x

x

x1 = 0.010 Mx2 = 0.0097 Mx3 = 0.0097 M = [H3O+]

pH = - log (0.0097) = 2.01

HF is a weak acid - its Ka is 6.7 x 10-

4. It doesn’t dissociate

completely.

Percent Dissocation (Ionization)

• Calculate the %-Dissociation of the HF molecules in the 0.15-M solution of HF.

Of the 0.15 mol/L HF molecules, we know that 0.0097 mol/L have dissociated. Thus…

% Dissociation = = 6.5% 100

15.00097.0

x

Does it make sensethat the %-dissociationis so small for HF?

% Dissociation - Effect of Concentration

Calculate the pH of 0.20M H2SO4 (aq)

H2SO4 + H2O H3O+ + HSO4-

0.20 M ---- ----

- 0.20 + 0.20 + 0.20

---- 0.20 M 0.20 M

H2SO4 is a diprotic acid - it dissociates in 2 steps.

Ka for the first dissociation is LARGE.

HSO4- + H2O H3O+ + SO4

2-

0.20 M 0.20 M ----

- x + x + x

0.20 - x 0.20 + x x

Note that the second step starts with both HSO4

- and H3O+ at 0.20 M.

The second dissociation is weak - Ka2 = 0.012.

pH of 0.20M H2SO4 (aq) (cont’d)

Since the Ka is small for the second dissociation,

we’ll assume that x << 0.20 M

Note that the assumption applies TWICE - in the

numerator and denominator.

][

]][[012.0

4

243

2

HSO

SOOHKa

xxx

20.0))(20.0(

012.0

120.0))(20.0(

012.0 xx

x2 = 0.011 M

This is the [H3O+] from the 2nd dissociation.

[H3O+] for both steps = 0.20 + 0.011 = 0.21 M

pH = - log (0.21) = 0.68

Find the TOTAL [H3O+] from BOTH steps of the dissociation in order to

find the pH.

Other Polyprotic Acids

• For most polyprotic acids (H2CO3, H3PO4 etc…) the first dissociation is the only one that contributes significantly to the [H3O+] of the solution.

• This is because Ka1 >> Ka2 , Ka3

• Thus, the first step is the only one that needs to be considered and the acids can be treated as weak monoprotic acids.

Find the pH of 0.10 M H3PO4

H3PO4 + H2O H3O+ + H2PO42-

0.10 M ---- ----

- x + x + x

0.10 - x x x

H3PO4 is a triprotic acid - it dissociates in 3 steps.

Ka for the first

dissociation is small.Ka1 = 7.5 x 10 -3

][]][[

105.743

4233

POHPOHOH

xKa

xx

x

10.0

105.72

3

10.0105.7

23 x

x

Since Ka is small assume that x << 0.10 M

x1 = 0.027 Mx2 = 0.023 Mx3 = 0.024 M = [H3O+]

Since Ka1 >> Ka2, Ka3

assume these steps are negligible

pH = - log (0.024) = 1.62

Strong Bases

• Strong bases are metal hydroxides:

• NaOH Na+ + OH-

• KOH K+ + OH-

• LiOH Li+ + OH-

• Ca(OH)2 Ca2+ + 2 OH-

• Sr(OH)2 Sr2+ + 2 OH-

• “Strong” refers to the dissociation of the bases.

• Strong bases dissociate completely into ions.

• They are good electrolytes

• Notice that hydroxide ions are produced in a basic solution

pH of Strong Base Solutions• Calculating the pH of a strong base solution is relatively

easy because strong bases completely dissociate in water.

• For example, calculate the pH of 0.010-M NaOH solution.

NaOH Na+ + OH-

In a 0.010-M solution, [Na+] = [OH-] = 0.010 M

Thus, pOH = - log [OH-] = - log (0.010) = 2.00

And pH = 14.00 - pOH = 14.00 - 2.00 = 12.00

Strong Basesdissociate 100 %

Weak BasesAs you can see, most of the weak bases that are commonly encountered are derived from ammonia, NH3.

Note that in addition to the weak bases listed below, every weak acid has a corresponding weak base.

NH3 + H2O NH4+ + OH-

0.25 M ---- ----

- x + x + x

0.25 - x x x

Calculate the pH of 0.25-M NH3 solution.

][]][[

108.13

45

NHOHNH

xKb

xx

x

25.0

108.12

5

Since Kb is small let’s assumethat x << 0.25 M

25.0108.1

25 x

x

x1 = 0.0021 Mx2 = 0.0021 M = [OH -]

pOH = - log (0.0021) = 2.68pH = 14.00 - pOH = 11.32

NH3 is a weak base -

its Kb is 1.8 x 10 -5.

It doesn’t dissociate completely.

Relationship between Ka and Kb

Consider the dissociation of HF and its conjugate base in water:

HF + H2O H3O+ + F- F- + H2O HF + OH-

][]][[ 3

HFFOH

Ka

][

]][[

F

OHHFKb

wba KOHOHF

OHHFHF

FOHKK

]][[

][

]][[·

][]][[

· 33

Ka · Kb = Kw

Relative Strengths of Conjugate Acid-Base Pairs

Acid-Base Properties of Ions

pH of Salt Solutions• Salts are made of

cations and anions.

