Acids - Ionisation All acids ionise in solution to give ………ions. HCl  H+ H+ H+ H+ + Cl - Cl - (100%) HCl (+ H 2 O) H 2 O)  H 3 O + (aq) H 3 O + (aq) + Cl - (aq) H 2 O H 2 O + H+ H+ H+ H+  H3O+H3O+H3O+H3O+ Nitric Acid: Acid: HNO 3 HNO 3 …………………… (nitrate anion) Sulphuric Acid: Acid: H 2 SO 4 H 2 SO 4 ……………………. (sulphate anion) Carbonic Acid: H 2 CO 3 H 2 CO 3 ……………………. (carbonate anion) Ethanoic acid: CH 3 COOH CH 3 COOH …………………….. (ethanoate anion)

Acids - Ionisation All acids ionise in solution to give H+ H+ H+ H+ ions. HCl  H+ H+ H+ H+ + Cl - Cl - (100%) HCl (+ H 2 O) H 2 O)  H 3 O + (aq) H 3 O + (aq) + Cl - (aq) H 2 O H 2 O + H+ H+ H+ H+  H3O+H3O+H3O+H3O+ Nitric Acid: Acid: HNO 3 HNO 3  H+ H+ H+ H+ + NO 3 - NO 3 - (nitrate anion) Sulphuric Acid: Acid: H 2 SO 4 H 2 SO 4  2H+ 2H+ 2H+ 2H+ + SO 4 2- SO 4 2- (sulphate anion) Carbonic Acid: H 2 CO 3 H 2 CO 3  2H+ 2H+ 2H+ 2H+ + CO 3 2- CO 3 2- (carbonate anion) Ethanoic acid: CH 3 COOH CH 3 COOH  H+ H+ H+ H+ + CH 3 COO - (ethanoate anion)

Bases Bases …………….acids forming a ……….+ WATER by accepting a Hydrogen ion. NaOH (s) (+ H 2 O)  …………. (aq) + OH - (aq) NH 3 + H 2 O  …………. + OH - NaOH + HCl  …………….. + H2OH2O NH 3 + HNO 3  ……………+ NO 3 -  Liberate ……………. ions in water. HF + KOH  ……… + H 2 O H 2 CO 3 + 2KOH  ……………+ 2H 2 O Ca(OH) 2(s) (+ H 2 O)  …………… (aq) + 2OH - (aq) Mg(OH) 2(s) (+ H 2 O)  …………. (aq) + 2OH - (aq)

Bases Bases neutralise acids forming a SALT + WATER by accepting a Hydrogen ion. NaOH (s) (+ H 2 O)  Na + (aq) + OH - (aq) NH 3 + H 2 O  NH OH - NaOH + HCl  NaCl + H2OH2O NH 3 + HNO 3  NH4+ NH4+ + NO 3 -  Liberate OH - ions in water. HF + KOH  KF + H 2 O H 2 CO 3 + 2KOH  K 2 CO 3 + 2H 2 O Ca(OH) 2(s) (+ H 2 O)  Ca 2+ (aq) + 2OH - (aq) Mg(OH) 2(s) (+ H 2 O)  Mg 2+ (aq) + 2OH - (aq)

Bases Soluble bases are called alkalis. Metal Hydroxides Metal Oxides Metal Carbonates Metal Hydrogen Carbonates Ammonia NaOH CaO CaCO 3 NaHCO 3 NH 3 Ca(OH) … Na … O Na …. CO 3 Ca(HCO 3 )... CH 3 NH 2 Zn(OH) … Al …….. Li …. CO 3 Mg ………… Al………. Fe (III).. Al …. (CO 3 ) …. Al ………… NH 4 ……. Mn. (4+) NH 4……….. NH 4…………

Bases Soluble bases are called alkalis. Metal Hydroxides Metal Oxides Metal Carbonates Metal Hydrogen Carbonates Ammonia NaOH CaO CaCO 3 NaHCO 3 NH 3 Ca(OH) 2 Na 2 O Na 2 CO 3 Ca(HCO 3 ) 2 CH 3 NH 2 Zn(OH) 2 Al 2 O 3 Li 2 CO 3 Mg(HCO 3 ) 2 Al(OH) 3 Fe 2 O 3 Al 2 (CO 3 ) 3 Al(HCO 3 ) 3 NH 4 OH MnO 2 (NH 4 ) 2 CO 3 NH 4 HCO 3

