Lecture Notes – Chem 222 – Monoprotic Acids and Bases – Rob Rossi
Based on S07 & S06 handwritten notes, references and focus updated to match 7th Ed of Harris
MONOPROTIC ACIDS AND BASES – HARRIS, CH 6 AND 9
Ostensibly, this should be review for almost all of you, but I'm going to go over it pretty thoroughly
anyway, because it is darn important. Key points we will cover:
• What are they? (acids and bases, that is)
• Definition of pH
• Strong vs. weak
• Ka, Kb, and their associated equilibrium expressions
• α fractions (Fraction of association/dissociation)
• Buffers – what they are, and how to prepare them (Buffer point, Henderson-Hasselbalch)
I'll also demonstrate how to apply the Systematic Treatment in working out some of the examples.
Acids and Bases, Defined – Harris 6-5
In this class, we will use the Brønsted-Lowry definitions of acids and bases:
Acid – Donates protons; in water, increases [H+] (H+ will be short for H3O+(aq))
Base – Accepts protons; in water, decreases [H+]
Conjugate – What's left after an acid or base donates or accepts a proton.
HCl (hydrochloric acid or hydrogen chloride) is an acid – it releases H+ into solution. Its conjugate
(base) is Cl-. NH3 (ammonia) is a base – it accepts H+ from solution. Its conjugate (acid) is NH4+.
pH – Harris 6-6 and 8-3
The exact definition of pH is
pH ≡ -log AH+ = -log[H+]γH+
However, it will be rare for us to actually worry about activities in calculating pH. In most cases, their
effect is too small to measure with normal instrumentation, and only adds a lot of needless calculation.
Strong Acids and Bases – Harris 6-7 and 9-1
Strong acids and bases dissociate/react "completely" in water. (Ka or Kb » 1) What this really means is
that for moderate concentrations (for solids, usually up to the solubility limit), the acid or base reaction
can be safely assumed to go to completion. I'm not much for memorization, but count on other
professors to expect you to "just know" that the acids in Table 6-2 (p. 108) are strong. We already
discussed the odd case of HF, which "should" be strong, but isn't; that exception aside, this general trend
can help you remember and/or figure out if an acid or base is strong: acids and bases built around ions
that generally lead to highly soluble salts (halides, alkali metals, nitrates, sulfates, perchlorates, and
quaternary ammonium salts amongst them) are almost always strong.
0.10 M HI → 0.10 M H+ + 0.10 M I-
pH ≈ -log10[H+] = 1.00
Calculate the pH of 5.0 × 10-8 M HI (in an ambient environment containing no pH-altering gases)
Presumption: 5.0 × 10-8 M H+, thus pH ≈ -log10[H+] = 7.30
Cool? Heck no, that's basic! HI is an acid. What up? Is it because we aren't accounting for activity?
Nope! γ's are less than one, A's are less than []; so pH would be higher still if we accounted for activity.
So what the heck is going on here? We're neglecting something that's usually negligible, but isn't in this
case, it turns out. Let's step back and take a systematic approach. What are the possible reactions that
can take place in a beaker of water you add HI to and then isolate from other reactive chemical species?
(A)
HI → H+ + I-
[Should go to completion, because HI is strong]
(B)
I- + H2O ← HI + OH-
[Forward reaction will not occur: HI is strong]
(C)
H2O
H+ + OH-
[Always possible in water, often negligible]
CAUTION: These three are NOT independent! Pick two reactions from the above if you ever decide you
need to consider them all simultaneously.
The trick here is that the amount of HI we've added is so tiny that the amount of H+ released by the
autoionization of water is not negligible. How do we figure out how much it is? Systematic treatment:
[H+] = [OH-] + [I-]
Charge Balance:
Atomic Balance (easiest one):
Equilibrium expression:
[I-] = FHI = 5.0 × 10-8 M
[H+][OH-] = 1.0 × 10-14
Variables: [I-], [H+], [OH-]
3 equations, 3 variables, so we are ready to go.
