ACIDS AND BASES ACID – A compound that produces hydrogen ions in a water solution HCl (g) → H + (aq) + Cl - (aq) BASE – A compound that produces hydroxide ions in a water solution NaOH (s) → Na + (aq) + OH - (aq) 1884 SVANTE ARRHENIUS Proposed the first definitions of acids and bases 1E-1 (of 25)

FORMATION OF ACIDS AND BASES Bases are produced from the reaction of metal oxides with water Na 2 O (s) + H 2 O (l) → 2NaOH (aq) Acids are produced from the reaction of nonmetal oxides with water CO 2 (g) + H 2 O (l) → H 2 CO 3 (aq) 1E-2 (of 25)

Na 2 OMgOAl 2 O 3 SiO 2 P 2 O 5 SO 3 Cl 2 O 7 NaOHMg(OH) 2 Al(OH) 3 Si(OH) 4 PO(OH) 3 SO 2 (OH) 2 ClO 3 (OH) H 4 SiO 4 H 3 PO 4 H 2 SO 4 HClO 4 AMPHOTERIC – A substance that can act as an acid or a base base base or acid acid strongest base strongest acid Products of elemental oxides with water: 1E-3 (of 25)

base Products of elemental oxides with water: The higher the element’s oxidation number, the more acidic its hydroxide Cr(OH) 2 Cr(OH) 3 Cr(OH) 6 amphoteric acid 1E-4 (of 25) Na 2 OMgOAl 2 O 3 SiO 2 P 2 O 5 SO 3 Cl 2 O 7 NaOHMg(OH) 2 Al(OH) 3 Si(OH) 4 PO(OH) 3 SO 2 (OH) 2 ClO 3 (OH) H 4 SiO 4 H 3 PO 4 H 2 SO 4 HClO 4 base base or acid acid

1923 Expanded the definitions of acids and bases THOMAS LOWRY 1E-5 (of 25)

ACID – A hydrogen ion (or proton) donor BASE – A hydrogen ion (or proton) acceptor 1923 Expanded the definitions of acids and bases JOHANNES BRØNSTED 1E-6 (of 25)

HCl + H 2 O → 1E-7 (of 25)

HCl + H 2 O →Cl - + H 3 O + acidbase HYDRONIUM ION – H 3 O +, formed when a hydrogen ion attaches to a water +- 1E-8 (of 25)

NH 3 + H 2 O → 1E-9 (of 25)

NH OH - NH 3 + H 2 O → baseacid Water is amphoteric +- 1E-10 (of 25)

Acids turn into bases, and bases turn into acids HCl + H 2 O Cl - + H 3 O + acidbase → 1E-11 (of 25)

conjugate base of HCl conjugate acid of H 2 O Acids turn into bases, and bases turn into acids HCl + H 2 O Cl - + H 3 O + acidbase ← NH 3 + H 2 O → NH OH - baseacid conjugate acid of NH 3 conjugate base of H 2 O 1E-12 (of 25)

conjugate base of NH 4 + conjugate acid of NO 2 - NH NO 2 - → NH 3 + HNO 2 acidbase ← The favored reaction direction turns strong acids and bases into weak acids and bases HNO 2 is a stronger acid than NH 4 + NH 3 is a stronger base than NO 2 - 1E-13 (of 25)

AUTOIONIZATION OF WATER Water ionizes itself to a small extent 1E-14 (of 25)

AUTOIONIZATION OF WATER Water ionizes itself to a small extent 2H 2 O (l) → H 3 O + (aq) + OH - (aq) - + Makes solutions acidicMakes solutions basic Equal amounts of H 3 O + and OH - make a solution NEUTRAL  pure water is neutral 1E-15 (of 25)

The ionization of water is an endothermic process 2H 2 O (l) ⇆ H 3 O + (aq) + OH - (aq) Write the equilibrium constant expression for the autoionization of water K eq = [H 3 O + ][OH - ] ION-PRODUCT CONSTANT FOR WATER (K w ) – The equilibrium constant for the ionization of water x x x Temp. (°C)K w energy + 1E-16 (of 25)

