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Chapter 8 Monoprotic Acid-Base Equilibria. 8-1 Strong Acids and Bases. HBr is a strong acid , so the reaction goes to completion. HBr + H 2 O H 3 O 3+ + Br - pH = -log[H + ] = -log(0.10) = 1.00. Box 8-1 Concentrated HNO 3 Is Only Slightly Dissociated 2.
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8-1 Strong Acids and Bases
HBr is a strong acid, so the reaction goes to completion.
HBr + H2O H3O3+ + Br-
pH = -log[H+] = -log(0.10) = 1.00
The figure shows a Raman spectrum of solutions of nitric acid of increasing concentration.
pH = -log[H+] = 13.00
Relation between pH pH + pOH = -logKW = 14.00 at 250C (9-1)
[H+] = KW/(1.0 X 10-8) = 1.0 X 10-6M pH = 6.00
Step 1Pertinent reactions. The only one is H2O = H+ + OH-.
Step 2Charge balance. The species in solution are K+, OH-, and H+. So,
[K+] + [H+] = [OH-] (9-2)
Step 3Mass balance. All K+ comes from the KOH, so [K+] = 1.0 X 10-8 M.
Step 4Equilibrium constant expression. KW = [H+][OH-] = 1.0 X 10-14.
Step 5Count. There are three equations and three unknowns ([H+], [OH-], [K+]), so we have enough information to solve the problem.
Step 6 Solve. Because we are seeking the pH, let’s set [H+] = x. Writing [K+] = 1.0 X 10-8 M in Equation 9-2, we get
[OH-] = [K+] + [H+] = 1.0 X 10-8 + x
[H+] = 9.6 X 10-8 M pH = -log[H+] = 7.02
When the concentration is “high” (≥10-6M), pH is calculated by just considering the added H+ or OH-. That is, the pH of 10-5.00M KOH is 9.00.
When the concentration is “low” (≤10-8M), the pH is 7.00. We have not added enough acid or base to change the pH of the water itself.
At intermediate concentrations of 10-6 to 10-8M, the effects of water ionization and the added acid or base are comparable. Only in this region is a systematic equilibrium calculation necessary.
Let’s review the meaning of the acid dissociation constant, Ka, for the acid HA:
A weak acid is one that is not completely dissociated. That is, Reaction 9-3 does not go to completion. For a base, B, the base hydrolysis constant, Kb, is defined by the reaction
A weak base is one for which Reaction 9-4 does not go to completion.
pK is the negative logarithm of an equilibrium constant:
pKW = -log KW
pKa = -log Ka
pKb = -log Kb
The acid HA and its corresponding base, A-, are said to be a conjugate acid-base pair, because they are related by the gain or loss of a proton.
Relation between Ka and Ka· Kb = Kw(9-5)
Kb for conjugate pair:
Weak Is Conjugate to Weak
The conjugate base of a weak acid is a weak base. The conjugate acid of a weak base is a weak acid.
Appendix G lists acid dissociation constants. Each compound is shown in its fully protonated form.
x2 + (1.07 X 10-3) x – 5.35 X 10-5 = 0
x = 6.80X 10-3 (negative root rejected)
[H+] = [A-] = x = 6.80X 10-3 M
[HA] = F – x = 0.0432 M
pH = -logx = 2.17
[H+] from HA dissociation = 6.8 X 10-3 M
[OH-] (from H2O dissociation) = 1.5 X 10-12 M
[H+] from H2O dissociation = 1.5 X 10-12 M
The fraction of dissociation, α, is defined as the fraction of the acid in the form A-:
Fraction of dissociation of an acid:
Weak electrolytes (compounds that are only partially dissociated) dissociate more as they are diluted.
The Essence of a Weak-Acid Problem
Equation for weak acids:
Dyes are colored molecules that can form covalent bonds to fabric. For example, Procion Brilliant Blue M-R is a dye with a blue chromophore (the colored part) attached to a reactive dichlorotriazine ring:
CelluloseㅡCH2OH = cellulose ㅡ CH2O- + H+
Fraction of dissociation
fraction of dissociation
A handy tip: Equation 9-11 can always be solved with the quadratic formula. However, an easier method worth trying first is to neglect x in the denominator.
