Topic 18- Acids and bases

18.1 Calculations involving acids and bases18.2 Buffer solutions18.3 Salt hydrolysis18.4 Acid-base titrations18.5 Indicators

The relation between oxonium ions and hydroxide ions

18.1 Calculations involving acids and bases

• pH = -log[H3O+] pOH = -log[OH-]]

• [H3O+] = 10-pH [OH-] = 10-pOH

• pH + pOH = 14

[H+]=[H3O+]

Calculate the pH in following solutions:

• 1M HCl pH = -log[1] = 0

• 0.001 M HNO3 pH = -log [0.001] = 3

• 0.5 M H2SO4 2H+ pH = -log [2*0.5]= 0

• 0.15 M NaOH pOH =-log[0.15]= 0.82pH = 14-0.82 = 13.18

Calculate the [H3O+] in following solutions

• pH = 5,5 [H3O+] = 10-5.5 = 3.2*10-6

• pH =- 1 [H3O+] =10-(-1) = 101 = 10 M

Autoprotolysis of water

H2O + H2O H3O+ + OH-

Kc= [H3O+] . [OH- ]

[H2O] . [H2O]The concentration of water is not changing- it is constant

Kw= K . [H2O] . [H2O] = [H3O+] . [OH-]

Kw = the dissociation constant of water

Kw = dissociation constant of water

H2O + H2O H3O+ + OH- In 25oC pure water:

[H3O+] = 10-7 mol/dm3

[OH-] = 10-7 mol /dm3

Kw = [H3O+]*[OH-] = 10-7*10-7 = 10-14 mol2/dm6

-log Kw = -log ([H3O+]*[OH-])= -log [H3O+]+ -log[OH-]=-log 10-14

pKw = pH + pOH = 14

Kw= 1,0 . 10 -14 vid 25 C

The effect of temperature on the water dissociation constant

Temperature (◦C)

Kw pKw

0 0,11 . 10 -14 14,96

15 0,45 . 10 -14 14,35

25 1,0 . 10 -14 14,00

50 5,5 . 10 -14 13,26

100 51 . 10 -14 12,29

H2O + H2O H3O+ + OH- DH> 0

HA + H2O H3O+ + A-

K= [H3O+] . [A- ]

[HA] . [H2O]

Ka=K . [H2O] = [H3O+] . [A- ]

[HA]

Weak acids

Ka = acid dissociation constantKa => acid strength => higher value => stronger acid.pKa = -log Ka => lower value => stronger acid

Ka pKa

Hydrochloric acid HCl 104 -4

Sulphuric acid H2SO4 103 -3

Nitric acid HNO3 1 -1

Oxalic acid H2C2O4 0.06 1.25

Phosphoric acid H3PO4 0.007 2.15

Salicylic acid C7H6O3 0.001 2.97

Citric acid C6H8O7 7.10-4 3.13

Ascorbic acid C6H4O2(OH)4 1.10-4 4.17

Acetic (etanoic) acid HAc CH3COOH 1.8.10-5 4.75

Carbonic acid H2CO3 4.2.10-7 6.37

Ammonium ion NH4+ 5.5.10-10 9.26

Some acid dissociation constants

BH + H2O BH2+ + OH-

K= [BH2+] . [OH-]

[BH] . [H2O]

Kb=K . [H2O] = [BH2+] . [OH-]

[BH]

Weak bases

Kb = base dissociation constantKb => base strength => higher value => stronger basepKb = -logKb => lower value => stronger base

Kb pKb

Litium hydroxide LiOH 2.3 -0.36

Sodium hydroxide NaOH 0.63 0.2

Ammonia NH3 1.8.10-5 4.74

Hydrogen carbonate ion

HCO3- 2.4.10-8 7.62

Acetate ion CH3COO- 5.5.10-10 9.25

Nitrate ion NO3- 10-15 15

Chloride ion Cl- 10-18 18

Bases

Kw connects Ka and Kb for a

corresponding acid/ base pair, such as CH3COOH/CH3COO-

Ka * Kb = Kw = 10-14

pKa + pKb = pKw = 14

(at 25 ºC)