Find [H 3 O + ] and [OH - ] in pure water at 25°C. xx 2H 2 O (l) ⇆ H 3 O + (aq) + OH - (aq) Initial M’s Change in M’s Equilibrium M’s 00 + x K w = [H 3 O + ][OH - ] 1.00 x M 2 = x x M = x= [H 3 O + ] = [OH - ] 1E-17 (of 25)

THE pH SCALE pH – The negative logarithm of [H 3 O + ] of a solution The common logarithm of a number is: the exponent to which 10 must be raised to equal the number = = = = log 100 = 2log 0.01 = -2log = -3log = pOH – The negative logarithm of [OH - ] of a solution 1E-18 (of 25)

Calculate the pH of orange juice if its [H 3 O + ] = 2.5 x M. = -log(2.5 x M) pH= -log[H 3 O + ] = -( …) 1E-19 (of 25) = 3.60 For logarithmic numbers, only the digits after the decimal point are significant figures, not the digits before = 3.6incorrect number of significant figures = -[log(2.5) + log(10 -4 )]

Calculate the pH and pOH of pure water at 25ºC. = -log(1.00 x M) pH= -log[H 3 O + ] For pure water: [H 3 O + ] = 1.00 x M = = -log(1.00 x M) pOH= -log[OH - ] [OH - ] = 1.00 x M = E-20 (of 25)

In any water solution: K w =[H 3 O + ][OH - ] K w =[OH - ] ________ [H 3 O + ] at 25°C 1.00 x =[OH - ] _____________ [H 3 O + ] 1E-21 (of 25) For orange juice: [H 3 O + ] = 2.5 x M 1.00 x M 2 = [OH - ] _________________ 2.5 x M = 4.0 x M

pH 7 = Neutral < 7 is Acidic > 7 is Basic [H 3 O + ] = M [OH - ] = M [H 3 O + ] = M [OH - ] = M [H 3 O + ] = M [OH - ] = M 15 [H 3 O + ] = 10 1 M [OH - ] = M 1E-22 (of 25)

In any water solution: K w =[H 3 O + ][OH - ] log K w =log ([H 3 O + ][OH - ]) -log K w =-log ([H 3 O + ][OH - ]) -log K w =- log[H 3 O + ] - log[OH - ] pK w =pH + pOH at 25°C, -log(1.00 x ) = ∴ = pH + pOH 1E-23 (of 25)

Calculate the pH and pOH of soda if its [H 3 O + ] = 1.6 x M. = -log(1.6 x M) pH= -log[H 3 O + ] [H 3 O + ] = 1.6 x M = 2.80 = pOH= pK w - pH pH + pOH = pK w = E-24 (of 25)

Calculate the [H 3 O + ] of blood, which has a pH of 7.4. = M pH= -log[H 3 O + ] -pH= log[H 3 O + ] antilog (-pH)= [H 3 O + ] antilog (-7.4)= [H 3 O + ] For logarithmic numbers, only the digits after the decimal point are significant figures, not the digits before = 3.98 x M = 4 x M 1E-25 (of 25)

IONIZATION OF ACIDS HF (aq) + H 2 O(l) ⇆ H 3 O + (aq) + F - (aq) Write the K eq expression for the ionization of hydrofluoric acid = [H 3 O + ][F - ] ____________ [HF] ACID IONIZATION CONSTANT (K a ) – The equilibrium constant for an acid ionizing K eq 1F-1 (of 19)

IONIZATION OF ACIDS HF (aq) + H 2 O(l) ⇆ H 3 O + (aq) + F - (aq) Write the K eq expression for the ionization of hydrofluoric acid = [H 3 O + ][F - ] ____________ [HF] ACID IONIZATION CONSTANT (K a ) – The equilibrium constant for an acid ionizing Ka Ka 1F-2 (of 19)

Acids are a) strong if every acid molecule gives up a hydrogen ion HCl, HBr, HI, any acid with 2 or more O’s than H’s K a = large b)weak if less than every acid molecule gives up a hydrogen ion all other acids K a = small 1F-3 (of 19) ()()

IONIZATION OF BASES NH 3 (aq) + H 2 O(l) ⇆ NH 4 + (aq) + OH - (aq) Write the K eq expression for the ionization of ammonia = [NH 4 + ][OH - ] ______________ [NH 3 ] BASE DISSOCIATION CONSTANT (K b ) – The equilibrium constant for a base ionizing K eq 1F-4 (of 19)

