ACID-BASE EQUILIBRIUM ERT 207 ANALYTICAL CHEMISTRY SEMESTER 1, ACADEMIC SESSION 2015/16

Overview  ACID-BASE THEORIES  ACID DISSOCIATION CONSTANT  pH SCALE  METHODS OF MEASURING pH  POLYPROTIC ACIDS  WEAK BASES  BASE DISSOCIATION CONSTANT  RELATIONSHIP BETWEEN K w, K a AND K b 2

Overview  BEHAVIOR OF SALTS IN WATER  SALT SOLUTIONS  ACIDS-BASE REACTIONS  BUFFER SOLUTIONS  HENDERSON-HASSELBALCH EQUATION  PREPARING A BUFFER 3

ACID-BASE THEORIES 4  Arrhenius (or Classical) Acid-Base Definition  An acid is a substance that contains hydrogen and dissociates in water to yield a hydronium ion : H 3 O +  A base is a substance that contains the hydroxyl group and dissociates in water to yield : OH –  Neutralization is the reaction of an H + (H 3 O + ) ion from the acid and the OH - ion from the base to form water, H 2 O. Arrhenius 1903 Nobel Prize

ACID-BASE THEORIES 5 H 3 O + = H + (aq) = proton in water

ACID-BASE THEORIES 6

ACID-BASE THEORIES 7  The neutralization reaction is exothermic and releases approximately 56 kJ per mole of acid and base.  Brønsted-Lowry Acid-Base Definition  An acid is a proton donor, any species that donates an H + ion.  An acid must contain H in its formula; HNO 3 and H 2 PO 4 - are two examples, all Arrhenius acids are Brønsted-Lowry acids.

ACID-BASE THEORIES 8  A base is a proton acceptor, any species that accepts an H + ion.  A base must contain a lone pair of electrons to bind the H + ion; a few examples are NH 3, CO 3 2-, F -, as well as OH -.  Brønsted-Lowry bases are not Arrhenius bases, but all Arrhenius bases contain the Brønsted-Lowry base OH -.  Therefore in the Brønsted-Lowry perspective, one species donates a proton and another species accepts it: an acid-base reaction is a proton transfer process.

ACID-BASE THEORIES 9 Figure 1: Acid-Base Theories

ACID-BASE THEORIES 10

ACID DISSOCIATION CONSTANT

 Strong Acids:100% dissociation  good H + donor  equilibrium lies far to right (HNO 3 )  generates weak base (NO 3 - )  Weak Acids:<100% dissociation  not-as-good H + donor  equilibrium lies far to left (CH 3 COOH)  generates strong base (CH 3 COO - )

ACID DISSOCIATION CONSTANT

Strength vs. K a ACID DISSOCIATION CONSTANT

17  [H 3 O + ] 1x10 0 to 1x in water  [OH - ] 1x to 1x10 0 in water [H 3 O + ] and [OH - ]:

ACID DISSOCIATION CONSTANT 18  Finding [H 3 O + ] and [OH - ]:

pH SCALE 19  pH:  pH is defined as the negative base-10 logarithm of the hydronium ion concentration. pH = –log [H 3 O + ]  The “p” in pH tells us to take the negative log of the quantity (in this case, hydronium ions).  Some similar examples are : pOH = –log [OH - ] pK w = –log K w

pH SCALE 20

pH SCALE 21  pH and pOH:  As [H 3 O + ] rises, [OH - ] falls  As pH falls, pOH rises

pH SCALE 22

pH SCALE 23  Some pH values:

METHODS OF MEASURING pH 24 pH indicatorpH meter  It measures the voltage in the solution.

METHODS OF MEASURING pH 25 Figure 1: pH indicators

METHODS OF MEASURING pH 26  Relationship between K a and pK a :

METHODS OF MEASURING pH 27 EXAMPLE 1: Calculating Ka from pH  The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is  Calculate K a for formic acid at this temperature.  We know that

METHODS OF MEASURING pH 28 EXAMPLE 2: Calculating pH from K a  Calculate the pH of a 0.30 M solution of acetic acid, C 2 H 3 O 2 H, at 25°C.  K a for acetic acid at 25°C is 1.8 

POLYPROTIC ACIDS 29  Polyprotic acids have more than one acidic proton.  If the difference between the K a for the first dissociation and subsequent K a values is 10 3 or more, the pH generally depends only on the first dissociation.