• A few cations may be acidic (NH4

+)

• Many anions may be basic (F-, SO4

2- etc…)

• Some ions are amphoteric (HCO3

-)

• Salt solutions will be acidic, basic or neutral depending on the ions that make up the salt.

e.g. Sodium Chloride

NaCl Na+ + Cl-

Na+ = neither acid nor base

= neutral

Cl- = conj. base of HCl (strong acid)

= Kb too small to affect pH

= neutral

Thus, NaCl is a neutral salt.

pH of Salt Solutions (cont’d)e.g. Ammonium nitrate, NH4NO3

NH4+ = conj. acid of ammonia,

NH3 (weak base)

= ACIDIC

NO3- = conj. base of nitric acid,

HNO3 (strong acid)

= Kb too small to affect pH

= NEUTRAL

Thus, NH4NO3 is ACIDIC

e.g. Sodium Bicarbonate NaHCO3

Na+ = NEUTRAL

HCO3- = AMPHOTERIC

Acid: Ka = 5.6 x 10-11

Base: Kb = Kw ÷ Ka (of H2CO3)

= 2.3 x 10-8

Since its Kb > Ka , HCO3- will

be BASIC

Thus, NaHCO3 is BASIC

Acid-Base Titrations• A titration is a volumetric technique

often involving acid-base neutralizations.

• It often involves reacting one solution of known concentration, with another of unknown concentration.

• Let’s take a look at how a titration is performed.

• Then we’ll look at calculations.

Titration Description• A titration is a procedure for quantitative analysis.

• In a titration two reagents are mixed, one with a known concentration and one with an unknown concentration.

• An acid-base indicator is added to determine when the two reagents have reacted essentially completely

• At the end of the titration the unknown solution's concentration (or molar mass) can be calculated.

Description• One reagent is a solution and

is added from a buret.

• This solution is called the titrant.

• The solution from the buret is added to a flask that contains either a measured volume of a solution or a weighed quantity of solid that has been dissolved.

Preparing a Buret• Rinse a clean buret three with

~ 5 mL portions of distilled water.

• Repeat the rinse using the titrant (the solution that will be added to the flask).

• Allow the rinse to drain through the buret stopcock to rinse the tip of the buret.

• Discard the rinse portions.

Fill the Buret• Clamp the buret into place

on the buret stand.

• Use a funnel to fill the buret with the titrant.

• Allow some titrant to drain out the tip. Remove any air bubbles in the stopcock tip.

Reading the Volume in the Buret• Make sure the volume

reading in the buret is at or below the 0.00 mL mark.

• Record the volume reading on the buret.

• Be sure to read the bottom of the meniscus.

• Record all certain digits plus one uncertain digit. suggest a better way to

fill the buret!

Prepare the Sample• If the sample to be titrated is a solution, pipet the desired

volume into an Erlenmeyer flask. Record the exact volume transferred. Dilute the sample with a small portion of distilled water (about 10 to 20 mL).

• If the sample is a solid, obtain the desired mass, and dissolve it in about 25 mL of distilled water in an Erlenmeyer flask. Record the exact mass of sample used.

• A few drops of chemical indicator is usually added to signal the endpoint of the titration with a colour change.

• The choice of indicator depends on the pH at the end of the neutralization reaction (consider the salt produced!)

Add the Titrant

• At the beginning of the titration, titrant may be added quickly since the indicator color disappears rapidly.

• When the color persists for longer periods of time, add titrant more slowly (a drop or less at a time).

Adding the Titrant (cont’d)• Be sure to mix the two reactant solutions thoroughly by

swirling the flask as the titrant is added.

• If solution splashes up to the side of the flask, you can use distilled water to wash it back into the solution.

• Placing the flask on a piece of white paper will often help you observe the first appearance of color change.

Determining the Endpoint• Overshooting the endpoint

of the titration by adding too much titrant is a common error.

• The endpoint for this titration is reached when you reach a pale color that persists throughout the solution for several seconds.

• Record the volume in the buret.

Endpoint Problems

Suggest what went wrong in each titration below!

Titration CalculationsStep 1: Write a balanced equation for the neutralization reaction.

Step 2: Write a mathematical equation using the mole-ratio for the acid and base.

Step 3: For solutions, use “CV” to represent moles. For solids, use “m/M” to represent moles.

Step 4: Substitute the given information into the equation.

Step 5: Solve for the unknown quantity.

Titration Calculations - exampleA 25.00-mL sample of carbonic acid, H2CO3 , required 22.34 mL of 0.126-M potassium hydroxide, KOH, for complete neutralization.

Calculate the concentration of the carbonic acid solution.

H2CO3 + 2 KOH K2CO3 + 2 H2O

moles KOH = 2 (moles H2CO3)

CbVb = 2 CaVa

320563.0)00.25(2

)34.22)(126.0(2

COHMmL

mLMVVC

Ca

bba

Titration Calculations - exampleAn unknown monoprotic acid, HA, was titrated using 0.0981-M NaOH. A 0.214-g sample of the acid required 32.70 mL of the base for complete neutralization.

Calculate the molar mass of the unknown acid.

HA + NaOH NaA + H2O

moles HA = moles NaOH

bba

a VCMm

molgLM

gVC

mM

bb

aa /7.66

)03270.0)(0981.0(214.0