IONIZATION OF BASES NH 3 (aq) + H 2 O(l) ⇆ NH 4 + (aq) + OH - (aq) Write the K eq expression for the ionization of ammonia = [NH 4 + ][OH - ] ______________ [NH 3 ] BASE DISSOCIATION CONSTANT (K b ) – The equilibrium constant for a base ionizing Kb Kb 1F-5 (of 19)

Bases are a) strong if every molecule/ion accepts a hydrogen ion alkali metal hydroxides, dilute Ca(OH) 2, Sr(OH) 2, Ba(OH) 2 solutions K b = large b)weak if less than every molecule/ion accepts a hydrogen ion all other bases, including NH 3 K b = small 1F-6 (of 19)

Calculate the pH of M nitric acid. 1F-7 (of 19) CALCULATING THE pH OF STRONG ACID OR BASE SOLUTIONS Assume complete ionization or dissociation HNO 3 (aq) + H 2 O (l) → H 3 O + (aq) + NO 3 - (aq) Initial M’s Change in M’s Final M’s x+ x 0 = -log(0.010 M) pH= -log[H 3 O + ] =

Calculate the pH of M sodium hydroxide. 1F-8 (of 19) NaOH (aq) → Na + (aq) + OH - (aq) Initial M’s Change in M’s Final M’s x+ x 0 = -log( M) pOH= -log[OH - ] = 3.00= pH= pK w - pOH pH + pOH = pK w =

the K a for HNO 2 can be looked up Calculate the pH of M nitrous acid. 1F-9 (of 19) CALCULATING THE pH OF WEAK ACID OR BASE SOLUTIONS Assume INCOMPLETE ionization or dissociation, reaching a state of equilibrium xx HNO 2 (aq) + H 2 O (l) ⇆ H 3 O + (aq) + NO 2 - (aq) Initial M’s Change in M’s Equilibrium M’s x + x x : 4.0 x K a = [H 3 O + ][NO 2 - ] ________________ [HNO 2 ] K a = x 2 _____________ (0.010 – x)

4.0 x = x 2 ______________ (0.010 – x) If the K a is < the acid does not ionize much, so you may assume that x is very small, and ignore it when it is subtracted from the initial molarity 4.0 x = x 2 _______ x = x For x values that are >1% of the original acid molarity, they should be put back in place of x in the denominator of the K a expression, and solved again 1F-10 (of 19)

4.0 x = x 2 _________________________ (0.010 – 2.00 x ) 1.79 x = x Repeat until the answer (to 2 sig fig’s) is the same twice in a row 4.0 x = x 2 _________________________ (0.010 – 1.79 x ) 1.81 x = x = -log(1.81 x M)pH= -log[H 3 O + ]= 2.74 = [H 3 O + ] 1F-11 (of 19)

Calculate the pH of 0.20 M acetic acid if its K a = 1.8 x xx HC 2 H 3 O 2 (aq) + H 2 O (l) ⇆ H 3 O + (aq) + C 2 H 3 O 2 - (aq) Initial M’s Change in M’s Equilibrium M’s x + x x K a = [H 3 O + ][C 2 H 3 O 2 - ] ____________________ [HC 2 H 3 O 2 ] 1.8 x = x 2 ____________ (0.20 – x) 1.8 x = x 2 ______ x = x 1F-12 (of 19)

1.8 x = x 2 ________________________ (0.20 – 1.90 x ) 1.89 x = x = -log(1.89 x M)pH = -log[H 3 O + ]= 2.72 Calculate the pH of 0.20 M acetic acid if its K a = 1.8 x = [H 3 O + ] Calculate the percent ionization of acetic acid x M _________________ 0.20 M x 100 = 0.95% 1F-13 (of 19)

Calculate the pH of a 0.10 M ammonia solution if its K b = 1.8 x xx NH 3 (aq) + H 2 O (l) ⇆ NH 4 + (aq) + OH - (aq) Initial M’s Change in M’s Equilibrium M’s x + x x K b = [NH 4 + ][OH - ] ______________ [NH 3 ] 1.8 x = x 2 ____________ (0.10 – x) 1.8 x = x 2 ______ x = x 1F-14 (of 19)

