Chapter 17
Acids-Base Equilibria and Solubility Equilibria
Buffer Solutions: is a solution of (1) a weak acid or a weak base and (2) its salt; both components must be present. The solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.
A buffer solution must contain a relatively large concentration of acid to react with any OH- ions that may be added to it and must contain a similar concentration of base to react with any added H+ ions. These requirements are satisfied by an acid-base conjugate base (a weak acid and its conjugate base or a weak base and its conjugate acid).
Consider an equal molar mixture of CH3COOH and CH3COONa
CH3COONa(aq) g CH3COO-(aq) + Na+(aq)
- If an acid is added, the H+ ions will be consumed by the conjugate base in the buffer, CH3COO-
CH3COO-(aq) + H+(aq) g CH3COOH(aq)
- If a base is added, the OH- ions will be neutralized by the acid in the buffer:
CH3COOH(aq) + OH-(aq) g CH3COO-(aq) + H2O(l)
The buffering capacity, that is, the effectiveness of the buffer solution, depends on the amounts of acid and conjugate base from which the buffer is made. The larger the amount, the greater the buffering capacity.
In general, a buffer system can be represented as salt/acid or conjugate base/acid.
Thus, the sodium acetate-acetic acid buffer system can be written as CH3COONa/CH3COOH or CH3COO-/CH3COOH.
Example 17.1, 17.2.
Another definition: A buffer contains a weak acid and its conjugate base, or a weak base and its conjugate acid.
Preparing a Buffer Solution with a Specific pH:
Referring to sodium acetate-acetic acid buffer system, we can write the equilibrium constant as:
Ka = [H+][CH3COO-]/[CH3COOH] [H+] = Ka [CH3COOH]/[CH3COO-]
- log [H+] = - log Ka - log [CH3COOH]/[CH3COO-]
- log [H+] = - log Ka + log [CH3COO-]/[CH3COOH]
pH = pKa + log [CH3COO-]/[CH3COOH]
This equation is called Henderson-Hasselbalch equation. In a more general form it can be expressed as:
pH = pKa + log [conjugate base]/[acid]
If the [acid] ≈ [conjugate base], then log [conjugate base]/[acid] = 0
or pH = pKa
Thus, to prepare a buffer solution, we choose a weak acid whose pKa is close to the desired pH. This choice not only gives the correct pH value of the buffer system, but also ensures that we have comparable amounts of the acid and its conjugate base present; both are prerequisites for the buffer system to function effectively.
Example 17.3.
A Closer Look at Acid-Base Titrations:
In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.
Equivalence point : the point at which the reaction is complete
Indicator : substance that changes color at (or near) the equivalence point.
Strong Acid-Strong Base Titrations: Consider the reaction between HCl and NaOH, which can be represented by:
HCl(aq) + NaOH(aq) g NaCl(aq) + H2O(l)
The net ionic equation is H+(aq) + OH-(aq) g H2O(l)
Consider the addition of 0.10 M NaOH solution to 25 mL of 0.10 M HCl.
In a strong acid-base titration both H+ and OH- concentrations are very small at the equivalence point (≈ 1.0*10-7M); consequently, the addition of a single drop of the base can cause a large increase in [OH-] and in the pH of the solution.
1- After the addition of 10.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl.
No. of moles of NaOH = 1.00*10-3 mol.
No. of moles of HCl = 2.50*10-3 mol.
· The amount of HCl left after partial neutralization = (2.50*10-3) – (1.00*10-3) = 1.50*10-3 mol.
· The concentration of H+ ions in 35.0 mL of solution = 1.50*10-3 mol/0.035 L = 0.0429 M.
2- After the addition of 25.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl.
It involves a complete neutralization reaction and the salt (NaCl) doesn’t undergo hydrolysis.
At the equivalence point [OH-] = [H+] = 1.0*10-7M and the pH of the solution = 7.
3- After the addition of 35.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl.
