Acids and Bases

Acids and Bases• Acids taste sour (citric acid, acetic acid)• Bases taste bitter (sodium bicarbonate)• There are 3 ways to define acids and bases, you will

learn 2 of these:Arrhenius:

- Acids form H3O+ in water (HCl + H2O H3O+ + Cl-)

- Bases form OH- in water (NaOH Na+ + OH-)Brønsted-Lowry (B-L):

- Acids donate H+ and Bases accept H+

HCl + NaOH H2O + NaCl

HCl is the acid, it donates H+ to OH- (the base)

B-L Acids and Formation of H3O+

• In an acid-base reaction, there is always an acid/base pair (the acid donates H+ to the base)

• H+ is not stable alone, so it will be transferred from one covalent bond to another

Example: formation of H3O+ from an acid in water

HBr + H2O H3O+ + Br-

Identifying B-L Acids and Bases• Compare the reactants and the products

- The reactant that loses an H+ is the acid- The reactant that gains an H+ is the base

Examples:HCl + H2O H3O+ + Cl-

Acid = HCl and Base = H2O (HCl gives H2O an H+)

NH3 + H2O NH4+ + OH-

Acid = H2O and Base = NH3 (H2O gives NH3 an H+)

CH3CO2H + NH3 CH3CO2- + NH4

+

Acid = CH3CO2H and Base = NH3 (CH3CO2H gives NH3 an H+)

Conjugate Acids and Bases• When a proton is transferred from the acid to the base

(in a B-L acid/base reaction), a new acid and a new base are formed:

HA + B A- + HB+

acid + base conjugate base + conjugate acid• The acid (HA) and the conjugate base (A-) that forms

when HA gives up an H+ are a conjugate acid/base pair

• The base (B) and the conjugate acid (HB+) that forms when B accepts an H+ are another conjugate acid/base pair

Identifying Conjugate Acid/Base Pairs

1. Identify the acid and base for the reactants 2. Identify the acid and base for the products3. Identify the conjugate acid/base pairs

acid conjugate base + +

base conjugate acid

HF H2O H3O+ F-

Acid and Base Strength• Strong acids give up protons easily and completely

ionize in water:HCl + H2O H3O+ + Cl-

• Weak acids give up protons less easily and only partially ionize in water:

CH3CO2H + H2O CH3CO2- + H3O+

• Strong bases have a strong attraction for H+ and completely ionize in water:

KOH(s) K+ (aq) + OH-(aq)NaNH2 + H2O NH3 + NaOH

• Weak bases have a weak attraction for H+ and only partially ionize in water:

HS- + H2O H2S + OH-

Direction of an Acid/Base Equilibrium• In general, there’s an inverse relationship between

acid/base strength within a conjugate pair:- strong acid weak conjugate base- strong base weak conjugate acid(and vice-versa)

• The equilibrium always favors the direction that goes from stronger acid to weaker acid

Example 1: HBr + H2O ? H3O+ + Br-

stronger acid (HBr) weaker acid (H3O+)(equilibrium favors products)

Example 2: NH3 + H2O ? NH4+ + OH-

weaker acid (H2O) stronger acid (NH4+)

(equilibrium favors reactants)

Dissociation Constants• Since weak acids dissociate reversibly in water, we can write an

equilibrium expression:HA + H2O H3O+ + A-

Keq = [H3O+][A-]/[HA][H2O]

• But, since [H2O] remains essentially constant we can write:

Ka = Keq x [H2O] = [H3O+][A-]/[HA]

• The acid dissociation constant (Ka) is a measure of how much the acid dissociates (A higher Ka = a stronger acid)

• Example: CH3CO2H + H2O CH3CO2- + H3O+

Ka = [H3O+][CH3CO2-]/[CH3CO2H] = 1.8 x 10-5

• Can also write dissociation constants for weak bases:NH3 + H2O NH4

+ + OH-

Kb = [NH4+][OH-]/[NH3] = 1.8 x 10-5

Ionization of Water• Since H2O can act as either a weak acid or a weak base,

one H2O can transfer a proton to another H2O:

H2O + H2O H3O+ + OH-

Keq = [H3O+][OH-]/[H2O][H2O]

• Since [H2O] is essentially constant, we can write:

Kw = Keq x [H2O]2 = [H3O+][OH-]

(where Kw = the ion-product constant for water)

• For pure water: [H3O+] = [OH-] = 1.0 x 10-7 M

So, Kw = [H3O+][OH-] = (1.0 x 10-7 M)2 = 1.0 x 10-14

(units are omitted for Kw as for Keq and Ka)

Using Kw

• If acid is added to water, [H3O+] goes up- for an acidic solution [H3O+] > [OH-]

• If base is added to water, [OH-] goes up- for a basic solution [OH-] > [H3O+]

• Kw is constant (1.0 x 10-14) for all aqueous solutions• Can use Kw to calculate either [H3O+] or [OH-] if given the

other concentration• Example: if [H3O+] = 1.0 x 10-4 M, what is the [OH-]?