1.8 x = x 2 ________________________ (0.10 – 1.34 x ) 1.33 x = x = -log(1.33 x M)pOH= -log[OH - ]= 2.88 Calculate the pH of a 0.10 M ammonia solution if its K b = 1.8 x = [OH - ] = pH= pK w - pOH pH + pOH = pK w = F-15 (of 19)

A M solution of the weak acid HA has a pH Calculate its K a. xx HA (aq) + H 2 O (l) ⇆ H 3 O + (aq) + A - (aq) Initial M’s Change in M’s Equilibrium M’s x + x x K a = [H 3 O + ][A - ] _____________ [HA] K a = x 2 ______________ (0.500 – x) x = [H 3 O + ] = antilog (-2.010) = antilog (-pH) = M 1F-16 (of 19) CALCULATING THE K a OF A WEAK ACID FROM pH

A M solution of the weak acid HA has a pH Calculate its K a. K a = ( M) 2 _____________________________ (0.500 M M) = 1.95 x What is the pK a of HA? = -log(1.947 x ) pK a = -log K a = The smaller the K a (or the bigger the pK a ) the weaker the acid 1F-17 (of 19) CALCULATING THE K a OF A WEAK ACID FROM pH

A M solution of the weak acid HX is 3.15% ionized. Calculate its K a. xx HX (aq) + H 2 O (l) ⇆ H 3 O + (aq) + X - (aq) Initial M’s Change in M’s Equilibrium M’s x + x x K a = [H 3 O + ][X - ] _____________ [HX] K a = x 2 ______________ (0.500 – x) 3.15 = x (100) _______ = x 1F-18 (of 19) CALCULATING THE K a OF A WEAK ACID FROM PERCENT IONIZATION

A M solution of the weak acid HX is 3.15% ionized. Calculate its K a. K a = ( M) 2 _____________________________ (0.500 M M) = 5.12 x What is the pK a of HX? = -log(5.123 x ) pK a = -log K a = F-19 (of 19) CALCULATING THE K a OF A WEAK ACID FROM PERCENT IONIZATION

1G-1 (of 13) POLYPROTIC ACIDS Polyprotic acids have K a values for the ionization of each H + H 2 CO 3 (aq) + H 2 O (l) ⇆ H 3 O + (aq) + HCO 3 - (aq) K a1 =[H 3 O + ][HCO 3 - ] __________________ [H 2 CO 3 ] K a1 =4.3 x HCO 3 - (aq) + H 2 O (l) ⇆ H 3 O + (aq) + CO 3 2- (aq) K a2 =[H 3 O + ][CO 3 2- ] _________________ [HCO 3 - ] K a2 =5.6 x Successive H + ’s are harder to remove ∴ H 2 CO 3 is a stronger acid than HCO 3 -

H 2 CO 3 (aq) + H 2 O (l) ⇆ H 3 O + (aq) + HCO 3 - (aq) K a1 =4.3 x HCO 3 - (aq) + H 2 O (l) ⇆ H 3 O + (aq) + CO 3 2- (aq) K a2 =5.6 x H 2 CO 3 (aq) + 2H 2 O (l) ⇆ 2H 3 O + (aq) + CO 3 2- (aq) K a1,2 =? K a1,2 = (4.3 x )(5.6 x )= 2.4 x G-2 (of 13) [H 3 O + ][HCO 3 - ] __________________ [H 2 CO 3 ] x [H 3 O + ][CO 3 2- ] _________________ [HCO 3 - ] [H 3 O + ] 2 [CO 3 2- ] __________________ [H 2 CO 3 ] = K a1 K a2 = K a1,2

Find the concentrations of each species in a 0.10 M H 2 CO 3 solution. xx H 2 CO 3 (aq) + H 2 O (l) ⇆ H 3 O + (aq) + HCO 3 - (aq) Initial M’s Change in M’s Equilibrium M’s x + x x K a1 = [H 3 O + ][HCO 3 - ] _________________ [H 2 CO 3 ] 4.3 x = x 2 ____________ (0.10 – x) x = 2.07 x [H 2 CO 3 ]=0.10 M – M= 0.10 M [H 3 O + ]= = M [HCO 3 - ]== M 1G-3 (of 13)

