Acids and Bases

Acid-Base Theory Acids Have a sour taste Can dissolve metals Acid in your stomach Produce H +1 in water.

Acid-BaseTheory Bases Have a bitter taste Feel slippery Produce OH -1 in water.

Acid-Base Theory Arrhenius Definition Acids produce H + Bases produce OH -

Brönsted-Lowrey Acid-Base Theory H +1 Acid - proton acceptor Acid  proton donor Base - Conjugate Base Base  Conjugate Acid HCl(aq) + H 2 O(l)  Cl −1 (aq) + H 3 O +1 (aq A CB CAB NaF(aq) + H 2 O(l)  HF(aq) + NaOH(aq) A CB BCA OH −1 (aq) + H 2 O(l)  H 2 O(l)+ OH −1 (aq) CBABCA NH 3 (aq)+ H 2 O(l)  NH 4 +1 (aq)+ OH −1 (aq) A CBB CA CO 3 -2 (aq)+ H +1 (l)  H 3 O +1 (l)  HCO 3 -1 (aq)+ H 2 O(l) A CBB CA

Acid Strength Acid strength describes position of the ionization reaction HA + H 2 O  H 3 O + + A - Strong Acid = equilibrium lies far to the right The conjugate base of a strong acid is a weak base. Weak Acid = equilibrium lies far to the left The conjugate base of a weak acid is a strong base.

Strong Acid Write the mass Action expression for the following acid reactions HF  H + + F - Is this strong or weak? Weak…Ka is small!! Lots or little HF? HI  H + + I - Is this strong or weak? Strong…Ka is Large!! Lots or little HI? = 7.2x10 -4 M =3.2x10 9 M

Common Strong Acids (K a = very large) Sulfuric Acid = H 2 SO 4 is diprotic acid (releases 2 proton) H 2 SO 4  H +1 + HSO 4 -1 HSO 4 -1  H +1 + SO 4 -2 Hydrochloric Acid = HCl is monoprotic acid (releases 1 proton) HCl  H +1 + Cl -1 Nitric Acid = HNO 3 is monoprotic acid (releases 1 proton) HNO 3  H +1 + NO 3 -1 Perchloric Acid = HClO 4 is monoprotic acid (releases 1 proton) HClO 4  H +1 + ClO 4 -1

Oxyacids versus Hydrohalic Acids Oxyacids = acids with the ionizable proton attached to oxygen HClO 4 HNO 3 H 2 SO 4 Hydrohalic Acids = acids with the ionizable proton attached to a halide HCl HBr HF HI H―Cl H―FH―F H―I H―Br

Common Weak Acids Phosphoric Acid = H 3 PO 4 (triprotic acid) H 3 PO 4  H + + H 2 PO 4 -1 Ka = 7.5 x H 2 PO 4 -1  H + + HPO 4 -2 Ka = 6.2 x HPO 4 -2  H + + PO 4 -3 Ka = 4.8 x Nitrous Acid = HNO 2 (monoprotic acid) HNO 2  H + + NO 2 -1 Ka = 4.0 x Organic Acids have a carboxyl group (COOH) Acetic Acid = CH 3 COOH (HC 2 H 3 O 2 ) CH 3 COOH  H + +CH 3 COO -1 Ka = 1.8 x10 -5 Benzoic Acid = C 6 H 5 COOH C 6 H 5 COOH  H + +C 6 H 5 COO -1 Ka = 6.4 x10 -5 Only the COOH hydrogen is acidic

Water as an Acid and Base An amphoteric substance can behave as an acid or a base. Autoionization of water (reaction with itself) H 2 O + H 2 O  H 3 O +1 + OH -1 What is the Mass Action Expression? K W = [H 3 O +1 ][OH -1 ] = [H +1 ][OH -1 ] For any water solution at 25 o C, [H +1 ][OH -1 ] K W = 1 x M 2 Neutral solutions have [H +1 ]=[OH -1 ] = 1 x M Acidic solutions: [H +1 ] > [OH - 1 ] Basic solutions: [ OH - 1 ] > [ H +1 ]

