Chapter 17 Acids and Bases

- we are all familiar with 'acids'- depicted on television as burning liquids- from foods (i.e. vinegar)- taste "sour" or "tart'

- less familiar with 'bases'- taste "bitter" - for example, beer or coffee

- several different definitions have been used for the terms "acid" and "base" over the years

17.1 The Arrhenius Theory

Svante Arrhenius was interested in "electrolytes" and "electrolysis" and he devised much of the early theory to explain ions in solution

- according to Arrhenius, acids dissolve in water to give protons while bases dissolve to give hydroxide ions and these react to give water and a salt

- this reaction is called "neutralization":acid + base → salt + water

for example,hydrochloric acid yields a proton in solutionsodium hydroxide generates hydroxide ions

hence:HCl(aq) + NaOH(aq) → NaCl(aq) + H2O

in ionic form:H+

(aq) + Cl-(aq) + Na+(aq) + OH-

(aq) → Na+(aq) + Cl-(aq) + H2O

or in "net" ionic form:H+

(aq) + OH-(aq) → H2O

- from the Arrhenius point of view, HCl is an acid because it participates in a neutralization reaction that gives water and a salt (in this case, "table salt") and NaOH is a base because it also participates in a neutralization reaction

that is, the reason that we know that 'HCl' and 'NaOH' are an acid and a base, respectively, is because they give us water - hence, they must yield up a proton and a hydroxide ion

this was a good "first step" but it was not sufficient- for example, it does not explain "ammonia"

7.2 Bronsted-Lowry Theory

redefines acids and bases in terms of their ability to donate or accept protons

an 'acid' is a 'proton donor'a 'base' is a 'proton acceptor'

hence, anything that increases the proton content of a solution is, by definition, an acid and anything that decreases it - by accepting a proton - is a base

- explains ammonia which accepts protons

but this definition is also slightly flawed because bases don't "accept" protons - they steal them!

- arrow pushing is how we show this:

Bronsted-Lowry gives us the concept of "conjugate acids and bases"

NH4+

(aq) + OH-(aq) → NH3 (aq) + H2O

acid base conj. conj.base acid

NH4+/NH3 and OH-/H2O both form acid/base pairs and

the partner of an acid in such a pair is called "the conjugate base" (and vice versa for a base)

Note: this all depends upon how the reaction is written!

also, most acid/base reactions are actually equilibria

for example, acetic acid

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

(aq)

acid base base acid

conjugates

we can write an equilibrium expression for this reaction

Ka = [CH3COO-][H3O+] / [CH3COOH]

where [H2O] is treated as a constant and incorporated into the value of Ka, the "acid ionization constant"

could equally well write an expression for Kb, the "base ionization constant"

Bronsted-Lowry is better than Arrhenius at accounting for substances that can both donate and accept protons such as water. Such substances are called "amphiprotic"

17.3 The (Auto)Ionization of Water and pH

- water is one of many substances that are capable of donating a proton to itself

that is, H2O + H2O W H3O+(aq) + OH-

(aq)

acid base conj. conj.acid base

this is more commonly written as,H2O W H+

(aq) + OH-(aq)

as an equilibrium, we can write the equilibrium constant expression,

K = [H+(aq)][OH-

(aq)] / [H2O]

but, since the concentration of water is a constant, this is more commonly expressed as:

Kw = [H+][OH-]

simply put, the concentrations of [H+] and [OH-] in any and all aqueous solutions are inversely related

furthermore,Kw = 1.0 x 10-14 = [H+][OH-]

or, in pure water,

[H+] = [OH-] = (1.0 x 10-14)½ = 1.0 x 10-7

This is our definition of "neutral". In pure water with no other species present, the concentrations of both H+

(aq)and OH-

(aq) are the same.

Note: Kw is an "ion product" as it is only concerned with the ionic components of the equilibrium.

Measuring the concentration of either the protons or hydroxide ions in a solution results in a measurement of the other.

We could use molarity and scientific notation to express the concentration of protons and hydroxide ions (for example, [H+] = 2.34 x10-5 M) but we more frequently use a "p-function"

By definition, a p-function is the "negative log10" of the quantity in question.

pH = -log [H+] i.e. pH = -log (2.34 x10-5) = 4.631

note that p-functions are not restricted to simply measuring pH

pOH = - log [OH-]pKa = - log Ka

pKb = - log Kb

pO2 = - log [O2]etcetera

The purpose of the p-functions is to change to a logarithmic scale but not to change the answers.