−14
1.0 × 10
[H+] =
+ 5.0 × 10-8
⇒
[H+]2 – 5.0 × 10-8 [H+] – 1.0 × 10-14 = 0
[H + ]
Solving the quadratic formula gives [H+] = 1.28 × 10-7 M
⇒
pH ≈ 6.893
Ah, now it's slightly acidic, as it should be. How much water auto-dissociates?
[H+] – [I-] = 1.28 × 10-7 M – 5.0 × 10-8 M = 7.8 × 10-8 M
That's less than the 1.0 × 10-7 M dissociation we would see in pure water, revealing an important truth:
the presence of any acid or base in solution suppresses the autoionization of water. Its magnitude is
never more than 1.0 × 10-7 M: that's an upper bound. So it doesn't contribute much to [H+] or [OH-] (a
negligible amount, except in a rare case such as this), but water may react in substantial amounts with
other acids or bases added to solution. (As when NH3 is added to water, NH3 + H2O → NH4+ + OH-.)
Weak Acids and Bases – Harris 6-7 and 9-2
Weak acids and bases dissociate/react incompletely in water. (Ka or Kb < 1) Their propensity to act as
an acid or base is quantified with a dissociation constant, Ka or Kb. (Note that the question of their
solubility is a separate matter! Mg(OH)2 is a strong base, but barely soluble in water [hence rarely used
as a base]. Ammonia is extremely soluble in water, but a weak base.)
[H + ][A − ]
+
HA → H + A
Keq = Ka =
[HA]
B + H2O → BH+ + OHA- + H2O → HA + OH-
Keq = Kb =
Keq = Kb =
BH+ → H+ + B
Ka⋅Kb =
[H + ][A − ] [HA][OH − ]
⋅
= [H+][OH-] = 1.0 × 10-14 = Kw
[HA]
[A − ]
[BH + ][OH − ]
[B]
[HA][OH − ]
[A − ]
Keq = Ka =
[H + ][B]
[BH + ]
General Solution of a Weak Acid Problem - Harris, 9-3
Let's consider the general case of a weak acid HA, with formal concentration FHA and acid dissociation
constant Ka. Applying the systematic treatment to such a problem, we get:
Equilibrium expression for HA:
[H + ][A − ]
= Ka
[HA]
Equilibrium expression for autoionization (Harris prefers the term autoprotolysis) of water:
[H+] [OH-] = Kw = 1.0 × 10-14
Charge Balance:
[H+] = [A-] + [OH-]
"Atomic" balance on A:
FHA = [HA] + [A-]
(A is not assembled or torn apart)
(Can't write true atomic balances on O or H until we know if they are present in A)
Let's apply this to determining the pH of common vinegar, which is 0.841 M acetic acid (CH3COOH) in
water. We'll use HA as a shorthand for acetic acid, and A- as short for its conjugate base, the acetate ion
(CH3COO-). Appendix G indicates that the pKa of acetic acid is 4.756. That means that (to demo pKa):
pKa = -log10 Ka = 4.756
⇒
Ka = 10 − pK a = 10-4.756 = 1.754 × 10-5
With this variable replaced with a number, the system of four equations above entails four variables:
[H+], [OH-], [HA], and [A-]; thus it is soluble if FHA is known (and not a variable). While this system
could be solved with brute force, "chemical intuition" can make things much easier on us. We know
acetic acid is an acid, with a not-ultra-puny Ka (» Kw) and at a respectable concentration. Thus we can
reasonably hope that the solution will be acidic, which will mean [OH-] will be quite small, probably
negligible relative to [A-] in the charge balance sum. We can assume anything we want, with little risk,
provided we check our assumptions. Think of it this way: if I'm writing a novel on my computer, and I
have an idea for a plot twist, I can write it all out. Then I read it, and see if it sucks. If it does, I'm out
only the time and energy it took me to write it. If it may makes the novel move along much more surely
and quickly than anything else I had in mind, it's a risk worth taking.