Calculate pKa of ethanoic acid, HAc (CH3COOH)We know that c= 0.01M We measure pH

HAc+H2O H3O+ +Ac-

Cstart 0.1 0 0

Ceq 0.1- 10-pH 10-pH 10-pH

Ka = [H3O+]*[Ac-] / [HAc] = 10-pH* 10-pH /0.1- 10-pH=

pKa= -log Ka =

Calculate pH in 0,1 M ethanoic acid, HAc

HAc H3O+ +Ac-

Cstart 0.1 0 0 pKa = 4.75

Ceq 0.1-X X X (see CDB)

Ka = [H3O+]*[Ac-] / [HAc] = 10-4.75 = X2/0.1-X ~ X2/0.1 if x is small

X2 = 0.1* 10-4.75 X = (0.1* 10-4.75 )½

pH = -log [X] = -log[(0.1* 10-4.75 )½] = 2.88

18.2 Buffer solutions

• In pure water pH= 7.• Addition of small amounts of acid or base

gives big changes in pH• That can be great problem, especially in biological systems. But there are

ways to make a solution that can be quite pH stable.

• A buffer resist changes in pH when a strong acid or base is added

A Buffer: a mixture of a weak acid and its corresponding base

The equilibrium: HA(aq) +H2O H3O+

(aq) + A-(aq)

If strong acid is added (fully dissociated, contains mainly H3O+) =>

reaction goes to the left (Le Chatelier’s principle) => little change in pH (-log [H3O+])

A Buffer: a mixture of a weak acid and its corresponding base

The equilibrium: HA(aq) +H2O H3O+(aq) + A-

(aq)

• If a strong base is added => OH- reacts with H3O+ to form water =>

• equilibrium goes to the right => Restore the [H3O+] => little change in pH

How to prepare a buffer:

• Mix a weak acid and its corresponding (conjugate) base e.g. CH3COOH and CH3COO-Na+.

• Mix a weak base and its corresponding (conjugate) e.g. NH3 and NH4Cl.

• Add strong base to an excess of weak acid.

• Add strong acid to an excess of weak base.

Add strong base to an excess of weak acid (HA)

The reaction of HA and water:HA(aq) +H2O H3O+

(aq) + A-(aq)

The reaction of HA and strong base (NaOH):HA + NaOH → H2O + Na+A-

When you add some NaOH to a HA-solution your mixture will consist of all the above particles, but in particular the weak acid and it’s corresponding base HA/A-

Add strong acid to an excess of weak base (BH)

The reaction of BH and water:

The reaction of B and strong acid (HCl):HCl + BH → BH2

+Cl-

When you add some HCl to a BH-solution your mixture will consist of all the above particles, but in particular the weak base and it’s corresponding acid BH/BH2

+

BH + H2O BH2+ + OH-

The Hydrogen ion concentration and pH of a buffer can be calculated from the expression for the acid dissociation constant:

[H3O+] = Ka* [HA] /[A-] take –log on both sides

-log [H3O+] = -log Ka+ -log [HA] /[A-] identify

pH = pKa -log([HA] /[A-])

You have to be able to derive the equation yourself!

Exercises1. Which combination will form a buffer solution?A. 100 cm3 of 0.10 mol dm–3 hydrochloric acid with 50 cm3 of 0.10 mol dm–3 sodium hydroxide.B. 100 cm3 of 0.10 mol dm–3 ethanoic acid with 50 cm3 of 0.10 mol dm–3 sodium hydroxide.C. 50 cm3 of 0.10 mol dm–3 hydrochloric acid with 100 cm3 of 0.10 mol dm–3 sodium hydroxide.D. 50 cm3 of 0.10 mol dm–3 ethanoic acid with 100 cm3 of 0.10 mol dm–3 sodium hydroxide.

2. Buffer solutions resist small changes in pH. A phosphate buffer can be made by dissolving NaH2PO4 and Na2HPO4 in water, in which NaH2PO4 produces the acidic ion and Na2HPO4 produces the conjugate base ion.(i) Deduce the acid and conjugate base ions that make up the phosphate buffer and state the ionic

equation that represents the phosphate buffer.

(ii) Describe how the phosphate buffer minimizes the effect of the addition of a strong base, OH–

(aq), to the buffer. Illustrate your answer with an ionic equation.

(iii) Describe how the phosphate buffer minimizes the effect of the addition of a strong acid, H+(aq), to the buffer. Illustrate your answer with an ionic equation.