Also, pH is not actually -log[H+] but rather it is the negative logarithm of the hydrogen ion activity.

That is, pH = -log(άH+)

The "activity" is the amount of a substance that is actually available to react - which is not necessarily equal to the amount present. This is a subtle point and not one that we will pursue further in this course.

Note:

Kw = [H+][OH-] = 1.0 x10-14

pKw = pH + pOH = 14

- as pH decreases, pOH increasesand vice versa

Acid Strength- the strength of an acid is not necessarily correlated with its concentration

- that is, a dilute solution of a strong acid might be more reactive than a concentrated solution of a dilute acid

- acid strength is a measure of how freely an acid will donate it proton

- for example, perchloric acid (HClO4) is a strong acid because it will readily give up its proton to any other base - hence, the perchlorate ion (ClO4

-) is a lousy base

- the stronger the acid, the weaker the conjugate base- that is, a strong acid will lose its proton to any other base and is incapable of stealing the proton back

- the strongest acid in water is the hydronium ion (H3O+)

- strong acids, when dissolved in water, will donate their proton to water

i.e. HCl + H2O → Cl- + H3O+

HClO4 + H2O → ClO4- + H3O+

HNO3 + H2O → NO3- + H3O+

- but any acid (or base) can only donate to their concentration

- that is, if the concentration of hydrochloric acid in an aqueous solution is 0.010 M, then the maximum concentration of H+ is 0.010 M.

(and it can only react with 0.010 M of anything else!)

- hence, a dilute solution of a strong acid is not particularly reactive but a concentrated solution is!

17.4 Strong Acids and Strong Bases

- a strong acid or a strong base will completely dissociate in aqueous solution - giving the concentration of H+ and OH- directly

- generally, can ignore the contribution of water as it will be very small compared to the contribution from the strong acid or base - except when the concentration of the strong acid or base is close to or less than 1 x10-7 M

- hence, calculating the pH, the pOH, or the pCl for the dissolution of 0.23M HCl is aqueous solution is straightforward:

0.23M HCl → 0.23M H+ + 0.23M Cl-

pH = -log(0.23) = 0.638

since, [H+] = [Cl-], pCl = pH = 0.638

pOH = 14 - pH = 14 - 0.638 = 13.362

Table 17-2 has the common "Strong acids and Bases"

"Are you wondering ….."These are sections of the book that seem a little tangental at times. However, the ones on pgs. 676 and 682 are important.

pg. 676 How to calculate [H3O+] in an extremely dilute solution of a strong acid.

Example: What is the pH of a solution of 1.0 x10-8 HCl?

If we assume [H+]total = 1.0 x10-8, then the answer would be pH = -log (1.0 x 10-8) = 8

- but clearly, an acid should have a pH more acidic than neutral

- must take auto-ionization of water into account

H2O + H2O W H3O+ + OH-

HCl + H2O W H3O+ + Cl-

[H3O+]total = 1.0 x10-8 + x (from water)

- hence, Kw = [H3O+][OH-] = 1.0 x10-14

(x + 1.0 x10-8)(x) = 1.0 x10-14

x2 + 1.0 x10-8x - 1.0 x10-14 = 0

solving the quadratic equation gives,

x = -(1.0 x10-8) ± [(1.0 x10-8)2 - 4(1)(1.0 x10-14)]½

2(1)

= -1.0 x10-8 ± 2.0024 x10-7

2x = 9.512 x10-8

and [H3O+]total = 1.0 x10-8 + x= 1.0 x10-8 + 9.952 x10-8

= 1.0512 x10-7 pH = 6.978

17.5 Weak acids and Weak bases

- the difference between a "strong acid" and a "weak acid" is that a weak acid is only partially dissociated

- some of the "parent" acid will be present in solution along with the conjugate basei.e.

strong acid: HClO4 + H2O W H3O+ + ClO4- Keq > 106

weak acid: CH3COOH + H2O W CH3COO- + H3O+

Keq = 1.8 x10-5

- the size of the equilibrium constants tell us that in the strong acid, the concentrations of H3O+ and ClO4

- are very large compared to the concentration of HClO4whereas in the weak acid, the concentrations of H3O+

and CH3COO- are small compared to CH3COOH.