Here, the assumption [OH-] « [A-] allows us to simplify the charge balance to just [H+] = [A-]. Then,
with this and the atomic balance on A the and HA equilibrium expression (we don't need the water
equilibrium unless we want to solve for the additional variable [OH-]), we can solve for [H+]:
FHA = [HA] + [A-] = [HA] + [H+] ⇒ [HA] = FHA - [H+]
[H + ][A − ]
[H + ][H + ]
= Ka =
⇒ Ka(FHA - [H+]) = [H+]2 ⇒ 0 = [H+]2 + Ka[H+] - KaFHA
FHA − [H + ]
[HA]
− K a ± K a + 4K a FHA
− b ± b 2 − 4ac
[H ] =
=
2a
2
2
Solving the quadratic:
+
− 1.75 × 10 −5 ± 5.90 × 10 −5
= -8.77 × 10-5 ± 3.84 × 10-3 = 3.83 × 10-3 or a negative value
2
1.0 × 10 −14
Checking validity of assumption: [OH-] =
= 2.61 × 10-12
Much smaller than [A-] = [H+]
[H + ]
=
∴the assumption is valid and the solution obtained by making it is correct, within significant figures.
The pH of vinegar is therefore pH ≈ -log10[H+] = -log10(3.83 × 10-3) = 2.417 (moderately acidic)
Important: When looking for approximations one might make, inspect the sums involved in a problem.
A very small quantity can be ignored in a sum, but not in a product.
In retrospect, we see we could have actually gotten away with an even more drastic simplifying
assumption in working this problem. Acetic acid is weak enough that (in the absence of other species)
the bulk of it remains associated in water. Thus [HA] » [A-], and we could have approximated [HA] as
just FHA, simplifying the rest of the math as well: (Yes, this is confusing: [A-] is tiny, but still »[OH ]!)
[H+] = [A-] + [OH-] ≈ [A-]
FHA = [HA] + [A-] ≈ [HA]
Ka =
[H + ][A − ]
[H + ][H + ]
≈
FHA
[HA]
[H+] =
So, in this case,
⇒
[H+]2 = KaFHA
⇒
[H+] =
K a FHA
(1.75 × 10 −5 )(0.841 ) = 3.836 × 10-3 M = [A-]
Since the putative [A-] is indeed negligibly small relative to FHA (0.0038 M vs 0.84 M), within the
significant figures of the given information, this assumption is also valid, and the results calculated on
the basis of it correct. An assumptions along this line is dicier to make, but worth a shot when the Ka
value of the acid is appreciably different from its formal concentration. (Either Ka » FHA, and we
assume that most of the acid is dissociated, or Ka « FHA, and we assume most of it remains as HA.)
General Solution of a Weak Base Problem - Harris 9-4
Solving a general weak base problem follows in the footsteps of doing so for a weak acid, as outlined in
Harris 9-4. Here we assume that we have a basic solution, and thus [H+] is negligible, however. We can
also try getting away with assuming the base is so weak that most of it remains as B, or strong enough
that the vast majority of it will be present as BH+.
Fraction of [Dissociation of an Acid/Association of a Base] (α) - Harris 9-3 and 9-4
A simple but very useful quantity in working on acid/base problems is the fraction of dissociation or
association, more commonly referred to as the "alpha fraction" of a given species:
αspecific form ≡
αA- =
Concentration of species present in specified form
Total concentration of species
[A − ]
(Harris: "Fraction of dissociation of HA")
[HA] + [A − ]
αBH+ =
[BH + ]
(Harris: "Fraction of association of B")
[BH + ] + [B]
Generalized equations for α in terms of pH can be derived, and so it's pretty easy to make an α(pH) plot.
Example: What's the alpha fraction of acetate ion in common vinegar, αA-? Using our work above,
α CH COO− = αA- =
3
[A − ]
[H + ] 3.83 × 10 −3 M
=
=
= 0.004554
FHA
0.841 M
[HA] + [A − ]
This underscores the fact that the vast majority of the acetic acid is undissociated, present as HA. The
fraction of dissociation of acetic acid in this solution is very small, as per our valid assumption above!
Spreadsheet demo: α fraction vs. pH for acetic acid