Bases Soluble bases are called Metal Hydroxides Metal Oxides Metal Carbonates Metal Hydrogen Carbonates Ammonia NaOH CaO CaCO 3 NaHCO 3 NH 3

Bases Soluble bases are called alkalis. Metal Hydroxides Metal Oxides Metal Carbonates Metal Hydrogen Carbonates Ammonia NaOH CaO CaCO 3 NaHCO 3 NH 3

Formula Class Test Write the name or formula of the following ions: 1. Carbonate ion 2. Nitrite ion 3. Sulphide ion 4. SO NH OH - 7. O 2- Write the name or formula of each of the following compounds: 1. Sodium chloride 2. Magnesium oxide 3. Calcium hydroxide 4. Potassium sulphate 5. Cu(NO 3 ) 2 6. NH 4 Cl 7. PbI

SALTS  A SALT is a compound formed when a metal is ionically bonded to a non-metal.  Salts are formed in many acid-base reactions. Acid + Metal  SALT + hydrogen Acid + Metal Oxide  SALT + water Acid + Metal hydroxide  SALT + water Acid + Metal carbonate  SALT + carbon dioxide + water Acid + Metal hydrogen carbonate  SALT + carbon dioxide + water SALTACIDBASEACIDBASESALT MgCl 2 HClMg(OH) 2 Fe(NO 3 ) 2 H 2 SO 3 NaOH MgCO 3 H 2 CO 3 CaO Al 2 (SO 4 ) 3 H 2 CO 3 CaCO 3 (NH 4 ) 3 PO 4 H 2 SO 4 Mg

SALTS  A SALT is a compound formed when a metal is ionically bonded to a non-metal.  Salts are formed in many acid-base reactions. Acid + Metal  SALT + hydrogen Acid + Metal Oxide  SALT + water Acid + Metal hydroxide  SALT + water Acid + Metal carbonate  SALT + carbon dioxide + water Acid + Metal hydrogen carbonate  SALT + carbon dioxide + water SALTACIDBASEACIDBASESALT MgCl 2HCLMgHClMg(OH) 2 MgCl 2 Fe(NO 3 ) 2 HNO 3 Fe(OH) 2 H 2 SO 3 NaOH Na 2 SO 3 MgCO 3 H 2 CO 3 MgO CaO CaCO 3 Al 2 (SO 4 ) 3 H 2 SO 4 Al(OH) 3 H 2 CO 3 CaCO 3 (NH 4 ) 3 PO 4 H 3 PO 4 NH 3 H 2 SO 4Mg SO 4 MgSO 4.

Cl - Acids - ionisation 2 H+H+ H+H+ H+H+ Cl - H+H+ H+H+ 20 HCl Dilute Strong acid all molecules dissociated few molecules / dm 3. Concentrated Strong Acid: All molecules dissociated Many molecules per dm 3 H Cl H+H+ Cl - H+H+ H+H+ H+H+ H+H+ H+H+ H+H+ H+H+ H+H+ H+H+ H+H+ H+H+ H+H+ H+H+ H Cl

(Water) Strong - Weak Acids 5 Strong acid all molecules dissociateall molecules dissociate Cl - H+H+ H+H+ Weak Acid  partially ionised - hardly any ions)  HF  F - + H +  hydrofluoric acid fluoride ion 5 Cl - H+H+ H+H+ H+H+ H Cl H+H+ F-F-F-F- H+H+ F-F-F-F- H+H+ F-F-F-F- H+H+ F-F-F-F- H+H+ F-F-F-F- H+H+ F-F-F-F-

Weak Acids - Ionisation HF --> F - + H + ( partially - hardly any) hydrofluoric acid fluoride ion Concentrated Weak acid  Hardly any dissociation  Lots of molecules/dm 3 2H-F F-F- HF H+H+ F-F- H+H+ HF - 20HF Dilute Weak acid Hardly any molecules dissociate Hardly any molecules dissociate few molecules / dm 3.few molecules / dm 3.