- why does this happen? the covalent bond between the proton and the acetate ion is much stronger than the one holding the proton to the perchlorate ion

Note: even the strongest covalent bond will eventually break, so even methane (CH4) is an acid - just a very, very weak acid!

- to talk about weak acids, we use pKa's as a method of presenting the acid equilibrium constant

working through questions follows the same steps that we used in solving other equilibrium problems:

example: determine the Ka of a weak acid if the total concentration of acid is 0.300 M and the pH is 2.46

step 1) write out equilibriumHA W H+ + A-

step 2) write out equilibrium constant expressionKa = [H+][A-] / [HA]

step 3) ICE tableHA W H+ + A-

initial 0.300 M 0.0 M 0.0 Mchange - x + x + xequilibrium 0.300 - x x x

but, in this case, we know "x" because we know the final pH - hence, x = [H+] = 10-2.46 = 3.467 x10-3 M

hence, [H+] = [A-] = 0.003467 M; [HA] = 0.2965 M

Ka = (0.003467)(0.003467)/(0.2965) = 4.054 x10-5

Example:What is the pH of a 1.0 M solution of a weak acid (Ka = 4.054 x10-5)?

step 1) equation: HA W H+ + A-

step 2) equilibrium: Ka = [H+][A-] / [HA]

step 3) ICE: HA W H+ + A-

initial 1.0 0.0 0.0change - x + x + x

equilibrium 1.0 - x x x

step 4) Ka = (x)(x) / (1.0 - x) = 4.054 x10-5

step 5) x2 = 4.054 x10-5(1.0 - x) = 4.05 x10-5 - 4.054 x10-5x

x2 + 4.054 x10-5x - 4.054 x10-5 = 0

step 6)x = -(4.054 x10-5) ± [(4.054 x10-5)2 - 4(1)(4.054x10-5)]½

2(1)x = -4.054 x10-5 ± 1.2734 x10-2 x = 0.00635

2

step 7) [H+] = x = 0.00635pH = -log(0.00635) = 2.197

step 8) can't really check the pH but we can check the equilibrium:

Ka = (0.00635)2 /(1 - 0.00635) = 4.058 x10-5

Also, does a pH of 2.197 sound reasonable? Yes. In the last problem, an acid concentration of 0.300 M resulted in a pH of 2.46. It makes sense that an increase in the acid concentration should result in a decrease in pH (an increase in the number of H+ ions in solution).

Example: A solution of vinegar is typically about 3% acetic acid in water, which corresponds to about 0.500 M. At this concentration, with Ka = 1.8 x10-5, what is the pH of vinegar?

step 1)

What about weak bases?Example:What is the pH of a 2.3 x102 M solution of Ammonia (Kb = 1.8 x10-5)?step 1)

book advocates using a simplifying assumption - that the amount of dissociation is small compared to the overall concentration of acid

in acetic acid example, we had

Ka = (x)(x) / (0.500 - x) = 1.8 x10-5

but if x << 0.500 then (0.500 - x) ≈ 0.500 and the equation would be:

x2 / 0.500 = 1.8 x10-5

x2 = 0.90 x10-5 x = 0.003 M

this is reasonably close to the value that we got using the quadratic, 2.99 x10-3 M, and certainly within rounding errors

- hence, we could have saved some time this way

however, consider the example of a weak base such as ammonia

Kb = (x)(x) / (0.023 - x) = 1.8 x10-5

after simplifying,x2 / 0.023 = 1.8 x10-5

x2 = 4.14 x10-5 x = 6.434 x10-4

compared to 6.345 x10-4

- while this is close, they are not exactly the same and could lead to errors.

Simplification does not always work. It is necessary to know the long route, no matter what.

As an approach, you can always try the short route first but if x is more than 5% of the concentration, then do the quadratic.

Note: quadratic formula always works but is slower.simplification doesn't always work but is fast.