Answers1. Which combination will form a buffer solution?B. 100 cm3 of 0.10 mol dm–3 ethanoic acid with 50 cm3 of 0.10 mol dm–3 sodium hydroxide.You have both ethanoic acid and sodium ethanoate

2. (i) Acid: H2PO4–; (Conjugate) base: HPO4

2–; H2PO4–(aq) H+(aq) + HPO4

2–(aq);

(ii) strong base/OH– replaced by weak base (H2PO42–, and effect minimized) /

strong base reacts with acid of buffer / equilibrium in (i) shifts in forwarddirection; OH–(aq) + H2PO4

–(aq) → H2O(l) + HPO42–(aq);

(iii) strong acid/H+ replaced by weak acid (H2PO4–, and effect minimized) /

strong acid reacts with base of buffer / equilibrium in (i) shifts inreverse direction;H+(aq) + HPO4

2–(aq) → H2PO4–(aq);

18.3 Salt hydrolysis

• The positive and negative ion in a salt can be neutral or act as an acid or a base

• Cations (positive ions) can act as acids and anions (negative ions) can act as bases

The acetate (ethanoate) ion

HAc + H2O H3O+ + Ac- pKa(HAc)= 4.75 acid base

Ac- + H2O HAc + OH- pKb(Ac-)= 9.25base acid

The ethanoate ion is salt of a WEAK acid (ethanoic acid) and thus basic

More basic ions

• CN- • HCO3

-

• CO32-

• PO43-

Salts of weak acids

The chloride ion

HCl + H2O H3O+ + Cl- pKa(HCl)= - 4 acid base

Cl- + H2O HCl + OH- pKb(Cl-)= 18base acid

The chloride ion is salt of a STRONG acid (hydrochloric acid) and thus so week so it is neutral (>pKw)

More ions with no acid/base character

• Na+ SO42-

• K+ ClO4-

• Ca2+

• NO3- Cl-

• Derives from strong acids and bases => no acid-base activity

NH3 + H2O NH4+ + OH- pKb(NH3)= 4.74

base acid

The ammonium ion

NH4+ + H2O H3O+ + NH3 pKa(NH4

+)= acid base

Ammonium ion is salt of a WEAK base (ammonia) and thus acidic

Metallic ions with high charge

• Metallic ions with high charge, e.g. Al3+ and Fe3+, form complexes with water:

• Al(H2O)63+ and Fe(H2O)6

3+. The electronegative effect of the ion weakens the O-H bond in water molecules:

[Fe(H2O)6]3+(aq) + H2O [Fe(OH)(H2O)5]3+

(aq)+ H3O+(aq)

An acidic solution

18.4 Acid-base titrations

• Strong acid – Strong base• Weak acid – Strong base• Strong acid – Weak base

Strong acid – Strong base HCl + NaOH NaCl + H2O

Start with 10 ml 0.1 M HCl (pH=1). Titrate with 0.1M NaOH

• When 90% of the base been added: HCl ~0.01 => pH = 2• When 99% of the base been added: HCl ~0.001 => pH = 3• When 101% of the base been added: [OH-] = 0.001 pH =11

Weak acid – strong baseCH3COOH + NaOH CH3COONa + H2O• When strong base is added the pH gradually increase.• At equivalence point all acid is consumed, pH increase rapidly.• The salt of the weak acid is a weak base => pH > 7 at

equivalence point.• At ½ equivalence point [HA] =[A-] => pH => pKa

Titration simulations athttp://chem-ilp.net/labTechniques/AcidBaseIdicatorSimulation.htm

18.5 Indicators

• A weak acid/base where the colours of the protonated and ionized forms are different

HIn H+ + In-

Red Blue

The colour depends both on pH and the pKa-value. => Different indicators change their colours at different pH

Structures of BTB at different pH

Increasing [OH-]/pH

pH 7.3

Indicators change colour around their pKa-values

How to choose indicator?• If you titrate CH3COOH with NaOH the pH will be above 7 at

the equivalence point => choose an indicator that change colour above 7 e.g. phenolphthalein (pKa =9.6), range 8.3 – 10.0. Rapid pH changes in that area.

- If you titrate NH3 with HCl the pH will be under 7 at the equivalence point => choose an indicator that change colour under 7 e.g. methyl orange(pKa = 3.7), range

3.1 – 4.4. Rapid pH changes in that area.