Weak Acids & Acid Ionization Constant Majority of acids are weak. Consider a weak monoprotic acid, HA: The equilibrium constant for the ionization would be: or

Acid Ionization Constant The magnitude of K a for an acid determines its strength Higher the Ka More [H + ] in solution Stronger the Acid Lower the Ka Lower [H + ] in solutio n Weaker the Acid

Determining pH from K a Calculate the pH of a 0.50 M HF solution at 25 C. The ionization of HF is given by:

Determining pH from K a Initial (M) Change (M) -x+x Equilibrium (M) 0.50 – xxx

Determining pH from K a This leads to a quadratic equation, so lets simplify: 0.50 – x ≈ 0.50 The expression becomes:

Determining pH from K a At equilibrium, [HF] = ( ) = 0.48 M [H + ] = M [F - ] = M This only works if x is less than 5% of 0.50 Why? K a are generally only accurate to ±5%

Was the approximation good? Consider if initial concentration of HF is M. Same process above, but x = 6.0 x M. This would not be valid. VALID! NOT VALID!

Weak Bases & Base Ionization Constants Same treatment as for acids. Given ammonia in water: Reminder: we solve using [OH - ], not [H + ]

Relationship of K a to their Conjugate Base K w = K a K b This leads us to conclude the stronger the acid (larger K a ), the weaker its conjugate base (smaller K b ) Mole Buck Opportunity! Mole Buck Opportunity!

Diprotic & Polyprotic Acids A more involved process due to stepwise dissociation of hydrogen ion Each step is like a monoprotic acid Be sure to think about what is present at each step! Note the conjugate base is used as the acid in the next step

Diprotic Acid Calculation Calculate all species present at equilibrium in a 0.10 M solution of oxalic acid (H 2 C 2 O 4 )

Diprotic Acid Calculation Initial (M) Change (M) -x+x Equilibriu m (M) 0.10 – xxx

Diprotic Acid Calculation Making the approximation 0.10 – x  0.10, we get: STOP Approx. good? NOT VALID! Quadratic!

Diprotic Acid Calculation x = M After the first stage, we have: [H + ] = M [HC 2 O 4 - ] = M [H 2 C 2 O 4 ] = (0.10 – 0.054) M = M Next step: Treat conj. base as acid for 2 nd step

Diprotic Acid Calculation Second dissociation would be: Initial (M) Change (M) -y+y Equilibriu m (M) – y yy

Diprotic Acid Calculation Applying the approximation (for both) we obtain: STOP Approx. good? VALID!

Diprotic Acid Calculation! Finally, at equilibrium: [H 2 C 2 O 4 ] = M [HC 2 O 4 - ] = (0.054 – 6.1 x ) = M [H + ] = ( x ) = M [C 2 O 4 2- ] = 6.1 x M

Conclusions on Polyprotic Acids The past example shows that for diprotic acids, K a1 >> K a2 From this, we can assume the majority of H + ions are produced in the first stage of ionization Secondly, concentration of conj. base is numerically equal to K a2

Molecular Structure and Strength of Acids

Strength of Binary Acids

Strength of Oxoacids

Oxoacids having different central atoms from the same group with the same oxidation number In this group, acid strength increases with increasing electronegativity of the central atom Higher polarity means easier to ionize HClO 3 > HBrO 3

Strength of Oxoacids

Acid-Base Properties of Salts Salt hydrolyis is the reaction of an anion or a cation of a salt (or both) with water Salts can be neutral, basic, or acidic and follow certain trends Acids mixed with bases forms salt + water! BEWARE!

Salts that Produce Neutral Sol’ns Alkali metal ion or alkaline earth metal ion (except Be) do not undergo hydrolysis Conjugate bases of strong acids (or bases) do not undergo hydrolysis i.e. Cl -, Br -, NO 3 - So NaNO 3 forms a neutral solution

Salts that Produce Basic Sol’ns Conjugate bases of a weak acid will react to form OH - ions For example, NaCH 3 COO forms Na + and CH 3 COO - in solution. Acetate ion is the conjugate base of acetic acid, and undergoes hydrolysis:

Basic Salt Hydrolysis Calculation Calculate the pH of a 0.15 M solution of sodium acetate (CH 3 COONa) Since the dissocation is 1:1 mole ratio, the concentration of the ions is the same

Basic Salt Hydrolysis Calc. Because acetate ion is the conj. base of a weak acid, it hydrolyzes as: Initial (M) Change (M) -x+x Equilibriu m (M) 0.15 – xxx

Basic Salt Hydrolysis Calc. Applying the approximation, STOP Approx. good? Also known as percent hydrolysis!

Acidic Salt Hydrolysis

Common Ion Effect Revisited Recall that the addition of a common ion causes an equilibrium to shift Earlier we related this to the solubility of a salt The idea is just an application of Le Chatelier’s Principle

pH changes due to Common Ion Effect Consider adding sodium acetate (NaCH 3 COO) to a solution of acetic acid: The addition of a common ion here (CH 3 COO - ) will increase the pH By consuming H + ions

Henderson-Hasselbalch Consider: Rearranging K a for [H + ]we get:

Henderson-Hasselbalch Take negative log of both sides: or:

Finding pH with common ion present Calculate the pH of a solution containing both 0.20 M CH 3 COOH and 0.30 M CH 3 COONa? The K a of CH 3 COOH is 1.8x Sodium acetate fully dissociates in solution

Finding pH with common ion present Can use I.C.E. table, or Henderson-Hasselbach

Effect of Common Ion on pH Consider calculating the pH for a 0.20 M acetic acid solution pH = 2.72 From our last example, its obvious the pH has increased due to the common ion

Buffer Solutions Buffers are a solution of (1) weak acid or weak base and (2) its salt Buffers resist changes in pH upon addition of acid or base Buffer Capacity: refers to the amount of acid or base a buffer can neutralize

Buffer Solutions Consider a solution of acetic acid and sodium acetate Upon addition of an acid, H+ is consumed: Upon addition of a base, OH- is consumed:

Buffer Animation emistry/flash/buffer12.swf

Buffer Problem (a) Calculate the pH of a buffer system containing 1.0M CH 3 COOH and 1.0 M CH 3 COONa. (b) What is the pH of the buffer system after the addition of 0.10 mole of gaseous HCl to 1.0 L of the solution? Assume volume of sol’n does not change.