Base Strength  Strong bases ionise completely.  Weak bases ionise only ……………. (Double arrow). NaOH (s) (+ H 2 O)  Na + (aq) + OH - (aq) NH 3 + H 2 O  ……………… + OH - Mg(OH) 2  ……….. + 2OH - Ca(OH) 2(s) (+ H 2 O)  ………………+ 2OH - (aq) KOH (s) (+ H 2 O)  K + (aq) + OH - (aq) 100 %

Base Strength  Strong bases ionise completely.  Weak bases ionise only partially (Double arrow). NaOH (s) (+ H 2 O)  Na + (aq) + OH - (aq) NH 3 + H 2 O  NH OH - Mg(OH) 2  Mg OH - Ca(OH) 2(s) (+ H 2 O)  Ca 2+ (aq) + 2OH - (aq) KOH (s) (+ H 2 O)  K + (aq) + OH - (aq)

Ionisation of water Water molecules can ionise as follows H 2 O  H + + OH - This does not happen to any great extent so the concentration of these ions is very small. [H + ] = [OH - ] mol.dm -3 (1 x ) at o C The reaction is reversible so can be written as H 2 O H + + OH - The equilibrium constant is written as K w = and has the value mol 2 dm -2 at 25 o C This value even if [H + ] or [OH - ] changes.

Ionisation of water Water molecules can ionise as follows H 2 O  H + + OH - This does not happen to any great extent so the concentration of these ions is very small. [H + ] = [OH - ] = mol.dm -3 (1 x ) at 25 o C The reaction is reversible so can be written as H 2 O  H + + OH - The equilibrium constant is written as K w = [H + ] [OH - ] and has the value 1 x mol 2 dm -2 at 25 o C This value does NOT change even if [H + ] or [OH - ] changes.

Acid strength & pH H 3 O + concentration (mol/dm 3 ) acids acids (1x10 -4 ) (1x10 -5 )  Diluting a strong acid by a factor of changes its pH by one unit. pH = (1x )

Acid strength & pH H 3 O + concentration (mol/dm 3 ) Strong acids Weak acids (1x10 -4 ) (1x10 -5 )  Diluting a strong acid by a factor of ten changes its pH by one unit. pH = - log 10 [H 3 O + ]

pH calculations 1 Equations pH = ………….[H + ] = ………….. Example: Calculate the pH for a specific [H + ]: Given [H + ] = 1.4 x M   pH = Calculate the [H + ] from pH: Given pH = 3.5 [H + ] =

pH calculations 1 Equations pH = -log 10 [H + ][H + ] = 10 -pH Example: Calculate the pH for a specific [H + ]: Given [H + ] = 1.4 x M   pH = log 10 [H + ] = log 10 (1.4 x ) = 4.85 Calculate the [H + ] from pH: Given pH = 3.5 [H + ] = 10 -pH = = 3.16 x mol.dm -3

pH calculations 2 Strong Acids To work from Acid concentration. Eg: Calculate the pH for a 0.15 mol.dm -3 HCl solution : Or Calculate the [H + ] from pH: Given pH of a HNO 3 solution is 2.5 calculate the acid concentration

pH calculations 2 Strong Acids To work from Acid concentration. Eg: Calculate the pH for a 0.15 mol.dm -3 HCl solution : strong acid HCl a strong acid.: 100% ionised so [H + ] = [HCl] = 0,15 mol.dm -3   pH = -log 10 [H + ] = -log 10 (0,15) = 0.82 Or Calculate the [H + ] from pH: Given pH of a HNO 3 solution is 2.5 calculate the acid concentration [HNO 3 ] = 10 -pH = = 3.16 x mol.dm -3

pH calculations 3 Strong Polyprotic Acids To work from Acid concentration. Polyprotic acids Eg: Calculate the pH for a mol.dm -3 H 2 SO 4 solution : Or Calculate the [H + ] from pH: Given pH of a H 3 PO 4 solution is 3.2 calculate the acid concentration

To work from Acid concentration. Polyprotic acids Eg: Calculate the pH for a mol.dm -3 H 2 SO 4 solution : strong acid H 2 SO 4 is a strong acid.: 100% ionised so 2x [H + ] = 2x[H 2 SO 4 ] = 0,05 mol.dm -3   pH = -log 10 [H + ] = -log 10 (0,05) = 1.3 Or Calculate the [H + ] from pH: Given pH of a H 3 PO 4 solution is 3.2 calculate the acid concentration [H + ] = 10 -pH = = 6.32 x mol.dm -3 since H 3 PO 4  3H + 1/3.: [H 3 PO 4 ] = 1/3 (6.32 x ) = 2.1 x mol.dm -3 pH calculations 3 Strong Polyprotic Acids