Percent Ionization

- the percent ionization gives the proportion of ionized molecules (or dissociated molecules)

% ionization = molarity of H+ from HA x 100%molarity of HA

- strong acids - percent ionization is virtually 100%- weak acids - percent ionization is less than 100%

but it changes with concentration

- the more dilute the acid, the greater the % ionizationi.e. for acetic acid

% ionization is 0.6% at 0.500 M% ionization is 4% at 0.010 M% ionization is 100% at 0.0000010 M

- this gives us a clue as to what sort of answer to expect as the more dilute the acid, the closer the concentration of H+ will be to the concentration of the acid

that is, [H+] goes to [HA] as the acid gets more dilute

Were you wondering ….. on page 682- math is obtuse and it is not something that we are going to deal with overtly at this point, however, it is a reminder that water is a good acid and a good base under dilute conditions

- also, the iterative process is a stepwise approach used in a lot of computer programming

- we will need to consider this when we talk about ions as acids and bases, and more importantly, the titration of weak acids and bases - but we will find a mathematically less obtuse answer!

17.6 Polyprotic Acids

- most of the acids that we have dealt with so far are "mono-protic" - they only give up a single proton upon dissociation (and the bases only accept a single proton)

- some acids are "polyprotic" as are some bases:

polyprotic acids: H3PO4, H2SO4, H2SO3, H2CO3

also the organic acids: oxalic, maleic, succinic

polyprotic bases: diaminoethane, morphine, DNA, RNA

- as examples, we will deal with two that are of equal importance to industry - phosphoric acid and sulfuric acid

Phosphoric acid, H3PO4, has three acid equilibria:

1) H3PO4 W H+ + H2PO4-

Ka1 = [H+][H2PO4-] / [H3PO4] = 7.1 x10-3

2) H2PO4- W H+ + HPO4

2-

Ka2 = [H+][HPO42-] / [H2PO4

-] = 6.3 x10-8

3) HPO42- W H+ + PO4

3-

Ka3 = [H+][PO43-] / [HPO4

2-] = 4.2 x10-13

as you might expect, Ka1 > Ka2 > Ka3

- it gets harder to remove the positively charged proton as the species gets more negatively charged

One way to illustrate what is going on is with a "speciation curve" - a plot of pH versus concentration:

Working questions with polyprotic acids means that we have to take each deprotonation into account.

Example,What is the pH of a 0.10M solution of phosphoric acid?

- need to do this step-wisefirst reaction:step 1) H3PO4 W H+ + H2PO4

-

step 2) Ka = [H+][H2PO4-] / [H3PO4] = 7.1 x10-3

step 3) ICE tableH3PO4 W H+ + H2PO4

-

initial 0.10 M 0.0 0.0change - x x x

equilibrium 0.10 - x x x

step 4) Ka = x2 /(0.10 - x) = 7.1 x10-3

check simplification: x2 / 0.10 = 7.1 x10-3

x = (7.1 x10-3 x 0.10)½

= 0.026 Mwhich is not small compared to 0.10 M

step 5) go the quadratic routex2 = 7.1 x10-3(0.10 - x)x2 + 7.1 x10-3x - 7.1 x10-4 = 0

step 6) x = -(0.0071) ± [(0.0071)2 - 4(1)(-7.1 x10-4)]½

x = -0.0071 ± 0.05376 x = 0.023332

step 7) [H+] = [H2PO4-] = 0.02333M; [H3PO4] = 0.07667M

step 8) check the equilibrium - it works

but what about the second step? does the second deprotonation now contribute?

step 1) write out equation: H2PO4- W H+ + HPO4

2-

step 2) Ka = [H+][HPO42-] / [H2PO4

-] = 6.3 x10-8

Note that we already have [H+] = 0.02333 M and [HPO4-] =

0.02333 M from the first equilibrium. This means that we are not starting from "scratch" here and the ICE table must reflect this.

step 3) ICE tableH2PO4

- W H+ + HPO42-

initial 0.02333M 0.02333M 0.0Mchange - x + x + x

equlibrium 0.02333 - x 0.02333 + x x

step 4) re-write Ka

Ka = (0.02333 + x)(x) / (0.02333 - x) = 6.3 x10-8

if we assume that 'x' is small compared to 0.02333, then we get: (0.02333)(x) / (0.02333) ≈ 6.3 x10-8

or: x = 6.3 x10-8 M

In this instance, the simplification works as 6.3 x10-8 M is very much smaller than 0.02333 M. Hence,

[H+] = 0.02333 + 6.3 x10-8 ≈ 0.02333 M

and pH = -log(0.02333) = 1.632

Note that since Ka3 is even smaller, it would make a very small contribution to [H+] - one that we can safely ignore.