Example Calculation What is the [H +1 ] and [OH -1 ] of a 0.1M solution of HCl? 1)What do we know? K W = [H +1 ][OH -1 ] = 1 x M 2 Almost all HA has ionized to H + and A - ([H + ] = [HCl]) in a strong acid. Therefore [H + ] = 0.1M 2) Calculate the [OH -1 ]. K W = [0.1M][OH -1 ] = 1 x M 2 [OH -1 ] = 1 x M

pH = -log[H +1 ] pH is a measure of the amount of hydrogen in a solution. It is based on the water product K w. It is based on the water product K w. pH Scale Acid Base Neutral 1M HCl 1M NaOH Ammonia Cleaner Blood Water Milk Stomach Acid Lemon Juice Vinegar pH +pOH =14

pH of strong acid Find the pH of a 0.15 M solution of Hydrochloric acidFind the pH of a 0.15 M solution of Hydrochloric acid pH = - log 0.15pH = - log 0.15 pH = - (- 0.82)pH = - (- 0.82) pH = 0.82pH = 0.82

pH of a strong Base What is the pH of the M NaOH solution?What is the pH of the M NaOH solution? pOH = - log (0.0010)pOH = - log (0.0010) pOH = 3pOH = 3 pH = 14 – 3 = 11pH = 14 – 3 = 11 pH +pOH =14

Notes Two Unit Eleven

Calculating pH (pH=-log[H +1 ]) The Ka for nitrous acid is 4.5 x Calculate the pH in a 0.15 M nitrous acid solution. HNO 2  NO H +1 [ HNO 2 ][ NO 2 -1 ][ H +1 ] Bef Eq Δ At Eq [HNO 2 ] [H +1 ] [NO 2 -1 ] Ka=Ka=Ka=Ka= -x +x +x 0.15-xx x K= [0.15] [x][x][x][x] [x][x][x][x] = 4.5 x X=[0.0082M] 1) Balanced Equation 2) Mass Action Expression 3) What do we know? 4) Calculate the [H +1 ] and pH. Very small = [H +1 ] pH=-log[ ] M pH=-log[H +1 ]) = 2.09

pH of A Weak Base Calculate the pH of a 0.50 M solution of ammonia, if K a = 1.8 x NH 3 + H 2 O ↔ OH -1 + NH 4 +1 NH 3 [ NH 3 ] OH -1 [ OH -1 ] NH 4 +1 [ NH 4 +1 ] Bef Eq Δ At Eq [NH 3 ] [NH 4 +1 ] [OH -1 ] Kb=Kb=Kb=Kb= -x +x +x 0.50-xx x K= [0.50] [x][x][x][x] [x][x][x][x] = 1.8 x X=[0.0095M] 1) Balanced Equation 2) Mass Action Expression 3) What do we know? 4) Calculate the pH. Very small = [OH - ] pOH=-log[ ] M pOH=-log[OH -1 ]) = 2.0 pH=12.0

Calculating pH (pH=-log[H +1 ]) K a = 1.8 x What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? K a = 1.8 x  HCOOH  + HCOO - + H +1 [HCOOH ][HCOO - ][ H +1 ] Bef Eq Δ At Eq [] [HCOOH] [H +1 ] [ ] [HCOO - ] Ka=Ka=Ka=Ka= -x +x +x 0.30-x 0.52+x x K= [0.30] [x][x][x][x] [0.52] = 1.8 x X= [1.0 x M] 1) Balanced Equation 2) Mass Action Expression 3) What do we know? 4) Calculate the pH. Very small = [H +1 ] pH=-log[ ] 1.0 x M pH=-log[H +1 ]) = 4.00

Henderson-Hasselbach Equation  HCOOH  + HCOO - + H +1 [] [HCOOH] [H +1 ] [ ] [HCOO - ] Ka=Ka=Ka=Ka= [] [HCOOH] =[H +1 ] [ ] [HCOO - ] KaKaKaKa [] [HCOOH] -log[H +1 ] [ ] [HCOO - ] -log(K a ) [] [HCOOH] [ ] [HCOO - ] [] [HCOOH] [ ] [HCOO - ] x x -log = [] [HCOOH] pH [ ] [HCOO - ] pK a -log = [] [HCOOH] pH [ ] [HCOO - ] pK a +log =

pH from Henderson-Hasselbach Equation K a = 3.74 What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? pK a = 3.74  HCOOH  + HCOO - + H +1 [HCOOH ][HCOO - ][ H +1 ] Bef Eq Δ At Eq -x +x +x 0.30-x 0.52+x x ) Balanced Equation 2) Henderson-Hasselbach Equation 3) What do we know? 4) Calculate the pH. Very small pH= [] [HCOOH] pH [ ] [HCOO - ] pK a +log = 0.30 pH log =

