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Chapter 8: Monoprotic Acid-Base Equilibria
Chapter 6: Strong acids (SA) and strong bases (SB) ionize completely in water (very large K) [H+] ions produced equals [S.A.] Example: What is the pH of 0.050 M HCl solution? HCl is S.A. so [HCl] = [H+]. Thus, pH = - log [H+] = - log (0.050); pH = 1. Similarly, [OH-] in solution will be equal to [S.B.] x number OH-^ per formula unit
Challenge: What is the pH of 1.0 x 10-8^ M HCl? What is the pH of 1.0 x 10-8^ M NaOH? pH = 8 for a 10-8^ M acid??? pH = 6 for a 10-8^ M base??? We are adding an acid to water (at a pH of 7) and then saying that the solution becomes more basic? We neglected the dissociation of water!
Summary: pH calculations of strong acids and strong bases
- At relatively high [S.A] or [S.B.], i.e. ≥ 10 -6^ M, pH is calculated from [S.A.] or [S.B.]
- In very dilute [S.A] or [S.B.], i.e. ≤ 10 -8^ M, dissociation of water is more important, so the pH is 7.
Question: How do we know if a given acid is strong or weak? Know the 6 S.A. and six S.B. by heart
Weak Acids and Weak Bases Weak acids (HA) and weak bases (B) do not dissociate completely. An equilibrium exists between reactants and products The equilibrium lies to the left (Ka for a weak acid is < 1) => mostly HA or B in solution
The dissociation (ionization) of a weak acid , HA , in water:
HA(aq) + H 2 O(l) A-^ (aq) + H 3 O+ (aq) Weak acid Conj. base (H+^ donor)
Ka, the acid dissociation constant , can be written as:
Similarly, the hydrolysis (ionization) of a weak base , B , in water can be written as:
Ka
Exercise : Using part of App. G above: (a) draw the structure of 4-methylaniline. (b) Is alanine an acid or a base? (c) Draw the dissociation of 4-methylaniline and write the corresponding K (Ka if acid or Kb if base) expression.
Q. If we are dealing with a base, how do find its Kb? (App. G only gives Ka)
Consider then two reactions: The dissociation of a weak acid, HA , and hydrolysis of its conjugate base, A-
If HOCl dissociates by an x amount, its actual concentration is not 0.10 M, right?
The given concentration of HOCl, 0.10 M, is called its formal concentration.
Formal concentration, F (or formality ), refers to the total number of moles of a compound dissolved per liter of solution , regardless of the number of species it forms upon dissociation.
Formality, F
Accounts for all forms of a species
Ex. Forms of H 2 CO 3 in solution = H 2 CO3, HCO 3 -^ and CO 3 2- Thus, FH2CO3 = [H 2 CO 3 ] + [HCO 3 - ] + [CO 3 2-]
Thus, for a weak monoprotic acid, HA , at equilibrium:
Exercise : Calculate the pH of a 1.50 x 10-2^ M formic acid, HCO 2 H. (pKa = 3.745). HINT : You must first determine Ka from pKa. How? Answer: pH = 2.
Weak Acid Equilibria: Fraction of Dissociation
The fraction of dissociation , α (Greek: alpha) , of a weak acid HA refers to the fraction of HA in the dissociated form A-, i.e.
Exercise : You calculated the pH of a 1.50 x 10-2^ M formic acid, HCO 2 H, in the previous exercise. What fraction of formic acid in this solution has dissociated?
Answer: α HA = 0.10 or 10 % dissociated
Exercise : Calculate the pH and fraction of association of 1.5 x 10-1^ M ethylamine. HINT : You must first recognize that ethylamine is a base. [NOTE: Amines are organic bases. Their names end in –ine ]
From Appendix G
Relative strengths of acids/bases
- The stronger the acid the weaker its conjugate base; the stronger the base the weaker _its conjugate acid. How?
- If Ka is large in the forward direction, the reverse reaction will have a low K_
Q. Which is a stronger base, NH 3 or H 2 PO 4 -? Which is a stronger acid, NH 4 +^ or H 2 PO 4 -? Answer: NH 3 is a stronger base; H 2 PO 4 -^ is a stronger acid than NH 4 +.
SALTS
The term salt refers to the ionic compound produced from acid-base reaction.
Q. What is the pH of an aqueous solution of a salt? The anion or the cation of a salt, or both, can react with water (= hydrolysis) The pH of the salt depends on the strengths of the original acids and bases
Exercise: Show that Na 2 CO 3 hydrolyzes in water to form a basic solution.
Solution: First, recognize that Na 2 CO 3 in water produces Na+^ and CO 3 2-
Hydrolysis of CO 3 2-
CO 3 2-^ + H 2 O HCO 3 - + OH
-
Basic!
Exercise: Show that ammonium chloride hydrolyzes in water to form an acidic solution.
NH 4 Cl (^) (aq) NH 4 +^ and Cl-
From S.B; does not react with water = spectator ion , S.I.
Will hydrolyze; acts as W.B.
S.I.
W.A.
Hydrolyzes to form H+^ = acidic!
Some common buffers
Importance of Buffers
Buffer Name Composition Useful pH (Approx.)
Acetate HC 2 H 3 O 2 /NaC 2 H 3 O 2 5
Carbonate H 2 CO 3 /NaHCO 3 7
Phosphate NaH 2 PO 4 /Na 2 HPO 4 7
Ammonia NH 4 Cl/NH 3 10
Calculating pH of buffers: The Henderson-Hasselbalch Equation
Relates the pH of a buffer solution to the concentration of buffer components
Simply a rearranged version of the Ka expression for the acid component
Example 1: Sodium hypochlorite (NaOCl), the active ingredient of almost all bleaches, was dissolved in a solution buffered to pH 6.80. Find the ratio [OCl]/[HOCl] in this solution.
Example 2: Find the pH of a 750-mL aqueous solution prepared by dissolving 15.50 g TRIS (MM 121.135) plus 5.23 g TRIS hydrochloride (MM 157.596).
The pH of a buffer is nearly independent of volume
Effect of Adding and Acid or a Base to a Buffer
Example 1: If we add 10.50 mL of 1.00 M HCl to the solution in the previous example, what will be the new pH?
Hint: Which buffer component will react with HCl? This is the time to set up an ICF table.
pH drops a little bit since [HA] increases while the numerator, [A-], decreases
Example 2: If we add 5.00 mL of 1.00 M NaOH to the original TRIS buffer solution in Example 2 above, what will be the new pH?
vs. original pH = 8.
Added base increases the buffer’s pH a little bit.
mol added OH = (0.00500 L)(^ -^ 1.00mol / L = 0.005 00 mol OH-
Will react w/ HA
[0.133]
pH = 8.07 + log [0.02819]; pH = 8.
Q. What is the pH of a buffer equal to when [HA] = [A-]?
Maximum buffer capacity
Thus, a buffer works best (max. buffer capacity) when pH = pKa
When choosing a buffer, select one whose pKa is as close as possible to the desired pH
Useful pH range of a buffer: pH = pKa ± 1
Exercise 1: Using Table 8-2, suggest the best buffer for each of the following pHs: (a) 4.00, (b) 5.50, (c) 7.30 and (d) 12.00.
Maximum buffer capacity when [HA] = [A
- ]
pH = pKa