Find the concentrations of each species in a 0.10 M H 2 CO 3 solution yy HCO 3 - (aq) + H 2 O (l) ⇆ H 3 O + (aq) + CO 3 2- (aq) Initial M’s Change in M’s Equilibrium M’s y + y y K a2 = [H 3 O + ][CO 3 2- ] _________________ [HCO 3 - ] 5.6 x = ( y) y __________________ ( – y) y = 5.6 x [H 2 CO 3 ]== 0.10 M [H 3 O + ]= x = M [HCO 3 - ]= – 5.6 x = M [CO 3 2- ]= = 5.6 x M 1G-4 (of 13)

pH OF SALT SOLUTIONS Acid HNO 3 HNO 2 Strong acids have non conjugate bases Weak acids have weak conjugate bases Strong Weak Conjugate Base NO 3 - NO 2 - Non Weak Base NH 3 NaOH NaOH(H 2 O) 5 Strong bases have non conjugate acids Weak bases have weak conjugate acids Weak Strong Conjugate Acid NH 4 + Na + Na(H 2 O) 6 + Weak Non 1G-5 (of 13)

from LiOH which is a strong base ∴ Li + is a non acid from H 3 PO 4 which is a weak acid ∴ PO 4 3- is a weak base Li 3 PO 4 ∴ pH > 7 1G-6 (of 13) Li + PO 4 3- PO 4 3- (aq) + H 2 O (l) ⇆ HPO 4 - (aq) + OH - (aq) HYDROLYSIS – The reaction of a dissolved ion with water

from NH 3 which is a weak base ∴ NH 4 + is a weak acid from HCl which is a strong acid ∴ Cl - is a non base NH 4 Cl ∴ pH < 7 1G-7 (of 13) NH 4 + Cl - NH 4 + (aq) + H 2 O (l) ⇆ NH 3 (aq) + H 3 O + (aq)

from KOH which is a strong base ∴ K + is a non acid from HNO 3 Which is a strong acid ∴ NO 3 - is a non base KNO 3 ∴ pH = 7 1G-8 (of 13) K+K+ NO 3 -

Find the pH of a 0.10 M potassium acetate solution x x C 2 H 3 O 2 - (aq) + H 2 O (l) ↔ HC 2 H 3 O 2 (aq) + OH - (aq) Initial M’s Change in M’s Equilibrium M’s x + x x K b = [HC 2 H 3 O 2 ][OH - ] ___________________ [C 2 H 3 O 2 - ] 5.6 x = x 2 ____________ (0.10 – x) x = 7.48 x Potassium ion is a non acid; acetate ion is a weak base Need the K b for acetate: 5.6 x G-9 (of 13) CALCULATING THE pH OF A SALT SOLUTION

7.48 x = x = -log(7.48 x M)pOH= -log[OH - ]= 5.13 = [OH - ] = – 5.13pH= pK w - pOH pH + pOH = pK w = 8.87 Find the pH of a 0.10 M potassium acetate solution 1G-10 (of 13) CALCULATING THE pH OF A SALT SOLUTION

HC 2 H 3 O 2 (aq) + H 2 O (l) ⇆ H 3 O + (aq) + C 2 H 3 O 2 - (aq) K a for acetic acid C 2 H 3 O 2 - (aq) + H 2 O (l) ⇆ HC 2 H 3 O 2 (aq) + OH - (aq) K b for acetate 2H 2 O (l) ⇆ H 3 O + (aq) + OH - (aq) KwKw K a K b = K w 1G-11 (of 13)

Find the pH of a 0.25 M ammonium chloride solution if the K b for ammonia is 1.8 x Ammonium ion is a weak acid; chloride ion is a non base Need the K a for ammonium Have the K b for ammonia : 1.8 x K a = K w ____ K b K a K b = K w = 1.00 x _______________ 1.8 x = 5.56 x G-12 (of 13)

Find the pH of a 0.25 M ammonium chloride solution if the K b for ammonia is 1.8 x xx NH 4 + (aq) + H 2 O (l) ⇆ H 3 O + (aq) + NH 3 (aq) Initial M’s Change in M’s Equilibrium M’s x + x x K a = [H 3 O + ][NH 3 ] ______________ [NH 4 + ] 5.56 x = x 2 ____________ (0.25 – x) = -log(1.18 x M) pH = -log[H 3 O + ]= 4.93 x = 1.18 x M 1G-13 (of 13)