Kw = [H3O+][OH-][OH-] = Kw/ [H3O+] = 1.0 x 10-14/1.0 x 10-4= 1.0 x 10-10 M

Is this an acidic or a basic solution?Since [H3O+] > [OH-], it’s an acidic solution

The pH Scale• pH is a way to express [H3O+] in numbers that are easy to

work with• [H3O+] has a large range (1.0 M to 1.0 x 10-14 M) so we use

a log scale:pH = - log [H3O+]

• The pH scale goes from 0 - 14• Each pH unit = a ten-fold change in [H3O+]

• pH 7 = neutral, pH < 7 = acidic, pH > 7 = basic• Can use an indicator dye (on paper or in solution) that

changes color with changes in pH, or a pH meter, to measure pH

Calculating pH and pOH• Can calculate pH from [H3O+]:

If [H3O+] = 1.0 x 10-3 M, what is the pH?

pH = - log [H3O+] = - log(1.0 x 10-3) = 3.00

• Note: sig. figs. in [H3O+] = decimal places in pH

• Can also calculate [H3O+] from pH:

If pH is a whole number, [H3O+] = 1 x 10-pH

So, if pH = 2, then [H3O+] = 1 x 10-2

• Can calculate pOH from [OH-]pOH = - log[OH-]

• Also, since Kw = [H3O+][OH-]

then pKw = - log Kw = - log (1.0 x 10-14) = 14.00

• And, pKw = pH + pOH = 14.00• So, if pH = 3.00, then pOH = 14.00 - 3.00 = 11.00

Reactions of Acids and Bases• Acids and bases are involved in a variety of chemical reactions (we’ll

study 3 types here)• Acids react with certain metals to produce metal salts and H2 gas, for

example:Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)

• Acids react with carbonates and bicarbonates to produce salts, H2O and CO2 gas, for example:

NaHCO3(aq) + HCl(aq) NaCl(aq) + H2O(l) + CO2(g)• Acids react with bases (neutralization reactions) to form salts and

H2O, for example:

HBr(aq) + LiOH(aq) LiBr (aq) + H2O(l)• Neutralization reactions are balanced with respect to moles of H+ and

moles of OH-, for example:H2SO4 + 2NaOH Na2SO4 + 2H2O

Acidity of Salt Solutions• Salts dissolved in water can affect the pH

• When salts dissolve, they dissociate into their ions

NaCl Na+ + Cl-

• If one of those ions can donate a proton to H2O, or

accept one from H2O, the pH will change:

Na2S 2Na+ + S2-

S2- + H2O HS- + OH-

S2- is a weak base that can accept an H+ from H2O

Since [OH-] is increased, the solution is basic

Salts that form Neutral Solutions• When a strong acid dissolves in water, a weak conjugate base is

formed that can’t remove a proton from water• When a strong base dissolves in water, the metal that dissociates

can’t form H3O+

• So, salts containing ions that come from strong acids and bases do not affect the pH of the solution

• Example:KBr K+ + Br- (KOH = strong base, HBr = strong acid)

(KOH + HBr KBr + H2O)

K+ has no proton to donate, so can’t form H3O+

Br- is too weak of a base to pull a proton off of H2O, so can’t form OH-

So, the solution remains neutral

Salts that form Basic Solutions• When a weak acid dissolves in water, the conjugate

base formed is usually strong enough to remove a proton from H2O to form OH-

• So, salts that contain ions that come from a weak acid and a strong base form basic solutions

• Example:NaCN Na+ + CN- (HCN is a weak acid)CN- + H2O HCN + OH-

(HCN + NaOH NaCN + H2O)

(Na+ doesn’t affect the pH, it’s from a strong base)

Salts that form Acidic Solutions

• When a weak base dissolves in water, the conjugate acid formed is usually strong enough to donate a proton to H2O to form H3O+

• So, salts that contain ions from a weak base and a strong acid form acidic solutions

• Example:

NH4Br NH4+ + Br- (from NH3 and HBr)