Example:What is the pH of a 0.10M solution of sulfuric acid?

- for sulfuric acid, the first deprotonation proceeds 100% as it is a "strong acid" and hence, we don't have to calculate this as the concentration of protons will equal the concentration of the acid. That is,

H2SO4 → H+ + HSO4-

0.10M 0.10M

Note that these are the initial concentrations for these species in the second deprotonation, so once again, for the second deprotonation we are not starting from 'scratch'.

the second deprotonation:step 1) HSO4

- W H+ + SO42-

step 2) Ka = [H+][SO42-] / [HSO4

-] = 0.013

step 3) ICE table:HSO4

- W H+ + SO42-

initial 0.10 0.10 0.0change - x x x

equilibrium 0.10 - x 0.10 + x x

step 4) Ka = (0.10 + x)(x) / (0.10 - x) = 0.013

step 5)x2 + 0.10x = 0.013(0.10 - x) = 0.0013 - 0.013x

x2 + 0.113x - 0.0013 = 0

step 6) x = -(0.113) ± [(0.113)2 - 4(1)(-0.0013)]½

2(1)= (-0.113 ± 0.134)/2

x = 0.0105 or x = -0.1235

step 7) [H+]e = 0.10 + x = 0.10 + 0.0105 = 0.1105MpH = -log(0.1105) = 0.957

less than 1, as expected, because 0.10M H+ = pH 1

17.7 Ions as Acids and Bases- ionic species that are components of acid/base equilibria will change the pH of a solution when dissolved

- further, "the product of the ionization constant of an acid and its conjugate base equals the ion product of water"

- that is, Ka x Kb = Kw

for example, NaF in water gives F- which binds H+

because it is the conjugate base of a weak acidF- + H+ W HF Ka

-1 = [HF]/[H+][F-]

Hydrolysis- dissolving NaCl in water gives Na+

(aq) and Cl-(aq) but no change in pH as Cl- is a weak base and doesn't pull protons from water and Na+ is a weak acid- dissolving NH4Cl in water gives NH4

+(aq) and Cl-(aq)

which undergoes a change in pH due to the hydrolysis of the ammonium ion

NH4+ + H2O W NH3 + H3O+

- the ionic component of a weak acid or base, when added to water as a salt, will result in the breaking up of water molecules and a shift in the pH from neutrality

Example:What is the pH of a 0.10M solution of NaF?

step 1) F- + H2O W HF + OH-

step 2) Kb = [HF][OH-] / [F-] = Kw/Ka = 1.52 x10-11

step 3) ICE: F- W HF + OH-

initial 0.10M 0.0 0.0change - x x x

equilibrium 0.10 - x x x

step 4) Kb = (x)(x) / (0.10 - x) = 1.52 x10-11

step 5) in this case, it is safe to assume that 'x' is going to be very much less than 0.10, hence,

x2 / 0.10 = 1.52 x10-11

x = [(1.52 x10-11)(0.10)]½

= 1.23 x10-6

step 7) [OH-]e = x = 1.26 x10-6 pOH = 5.909pH = 14 -5.909 = 8.09

- the fluoride ion shifts the pH from neutral to basic

17.8 Molecular Structure and Acid-Base Behaviour

- this section deals with questions around why some acids are strong and some are weak - and the simple answer is the strength of the bond holding the proton

- however, general trends are:acidity increases going down a group

HF Ka = 6.6 x10-4

HCl Ka = 1.3 x106

HBr Ka = ≈108

HI Ka = ≈109

oxoacids - acidity increases with the number of oxygensKHClO2 < KHClO3 < KHClO4

- decreases for different central atoms as one goes down a groupKHClO4 > KHBrO4 > KHIO4

strength of organic acids- the length of the organic chain has little effect- presence of an electron withdrawing group has a

major effect on Ka

strength of amine bases - leave for organic chemistry

17.9 Lewis Acids and Bases- this is the most useful and general definition of acids and bases- it encompasses all of the other definitions and then some

- a "Lewis acid" is an electron pair acceptor- a "Lewis base" is an electron pair donor

i.e. H+ accepts a pair of electrons from OH-

O2 donates a pair of electrons to the iron in hemoglobin