POLYPROTIC ACIDS 30

WEAK BASES 31  Strength of Bases:  Strong:  100% dissociation  OH - supplied to solution  NaOH (s)  Na + (aq) + OH - (aq)  Weak:  <100% dissociation  OH - by reaction with water  CH 3 NH 2(aq) + H 2 O (l) CH 3 NH 2(aq) + OH - (aq)

WEAK BASES 32  Sustainable  Sustainability.

BASE DISSOCIATION CONSTANT 33  Bases react with water to produce hydroxide ion.  Sustainability.

BASE DISSOCIATION CONSTANT 34  K b can be used to find [OH – ] and, through it, pH.

BASE DISSOCIATION CONSTANT 35

BASE DISSOCIATION CONSTANT 36  EXAMPLE 3: Calculating pH of Basic Solutions  What is the pH of a 0.15 M solution of NH 3 ? [NH 4 + ] [OH − ] [NH 3 ] K b = = 1.8  10 -5

RELATIONSHIP BETWEEN K w, K a AND K b 37

BEHAVIOR OF SALTS IN WATER 38

BEHAVIOR OF SALTS IN WATER 39  Properties of salt solutions:  The spectator ions in acid-base reactions form salts. Salts completely ionize in their aqueous solutions.  A salt NH 4 A, the ionizes NH 4 A ↔ NH A -  The ions of salts interact with water (called hydration), A - + H 2 O ↔ HA + OH -

BEHAVIOR OF SALTS IN WATER 40  If the anions are stronger base than H 2 O, the solution is basic. NH H 2 O = NH 3 + H 3 O +  If the cations are stronger acid than H 2 O, the solution is acidic.  These competitive reactions make the solution acidic or basic depending on the strength of the acids and bases.

SALT SOLUTIONS 41  Neutral Salt Solutions:  Salts consisting of the Anion of a Strong Acid and the Cation of a Strong Base yield a neutral solutions because the ions do not react with water.  Nitrate (NO 3 - ) is a weaker base than water (H 2 O) ; reaction goes to completion as the NO 3 - becomes fully hydrated; does not react with water

SALT SOLUTIONS 42  Na + & NO 3 - do not react with water, leaving just the “autoionization” of water, i.e., a neutral solution.  Salts that produce Acidic Solutions:  A salt consisting of the anion of a strong acid and the cation of a weak base yields an acidic solution.  The cation acts as a weak acid

SALT SOLUTIONS 43  The anion does not react with water.  In a solution of NH 4 Cl, the NH 4 + ion that forms from the weak base, NH 3, is a weak acid.  The Chloride ion, the anion from a strong acid does not react with water

SALT SOLUTIONS 44  Salts that produce Basic Solutions:  A salt consisting of the anion of a weak aid and the cation of a strong base yields a basic solution.  The anion acts as a weak base.  The cation does not react with water.  The anion of the weak acid accepts a proton from water to yield OH - ion, producing a “Basic” solution.

SALT SOLUTIONS 45  Salts of Weakly Acidic Cations and Weakly Basic Anions:  Overall acidity of solution depends on relative acid strength (K a ) or base strength (K b ) of the separated ions.  Eg. NH 4 CN - Acidic or Basic?  Write equations for any reactions that occur between the separated ions and water

SALT SOLUTIONS 46  Compare K a of NH 4 + & K b of CN -  Magnitude of K b (K b > K a ) = (1.6 x / 5.7 x = 3 x 10 4 ), K b of CN - >> K a of NH 4 + (Solution is Basic)  Acceptance of proton from H 2 O by CN - proceeds much further than the donation of a proton to H 2 O by NH 4 +.

SALTS OF WEAKLY ACIDIC CATIONS AND WEAKLY BASIC ANIONS 47

ACIDS-BASE REACTIONS 48 four  There are four classifications or types of reactions: i. strong acid with strong base, ii. strong acid with weak base, iii. weak acid with strong base, and iv. weak acid with weak base. NOTE:  For all four reaction types the limiting reactant problem is carried out first.  Once this is accomplished, one must determine which reactants and products remain and write an appropriate equilibrium equation for the remaining mixture.

ACIDS-BASE REACTIONS 49 (i) STRONG ACID WITH STRONG BASE:  The net reaction is:  The product, water, is neutral. (ii) STRONG ACID WITH WEAK BASE:  The net reaction is:  The product is HB + and the solution is acidic.

ACIDS-BASE REACTIONS 50 (iii) WEAK ACID WITH STRONG BASE:  The net reaction is:  The product is A - and the solution is basic. (iv) WEAK ACID WITH WEAK BASE:  The net reaction is:  Notice that K net may even be less than one.  This will occur when K a HB + > K a HA.