Strong Acid-Strong Base Reaction HCl(aq) + NaOH(aq)  H +1 +OH -1  H 2 O(l) Net Reaction H +1 +Cl -1 + Na +1 +OH -1  H 2 O(l)+ Cl -1 +Na +1

Strong Acid-Weak Base Reactions H 2 SO 4 (aq) + NH 3 (aq)  H +1 + NH 3  NH 4 +1 Net Reaction H +1 + HSO NH 3  NH HSO 4 -1

Weak Acid-Strong Base Reation CH 3 COOH(aq) + NaOH(aq)  CH 3 COOH +OH -1  H 2 O(l) + CH 3 COO -1 Net Reaction CH 3 COOH + Na +1 + OH -1  H 2 O(l)+ CH 3 COO -1 + Na +1

Weak Base-Weak Acid Reaction H 2 CO 3 (aq) + Cu(OH) 2 (s)  H 2 O(l)+CuCO 3 H 3 PO 4 (aq) + Al(OH) 3 (s)  3H 2 O(l)+AlPO 4

Weak Acid-Strong Base Titrations CH 3 COOH (aq) + NaOH (aq) CH 3 COONa (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (l) CH 3 COO - (aq) + H 2 O (l) OH - (aq) + CH 3 COOH (aq) At equivalence point (pH > 7): 16.4

Notes Three Unit Eleven

I.Using the Bayer Process, where Bauxite is added to a hot solution of Sodium Hydroxide to remove the Alumina as an Aluminium Hydroxide complex. This is then cooled to precipitate the Aluminium Hydroxide; when then heated to more than 1,000 degrees C, the Hydroxide decomposes to Alumina, which is then the starting point for the production of Aluminium. II.In oil drilling, Sodium Hydroxide is used as an additive to the drilling fluid, not only to increase its viscosity, but also to neutralise any acidic gas pockets encoutered during the drilling process. III.Paper Pulp Manufacture, where Sodium Hydroxide is initially used as a solution to separate the cellulose fibres from the lignin, as seen in the Kraft process. Sodium Hydroxide is also used later in the paper making process, where a strongly alkaline environment is necessary. IV.In the production of Biodiesel, which is produced by the trans-esterification of Methanol and Triglycerides. Anhydrous Sodium Hydroxide is used as a catalyst for this reaction to take place. V.Sodium Hydryxode is used in many industrial cleaning processes because Sodium Hydroxide readily dissolves oil, fat, grease or protein-based materials. It is used in large scale industrial cleaning processes and to clean process equipment, such as storage tanks. Usually the solution is heated and a surfactant is added to eliminate the dissolved material coming out of solution. VI.Sodium Hydroxide uses occur widely in the food industry, not only for product preparation, but also for cooking. Fruit and vegetables can be peeled, cocoa processed, poultry scalded and ice-cream thickened using suitable Sodium Hydroxide solutions. Meanwhile, pretzels, German lye rolls and the Scandanavian Luterfish (lye-fish) all use food grade Sodium Hydroxide in their preparation.

Acid-Base Equilibria Chapter 16

The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. The presence of a common ion suppresses the ionization of a weak acid or a weak base. Consider mixture of CH 3 COONa (strong electrolyte) and CH 3 COOH (weak acid). CH 3 COONa (s) Na + (aq) + CH 3 COO - (aq) CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) common ion 16.2

A buffer solution is a solution of: 1. 1.A weak acid or a weak base and 2. 2.The salt of the weak acid or weak base Both must be present! A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) Consider an equal molar mixture of CH 3 COOH and CH 3 COONa Adding more acid creates a shift left IF enough acetate ions are present

Which of the following are buffer systems? (a) KF/HF (b) KCl/HCl, (c) Na 2 CO 3 /NaHCO 3 (a) KF is a weak acid and F - is its conjugate base buffer solution (b) HCl is a strong acid not a buffer solution (c) CO 3 2- is a weak base and HCO 3 - is it conjugate acid buffer solution 16.3

What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? HCOOH (aq) H + (aq) + HCOO - (aq) Initial (M) Change (M) Equilibrium (M) x-x+x+x x x+x x x 16.2 Mixture of weak acid and conjugate base! K a for HCOOH = 1.8 x [H + ] [HCOO - ] K a = [HCOOH] x = X pH = 3.98