B + H2O = BH+ + OH-
[B] = F – [BH+] = F-x
Equation for weak base:
[H+] = KW/[OH-] = 1.0 X 10-14/3.1 X 10-4 = 3.2 X 10-11
pH = -log[H+] = 10.49
What fraction of cocaine has reacted with water? We can formulate α for a base, called the fraction of association:
Fraction of association of a base:
Earlier, we noted that the conjugate base of a weak acid is a weak base, and the conjugate acid of a weak base is a weak acid.
Isomer of hydroxyl benzoic acid Ka Kb(= Kw/Ka)
ortho 1.07X 10-3 9.3 X 10-12
para 2.9 X 10-5 3.5 X 10-10
pH of 0.050 0 M o-hydroxybenzoate = 7.83
pH of 0.050 0 M p-hydroxybenzoate = 8.62
A buffered solution resists changes in pH when acids or bases are added or when dilution occurs. The buffer is a mixture of an acid and its conjugate base.
Mixing a Weak Acid and Its Conjugate Base
If you mix A moles of a weak acid with B moles of its conjugate base, the moles of acid remain close to A and the moles of base remain close to B.
HA = H+ + A- pKa = 4.00
Fraction of dissociation
Fraction of dissociation
HA dissociates very little, and adding extra A- to the solution will make the HA dissociate even less. Similarly, A- does not react very much with water, and adding extra HA makes A- react even less.
The central equation for buffers is the Henderson-Hasselbalch equation, which is merely a rearranged form of the Ka equilibrium expression.
Henderson-Hasselbalch equation for an acid:
Henderson-Hasselbalch equation for a base:
Show that, when activities are included, the Henderson-Hasselbalch equation is
Properties of the Henderson-Hasselbalch Equation
Regardless of how complex a solution may be, whenever pH = pKa, [A‑] must equal [HA]. This relation is true because all equilibria must be satisfied simultaneously in any solution at equilibrium. If there are 10 different acids and bases in the solution, the 10 forms of Equation 9-16 must all give the same pH, because there can be only one concentration of H+ in a solution.
Notice that the volume of solution is irrelevant, because volume cancels in the numerator and denominator of the log term:
moles of B/L of solution
moles of BH+/L of solution
moles of B
moles of BH+
B + H+ = BH+
OH- + HA = A- + H2O
H+ + OH- = H2O
Procedure: All solutions should be fresh. Prepare a solution of formaldehyde by diluting 9 mL of 37 wt% formaldehyde to 100 mL. Dissolve 1.5 g of NaHSO310 and 0.18 g of Na2SO3 in 400 mL of water, and add ~1 mL of phenolphthalein solution (Table 11-4). Add 23 mL of formaldehyde solution to the well-stirred buffer solution to initiate the clock reaction. The time of reaction can be adjusted by changing the temperature, concentrations, or volume.
We see that the pH of a buffer does not change very much when a limited amount of strong acid or base is added.
But why does a buffer resist changes in pH? It does so because the strong acid or base is consumed by B or BH+.
Buffer capacity, β, is a measure of how well a solution resists changes in pH when strong acid or base is added. Buffer capacity is defined as
Buffer capacity reaches a maximum when pH = pKa.
That is, a buffer is most effective in resisting changes in pH when pH = pKa (that is, when [HA] = [A-]).
In choosing a buffer, seek one whose pKa is as close as possible to the desired pH.
The useful pH range of a buffer is usually considered to be pKa±1 pH unit.
Buffer pH depends on Ionic Strength and Temperature
The correct Henderson-Hasselbalch equation, 9-18, includes activity coefficients.
Mass balance: FHA + FA- = [HA] + [A-]
Charge balance: [Na+] + [H+] = [OH-] + [A-]
[HA] = FHA – [H+] + [OH-] (9-20)
[A-] = FA- + [H+] – [OH-] (9-21)
The Henderson-Hasselbalch equation (with activity coefficients) is always true, because it is just a rearrangement of the Ka equilibrium expression.