ACIDS-BASE REACTIONS 51 Example 4:  You titrate 100 mL of a M solution of benzoic acid with M NaOH to the equivalence point (mol HBz = mol NaOH).  What is the pH of (a) the final solution (b) half way point? Note: Note: HBz and NaOH are used up! HBz + NaOH ---> Na + + Bz - + H 2 O K a = 6.3 × K b = 1.6 × C 6 H 5 CO 2 H = HBz Benzoate ion = Bz -

BUFFER SOLUTIONS 52 HCl is added to pure water. HCl is added to a solution of a weak acid H 2 PO 4 - and its conjugate base HPO 4 2-.

BUFFER SOLUTIONS 53  The function of a buffer is to resist changes in the pH of a solution.  Buffers are just a special case of the common ion effect.  Buffer Composition Weak Acid+Conj. Base HC 2 H 3 O 2 +C 2 H 3 O 2 - H 2 PO 4 - +HPO 4 2- Weak Base+Conj. Acid NH 3 +NH 4 +

BUFFER SOLUTIONS 54  Consider HOAc/OAc - to see how buffers work.  ACID USES UP ADDED OH -.  We know that OAc - + H 2 O HOAc + OH - has K b = 5.6 x  Therefore, the reverse reaction of the WEAK ACID with added OH - has K reverse = 1/ K b = 1.8 x 10 9  K reverse is VERY LARGE, so HOAc completely uses up the OH - !!!! Acetic acid (HOAc) & a salt of the acetate ion (OAc)

BUFFER SOLUTIONS 55  Consider HOAc/OAc - to see how buffers work.  CONJUGATE BASE USES UP ADDED H + HOAc + H 2 O OAc - + H 3 O + has K a = 1.8 x  Therefore, the reverse reaction of the WEAK BASE with added H + has K reverse = 1/ K a = 5.6 x 10 4  K reverse is VERY LARGE, so OAc - completely uses up the H + !

BUFFER SOLUTIONS 56 Example 5  What is the pH of a buffer that has [HOAc] = M and [OAc - ] = M? HOAc + H 2 O OAc - + H 3 O + K a = 1.8 x K a = 1.8 x 10 -5

BUFFER SOLUTIONS  Notice that the expression for calculating the H + concentration of the buffer is  This leads to a general equation for finding the H + or OH - concentration of a buffer.  Notice that the H + or OH - concentrations depend on K and the ratio of acid and base concentrations. 57 [H 3 O + ] = Orig. conc. of HOAc Orig. conc. of OAc K a [OH  ]  [Base] [Conj. acid] K b [H 3 O  ]  [Acid] [Conj. base] K a

HENDERSON-HASSELBALCH EQUATION 58  Take the negative log of both sides of this equation: or  This is called the Henderson-Hasselbalch equation. [H 3 O + ] = [Acid] [Conj. base] K a pH = pK a + log [Conj. base] [Acid] pH = pK a - log [Acid] [Conj. base]

HENDERSON-HASSELBALCH EQUATION 59  This shows that the pH is determined largely by the pK a of the acid and then adjusted by the ratio of acid and conjugate base.  Sustainability. pH  pK a + log [Conj. base] [Acid]

HENDERSON-HASSELBALCH EQUATION 60 EXAMPLE 6  What is the new pH when 1.00 mL of 1.00 M HCl is added to: a) 1.00 L of pure water (before HCl, pH = 7.00) b) 1.00 L of buffer that has [HOAc] = M and [OAc-] = M (pH = 4.68)

PREPARING A BUFFER 61  You want to buffer a solution at pH =  This means [H 3 O + ] = 10 -pH = 5.0 x M  It is best to choose an acid such that [H 3 O + ] is about equal to K a (or pH ­ pK a ).  You get the exact [H 3 O + ] by adjusting the ratio of acid to conjugate base.  Buffer is prepared from:  HCO 3 - weak acid  CO 3 2- conjugate base

PREPARING A BUFFER 62  You want to buffer a solution at pH = 4.30 or [H 3 O + ] = 5.0 x M POSSIBLE ACIDSK a HSO 4 - / SO x HOAc / OAc x HCN / CN x  Best choice is acetic acid / acetate.