OR…… Use the Henderson-Hasselbach equation Consider mixture of salt NaA and weak acid HA. HA (aq) H + (aq) + A - (aq) NaA (s) Na + (aq) + A - (aq) K a = [H + ][A - ] [HA] [H + ] = K a [HA] [A - ] -log [H + ] = -log K a - log [HA] [A - ] -log [H + ] = -log K a + log [A - ] [HA] pH = pK a + log [A - ] [HA] pK a = -log K a Henderson-Hasselbach equation 16.2 pH = pK a + log [conjugate base] [acid]

What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? HCOOH (aq) H + (aq) + HCOO - (aq) Initial (M) Change (M) Equilibrium (M) x-x+x+x x x+x x x Common ion effect 0.30 – x  x  0.52 pH = pK a + log [HCOO - ] [HCOOH] HCOOH pK a = 3.77 pH = log [0.52] [0.30] = Mixture of weak acid and conjugate base!

HCl H + + Cl - HCl + CH 3 COO - CH 3 COOH + Cl

pH= 9.18 Calculate the pH of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. What is the pH after the addition of 20.0 mL of M NaOH to 80.0 mL of the buffer solution? NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) Initial End x xx 16.3 [NH 4 + ] [OH - ] [NH 3 ] K b = Change- x + x = 1.8 X (.36 + x)(x) (.30 – x) 1.8 X = 1.8 X  0.36x 0.30 x = 1.5 X pOH = 4.82

pH = 9.20 Calculate the pH of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. What is the pH after the addition of 20.0 mL of M NaOH to 80.0 mL of the buffer solution? NH 4 + (aq) + OH - (aq) H 2 O (l) + NH 3 (aq) start (M) end (M) final volume = 80.0 mL mL = 100 mL 16.3 NH M x L = mol /.1 L = 0.29 M OH x L = mol /.1 L = 0.01M NH M x = mol /.1 L = 0.24M K a = = 5.6 X [H + ] [NH 3 ] [NH 4 + ] = 5.6 X [H + ] [H + ] = 6.27 X

= 9.20 Calculate the pH of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. What is the pH after the addition of 20.0 mL of M NaOH to 80.0 mL of the buffer solution? NH 4 + (aq) H + (aq) + NH 3 (aq) pH = pK a + log [NH 3 ] [NH 4 + ] pK a = 9.25 pH = log [0.30] [0.36] = 9.17 NH 4 + (aq) + OH - (aq) H 2 O (l) + NH 3 (aq) start (M) end (M) pH = log [0.25] [0.28] final volume = 80.0 mL mL = 100 mL 16.3

Chemistry In Action: Maintaining the pH of Blood 16.3

Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at the endpoint (hopefully close to the equivalence point) Slowly add base to unknown acid UNTIL The indicator changes color (pink) 4.7

Strong Acid-Strong Base Titrations NaOH (aq) + HCl (aq) H 2 O (l) + NaCl (aq) OH - (aq) + H + (aq) H 2 O (l) % ionization! No equilibrium

Weak Acid-Strong Base Titrations CH 3 COOH (aq) + NaOH (aq) CH 3 COONa (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (l) CH 3 COO - (aq) + H 2 O (l) OH - (aq) + CH 3 COOH (aq) At equivalence point (pH > 7): 16.4

Strong Acid-Weak Base Titrations HCl (aq) + NH 3 (aq) NH 4 Cl (aq) NH 4 + (aq) + H 2 O (l) NH 3 (aq) + H + (aq) At equivalence point (pH < 7): 16.4 H + (aq) + NH 3 (aq) NH 4 Cl (aq)

Acid-Base Indicators 16.5

pH 16.5

The titration curve of a strong acid with a strong base. 16.5

Which indicator(s) would you use for a titration of HNO 2 with KOH ? Weak acid titrated with strong base. At equivalence point, will have conjugate base of weak acid. At equivalence point, pH > 7 Use cresol red or phenolphthalein 16.5

Finding the Equivalence Point (calculation method) I.Strong Acid vs. Strong Base A.100 % ionized! pH = 7 No equilibrium! II.Weak Acid vs. Strong Base A.Acid is neutralized; Need K b for conjugate base equilibrium III.Strong Acid vs. Weak Base A.Base is neutralized; Need K a for conjugate acid equilibrium IV.Weak Acid vs. Weak Base A.Depends on the strength of both; could be conjugate acid, conjugate base, or pH 7