Buffer Problem To calculate pH of buffer, I.C.E. or H.H. With addition of HCl, we are adding 0.10 M H +, which will react completely with acetate ion

Buffer Problem Now acetic acid will still dissociate and amount of H + formed is the pH of sol’n Initial (mol) Change (mol) Equilibrium (mol)

Buffer Problem Using H.H. (or I.C.E.)

Finding Buffers of Specific pH If concentrations of both species are equal this means: Using H.H., to find specific pH, search for pK a  pH

Finding Buffers of Specific pH Describe how you would prepare a “phosphate buffer” with a pH of about 7.40

Finding Buffers of Specific pH Using the HPO 4 2- /H 3 PO 4 - buffer: This means a mole ratio of 1.5 moles disodium hydrogen phosphate : 1.0 mole Monosodium dihydrogen phosphate will result in a buffer solution with a 7.4 pH

Finding Buffers of Specific pH To obtain this solution, disodium hydrogen phosphate (Na 2 HPO 4 ) and sodium dihydrogen phosphate (NaH 2 PO 4 ) is in 1.5:1.0 ratio Meaning it is 1.5 M Na 2 HPO 4 and 1.0 M NaH 2 PO 4 are combined per liter of solution

Acid-Base Titrations Three situations will be considered: Strong Acid/ Strong Base Weak Acid/ Strong Base Strong Acid/ Weak Base Titrations of weak acid/base are complicated by hydrolysis

Strong Acid-Strong Base Titration Reacting HCl and NaOH, the net ionic equation would be: For our example, M HCl is being titrated by M NaOH Calculating pH changes depends on the stage of the titration

Strong Acid-Strong Base Titration Scenario #1: After addition of 10.0 mL of M NaOH to 25.0 mL of M HCl. Total volume = 35.0 mL

Strong Acid-Strong Base Titration Scenario #1

Strong Acid-Strong Base Titration Scenario #2: After addition of 25.0 mL of M NaOH to 25.0 mL of M HCl (aka equivalence point) Because equivalent moles of acid/base

Strong Acid-Strong Base Titration

Scenario #3: After addition of 35.0 mL of M NaOH to 25.0 mL of M HCl Total volume = 60.0 mL

Strong Acid-Strong Base Titration Scenario #3

Weak Acid-Strong Base Titration Consider the neutralization between acetic acid and sodium hydroxide: The net ionic is: Let’s calculate the pH for this reaction at different stages

Weak Acid-Strong Base Titration Scenario #1: 25.0 mL of M acetic acid is titrated with 10.0 mL of M NaOH Total volume = 35.0 mL

Weak Acid-Strong Base Titration

Scenario #1 Buffer system exists here (CH 3 COONa/CH 3 COOH) Calculate moles of each

Weak Acid-Strong Base Titration Initial (mol) 2.5x x Change (mol) -1.00x x10 -3 Equilibriu m (mol) 1.50x x10 -3 Buffer System Wooo!!

Weak Acid-Strong Base Titration

Scenario #2: Adding 25.0 mL of M NaOH to 25.0 mL of M acetic acid Equivalence point is not at 7!

Weak Acid-Strong Base Titration Initial (mol) 2.5x x Change (mol) -2.50x x10 -3 Equilibrium (mol) x10 -3

Weak Acid-Strong Base Titration Acid/Base concentration being zero at the equivalence point, pH is determined by hydrolysis of salt

Weak Acid-Strong Base Titration I.C.E. it! Equation becomes:

Weak Acid-Strong Base Titration Scenario #3: Adding 35.0 mL of M NaOH to 25.0 mL of M acetic acid

Weak Acid-Strong Base Titration Initial (mol) 2.5x x Change (mol) -2.50x x10 -3 Equilibriu m (mol) 01.00x x10 -3

Weak Acid-Strong Base Titration At this point, both OH - and hydrolysis of CH 3 COO - affects pH Since OH - is much stronger, we can ignore the impact of the hydrolysis

Strong Acid-Weak Base Titration Consider the titration of HCl with NH 3 : Or Let’s calculate the pH for this titration at different stages

Strong Acid-Weak Base Titration Key main differences from Weak Acid/ Strong Base: Equivalence point is less then 7 due to salt hydrolysis Calculation of pH after equivalence point is based on remaining [H + ] only Even though salt still exists, effect on pH minimal First part of titration is the same (buffer!)

Strong Acid-Weak Base Titration Calculate the pH at the equivalence point when 25.0 mL of M NH 3 is titrated by a M HCl solution

Strong Acid-Weak Base Titration At equivalence point, moles of acid equal moles of base

Strong Acid-Weak Base Titration Initial (mol) 2.5x x Change (mol) -2.50x x10 -3 Equilibriu m (mol) x10 -3

Strong Acid-Weak Base Titration Initial (mol) Change (mol) -x-x+x+x+x Equilibriu m (mol) xxx

Strong Acid-Weak Base Titration Approx. good (0.01%)