PREPARING A BUFFER 63  You want to buffer a solution at pH = 4.30 or [H 3 O + ] = 5.0 x M  Solve for [HOAc]/[OAc - ] ratio = 2.78/1  Therefore, if you use mol of NaOAc and mol of HOAc, you will have pH = [H 3 O  ]  5.0 x = [HOAc] [OAc - ] (1.8 x )

PREPARING A BUFFER 64  The concentration of the acid and conjugate base are not important.  It is the RATIO OF THE NUMBER OF MOLES of each.  This simplifying approximation will be correct for all buffers with 3<pH<11, since the [H] + will be small compared to the acid and conjugate base.

PREPARING A BUFFER 65  Preparing Buffers 1) Solid/Solid: mix two solids. 2) Solid/Solution: mix one solid and one solution. 3) Solution/Solution: mix two solutions. 4) Neutralization: Mix weak acid with strong base or weak base with strong acid.

PREPARING A BUFFER 66 Example 7:  Preparing buffer: solid/solid  Calculate the pH of a solution made by mixing 1.5 moles of phthalic acid and 1.2 moles of sodium hydrogen phthalate in 500. mL of solution.  It is given that K a = 3.0 x 10 -4

PREPARING A BUFFER 67 Example 8:  Preparing buffer: Solid/solution  How many moles of sodium acetate must be added to 500. mL of 0.25 M acetic acid to produce a solution with a pH of 5.50?

EXAMPLE 1 To calculate K a, we need all equilibrium concentrations. We can find [H 3 O + ], which is the same as [HCOO − ], from the pH. pH = –log [H 3 O + ] – 2.38 = log [H 3 O + ] = 10 log [H 3 O + ] = [H 3 O + ] 4.2  = [H 3 O + ] = [HCOO – ] 68

EXAMPLE 1 In table form: 69

EXAMPLE 2 The equilibrium constant expression is: Use the ICE (Initial Change and Equilibrium) table: 70

EXAMPLE 2 Simplify: x is relatively same, 71

EXAMPLE 2 72 (1.8  ) (0.30) = x  = x  = x pH = –log [H 3 O + ] pH = – log (2.3  10 −3 ) pH = 2.64

EXAMPLE 3 Tabulate the data. Simplify: 73

EXAMPLE 3 Therefore, [OH – ] = 1.6  M pOH = –log (1.6  ) pOH = 2.80 pH = – 2.80 pH = (1.8  ) (0.15) = x  = x  = x (x) 2 (0.15) 1.8  =

EXAMPLE 4 (a) The product of the titration of benzoic acid, the benzoate ion, Bz -, is the conjugate base of a weak acid.  The final solution is basic K b = 1.6 x

EXAMPLE 4 (a) This is a two-step problem: 1 st :stoichiometry of acid-base reaction 2 nd :equilibrium calculation 1. Calculate moles of NaOH required. (0.100L HBz)(0.025M) = mol HBz mol NaOH This requires mol NaOH 2. Calculate volume of NaOH required mol (1 L / mol) = L 25 mL of NaOH required 76

EXAMPLE 4 (a) 3.Moles of Bz - produced = moles HBz = mol Bz - 4.Calculate concentration of Bz -.  There are mol of Bz - in a TOTAL SOLUTION VOLUME of 125 mL [Bz - ] = mol / L = M 77

EXAMPLE 4 (a) Solving: x = [OH - ] = 1.8 x 10 -6, pOH = 5.75, pH = K b  1.6 x = x x

EXAMPLE 4 (b) [H 3 O + ] = { [HBz] / [Bz - ] } K a At the half-way point, [HBz] = [Bz - ], so [H 3 O + ] = K a = 6.3 x pH =

EXAMPLE 5 Assuming that x << and 0.600, we have Assuming that x << and 0.600, we have [H 3 O + ] = 2.1 x and pH =

EXAMPLE 6 (a) Calculate [HCl] after adding 1.00 mL of HCl to 1.00 L of water M 1 V 1 = M 2 V 2 M 2 = 1.00 x M pH = 3.00 (b) Step 1 — do the stoichiometry H 3 O + (from HCl) + OAc - (from buffer) ---> HOAc (from buffer)  The reaction occurs completely because K is very large. 81

EXAMPLE 6 Step 2—Equilibrium. 82

EXAMPLE 6 Because [H 3 O + ] = 2.1 x M BEFORE adding HCl, we again neglect x relative to and [H 3 O + ] = 2.1 x M > pH = 4.68  The pH has not changed significantly upon adding HCl to the buffer! 83 [H 3 O  ]  [HOAc] [OAc - ] K a  (1.8 x )

EXAMPLE 7 84 Conjugates do not react!!

EXAMPLE 8 Let X = moles NaC 2 H 3 O 2, 85 Conjugates do not react!!