Exactly 100 mL of 0.10 M HNO 2 are titrated with 100 mL of a 0.10 M NaOH solution. What is the pH at the equivalence point ? HNO 2 (aq) + OH - (aq) NO 2 - (aq) + H 2 O (l) start (moles) end (moles) NO 2 - (aq) + H 2 O (l) OH - (aq) + HNO 2 (aq) Initial (M) Change (M) Equilibrium (M) x-x+x+x x x+x xx [NO 2 - ] = = 0.05 M Final volume = 200 mL K b = [OH - ][HNO 2 ] [NO 2 - ] = x2x x = 2.2 x – x  0.05x  1.05 x = [OH - ] pOH = 5.98 pH = 14 – pOH = 8.02

Complex Ion Equilibria and Solubility A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. Co 2+ (aq) + 4Cl - (aq) CoCl 4 (aq) 2- Co(H 2 O) 6 2+ CoCl

Complex Ion Formation I.These are usually formed from a transition metal surrounded by ligands (polar molecules or negative ions). II.As a "rule of thumb" you place twice the number of ligands around an ion as the charge on the ion... example: the dark blue Cu(NH 3 ) 4 2+ (ammonia is used as a test for Cu 2+ ions), and Ag(NH 3 ) 2 +. III.Memorize the common ligands.

Common Ligands LigandsNames used in the ion H2OH2Oaqua NH 3 ammine OH-hydroxy Cl-chloro Br-bromo CN-cyano SCN-thiocyanato (bonded through sulphur) isothiocyanato (bonded through nitrogen)

Names I.Names: ligand first, then cation Examples: A.tetraamminecopper(II) ion: Cu(NH 3 ) 4 2+ B.diamminesilver(I) ion: Ag(NH 3 ) 2 +. C.tetrahydroxyzinc(II) ion: Zn(OH) 4 2- I.The charge is the sum of the parts (2+) + 4(-1)= -2.

When Complexes Form I.Aluminum also forms complex ions as do some post transitions metals. Ex: Al(H 2 O) 6 3+ II.Transitional metals, such as Iron, Zinc and Chromium, can form complex ions. III.The odd complex ion, FeSCN 2+, shows up once in a while IV.Acid-base reactions may change NH 3 into NH 4 + (or vice versa) which will alter its ability to act as a ligand. V.Visually, a precipitate may go back into solution as a complex ion is formed. For example, Cu 2+ + a little NH 4 OH will form the light blue precipitate, Cu(OH) 2. With excess ammonia, the complex, Cu(NH 3 ) 4 2+, forms. VI.Keywords such as "excess" and "concentrated" of any solution may indicate complex ions. AgNO 3 + HCl forms the white precipitate, AgCl. With excess, concentrated HCl, the complex ion, AgCl 2 -, forms and the solution clears.

Coordination Number I.Total number of bonds from the ligands to the metal atom. II.Coordination numbers generally range between 2 and 12, with 4 (tetracoordinate) and 6 (hexacoordinate) being the most common.

Some Coordination Complexes molecular formulaLewis base/ligandLewis aciddonor atomcoordination number Ag(NH 3 ) 2 + NH 3 Ag + N2 [Zn(CN) 4 ] 2- CN-Zn 2+ C4 [Ni(CN) 4 ] 2- CN-Ni 2+ C4 [PtCl 6 ] 2- Cl-Pt 4+ Cl6 [Ni(NH 3 ) 6 ] 2+ NH 3 Ni 2+ N6

pH of A Weak Acid Calculate the pH of a M solution of formic acid, if K a = 1.8 x Calculate the pH of a M solution of formic acid, if K a = 1.8 x HCOOH + H 2 O ↔ HCOO -1 + H 3 O +1HCOOH + H 2 O ↔ HCOO -1 + H 3 O x x x0.010-x x x If [HCOOH ] is at least 100 times larger than K a then ignore x.If [HCOOH ] is at least 100 times larger than K a then ignore x. x 2 =(0.010)( 1.8 x )x 2 =(0.010)( 1.8 x ) x=0.0013x= pH=-log[0.0013]=2.9pH=-log[0.0013]=2.9