(NH3 + HBr NH4Br)

NH4+ + H2O NH3 + H3O+

(Br- doesn’t affect the pH)

Buffer solutions• A small amount of strong acid or base added to pure water will

cause a very large change in pH• A buffer is a solution that can resist changes in pH upon

addition of small amounts of strong acid or base• Body fluids, such as blood, are buffered to maintain a fairly

constant pH• Buffers are made from conjugate acid/base pairs (either a weak

acid and a salt of its conjugate base or a weak base and a salt of its conjugate acid)

• Thus, they contain an acid to neutralize any added base, and a base to neutralize any added acid

• Buffers can’t be made from strong acids or bases and the salts of their conjugates since they completely ionize in H2O

How to Make a Buffer Solution• An acetate buffer is made from acetic acid and a salt of its conjugate

base:CH3CO2H and CH3CO2Na

• The salt is used to increase the concentration of CH3CO2- in the buffer

solution• Recall: CH3CO2H + H2O CH3CO2

- + H3O+

(the equilibrium favors reactants, so the concentration of CH3CO2- is

low)But, CH3CO2Na CH3CO2

- + Na+

CH3CO2H + CH3CO2Na + H2O 2 CH3CO2- + H3O+ + Na+

• If acid is added: CH3CO2- + H3O+ CH3CO2H + H2O

• If base is added: CH3CO2H + OH- CH3CO2- + H2O

• Buffer capacity = how much acid or base can be added and still maintain pH (depends on buffer type and concentration)

Calculating pH of a Buffer• The pH of a buffer solution can be calculated from the

acid dissociation constant (Ka)

• Example (for acetate buffer):

CH3CO2H + H2O CH3CO2- + H3O+

Ka = [H3O+][CH3CO2-]/[CH3CO2H] = 1.8 x 10-5

[H3O+] = Ka x [CH3CO2H]/[CH3CO2-]

• What is the pH of an acetate buffer that is 1.0 M CH3CO2H and 0.50 M CH3CO2Na?

[H3O+] = 1.8 x 10-5 x 1.0 M/0.50 M = 3.6 x 10-5 M

pH = - log[H3O+] = - log(3.6 x 10-5) = 4.44

Dilutions• Often solutions are obtained and stored as highly

concentrated stock solutions that are diluted for use (i.e. cleaning products, frozen juices)

• When a solution is diluted by adding solvent, the volume increases, but amount of solute stays the same, so the concentration decreases:

Mol solute = concentration (mol/L) x V (L) = constant

So, C1V1 = C2V2

• For molarity, it becomes: M1V1 = M2V2

Dilution Calculations• Example 1:

What volume (in mL) of 8.0 M HCl is needed to prepare 1.0 L of 0.50 M HCl?

M1V1 = M2V2 V1 = M2V2/ M1

V1 = 0.50 M x 1.0 L/ 8.0 M = 0.0625 L = 63 mL

• Example 2:How many L of water do you need to add to dilute 0.50 L of a 10.0 M NaOH solution to 1.0 M ?

V2 = M1V1/ M2

V2 = 10.0 M x 0.50 L/ 1.0 M = 5.0 L

volume of water needed = 5.0 L - 0.50 L = 4.5 L

Acid-Base Titration• Molarity of an acid or base solution of unknown

concentration can be determined by titration:1. A measured volume of the unknown acid or base is

placed in a flask and a few drops of indicator dye (such as phenolpthalein) are added

2. A buret is filled with a measured molarity of known base or acid (the “titrant”) and small amounts are added until the solution changes color (neutralization endpoint)

- At neutralization endpoint [H3O+] = [OH-]

3. Molarity of unknown is calculated from moles of titrant added (mole ratio comes from balanced chemical equation)

Example: Titration of H2SO4 with NaOH

• What is the molarity of a 10.0 mL sample of H2SO4 if the neutralization endpoint is reached after adding 15.0 mL of 1.00 M NaOH?

• Calculate moles NaOH added:

• 15.0 mL x (1 L/ 1000 mL) x (1.00 mol/ 1 L) = 0.0150 mol NaOH

• Write the balanced chemical equation:

H2SO4 + 2NaOH Na2SO4 + 2H2O

• Calculate moles H2SO4 neutralized:

0.0150 mol NaOH x 1 mol H2SO4/ 2 mol NaOH = 0.00750 mol H2SO4

• Calculate molarity of H2SO4:

0.00750 mol H2SO4/ 0.0100 L = 0.750 M H2SO4