[H + ]  pH Calculate the pH of a mol.dm -3 HNO 3 solution Calculate the pH of a solution of ethanoic acid when the [H + ] concentration is 1.4 x mol.dm Calculate the pH of a 0,15 Molar solution of sulphuric acid Calculate the pH of a 1 M solution of H 2 SO Calculate the pH of a 0.5 mol.dm -3 solution of H 3 PO 4.

[H + ]  pH Calculate the pH of a mol.dm -3 HNO 3 solution Calculate the pH of a solution of ethanoic acid when the [H + ] concentration is 1.4 x mol.dm Calculate the pH of a 0,15 Molar solution of sulphuric acid Calculate the pH of a 1 M solution of H 2 SO Calculate the pH of a 0.5 mol.dm -3 solution of H 3 PO 4.

pH  [H + ] Calculate the [H + ] of a solution which has a pH of What concentration of HCl would give you a pH of What concentration of H 2 SO 4 would have a pH of 1? What mass of HCl would need to be dissolved in 250 cm 3 to give a pH of 2? What would be the concentration of sulphuric acid in car batteries if the pH of the solution is 0,6?

pH  [H + ] Calculate the [H + ] of a solution which has a pH of What concentration of HCl would give you a pH of What concentration of H 2 SO 4 would have a pH of 1? What mass of HCl would need to be dissolved in 250 cm 3 to give a pH of 2? What would be the concentration of sulphuric acid in car batteries if the pH of the solution is 0,6?

pH calculations Strong Base Equations K w = [H + ] [OH - ] = 1 x mol 2 dm -2 at 25 o C pH = -log 10 [H + ][H + ] = 10 -pH K w = [H + ] [OH - ] = 1 x mol 2 dm -2 at 25 o C Example: Calculate the pH for a specific [OH - ]: Given [OH - ] = 1.4 x o C  K w = [H + ] [OH - ] = 1 x  Since K w = [H + ] [OH - ] = 1 x .: [H + ] = 1x / [OH - ] = 1x / 1.4x10 -5 = 7.1 x M  7.1 x  pH = log 10 [H + ] = log 10 (7.1 x ) = 9.15  If the base was a strong base (say NaOH) then the base concentration would have been equal to the original OH - concentration.  I.e for STRONG BASES [BASE] = [OH - ]

In the Lowry-Brønsted model, an acid is a donor and a base is a proton acceptor. acid conj Why do we say the A - is a base? HA H + + A -

In the Lowry-Brønsted model, an acid is a proton donor and a base is a proton accep- tor. acid conj. base proton Why do we say the A - is a base? It ACCEPTS a proton HA H + + A -

The HA and A - is called an acid-base pair. Examples:

The HA and A - is called an acid-base pair. Examples:

Take note that a strong acid has a weak conjugate base and a strong base has a weak conjugate acid and vice versa. HCl H + + Cl acid conj. base H 2 O H + + OH acid conj. base

Take note that a strong acid has a weak conjugate base and a strong base has a weak conjugate acid and vice versa. HCl H + + Cl - strong acid weak conj. base H 2 O H + + OH - weak acidstrong conj. base

Protolytic Reactions Identify the acid base pairs in the following:  HNO 3 + KOH  KNO 3 + H 2 O  H 2 CO 3 + MgO  MgCO 3 + CO 2 + H 2 O

Protolytic Reactions  HNO 3 + KOH  KNO 3 + H 2 O a 1 b 2 b 1 a 2  H 2 CO 3 + MgO  MgCO 3 + CO 2 + H 2 O  a 1 b 2 b 1 a 2

H 2 PO 4 -  HPO H + AMPHOLYTES H 2 PO H +  H 3 PO 4

H 2 PO 4 -  HPO H + AMPHOLYTES H 2 PO H +  H 3 PO 4 HCO H +  H 2 CO 3 HCO 3 -  CO H + H 2 O  H + + OH - H 2 O + H +  H 3 O +