No. of moles of NaOH = 3.50*10-3 mol.
No. of moles of HCl = 2.50*10-3 mol.
· After complete neutralization, the number of moles of NaOH left = (3.50*10-3) - (2.50*10-3) = 1.00*10-3 mol.
· The conc. of NaOH ions in 60.0 mL of solution = 1.00*10-3 mol/0.060 L = 0.0167 M.
Thus the [OH-] = 0.0167 M and the pOH = 1.78.
pH = 14.00 – 1.78 = 12.22.
Weak Acid-Strong Base Titrations: Consider the reaction between CH3COOH and NaOH, which can be represented by:
CH3COOH(aq) + NaOH(aq) g CH3COONa(aq) + H2O(l)
The equation can be simplified to:
CH3COOH(aq) + OH-(aq) g CH3COO-(aq) + H2O(l)
The acetate undergo hydrolysis as follows:
CH3COO-(aq) + H2O(l) D CH3COOH(aq) + OH-(aq)
Therefore, at the equivalence point, when we only have CH3COONa present, the pH will be greater than 7 as a result of the excess OH- ion formed (Figure 17.4).
Example 17.4.
In the titration of a weak acid with a strong base when you are half way to the equivalence point:
pH = pKa [H+] = Ka
Strong Acid-Weak Base Titrations: Consider the titration of HCl, a strong acid, with NH3, a weak base. HCl(aq) + NH3(aq) g NH4Cl(aq)
or simply H+(aq) + NH3(aq) g NH4+(aq)
At the equivalent point, pH < 7 due to hydrolysis of NH4+ ion:
NH4+(aq) + H2O(l) D NH3(aq) + H3O+(aq)
or simply NH4+(aq) D NH3(aq) + H+(aq)
Figure 17.5 shows the titration curve for this experiment.
Example 17.5.
Acids-Base Indicators:
The equivalence point: is the point at which the number of moles of OH-ions added to a solution is equal to the number of moles of H+ ions originally present.
The end point of a titration occurs when the indicator changes color.
An acid-base indicator: is a weak organic acid or base that has distinctly different colors in its non ionized and ionized form.
Consider a weak monoprotic acid (HIn), and its conjugate base is In-
HIn(aq) D In-(aq) + H+(aq)
· If the indicator is in a sufficiently acidic medium, the equilibrium shifts to the left and the predominant color of the indicator is that of HIn form.
· If the indicator is in a sufficiently basic medium, the equilibrium shifts to the right and the color will be due to that of conjugate base In-.
1. [HIn]/[In-] ≥ 10 color of acid (HIn) predominates.
2. [HIn]/[In-] ≤ 0.1 color of conjugate base (In-) predominates.
3. If [HIn] ≈ [In-] then the indicator color is a combination of the colors of HIn and In-.
However, not all the indicators change color at the same pH, so the choice of indicator for a particular titration depends on the nature of the acid and base used in the titration (that is, whether they are weak or strong).
Some Common Acid-Base Indicators
Color
Indicator
/In Acid
/In Base
/ pH RangeThymol blue / Red / Yellow / 1.2-2.8
Bromophenol blue / Yellow / Bluish purple / 3.0-4.6
Methyl orange / Orange / Yellow / 3.1-4.4
Methyl red / Red / Yellow / 4.2-6.3
Chlorophenol blue / Yellow / Red / 4.8-6.4
Bromothymol blue / Yellow / Blue / 6.0-7.6
Cresol red / Yellow / Red / 7.2-8.8
Phenolphathalein / Colorless / Reddish pink / 8.3-10.0
The pH range is the range over which the indicator changes from the acid color to the base color. Example 17.6.
Solubility Equilibria:
Solubility Product:
Consider a saturated solution (which the undissolved solute is in dynamic equilibrium with the solution) of AgCl that is in contact with solid AgCl. The solubility equilibrium can be represented as:
AgCl(s) D Ag+(aq) + Cl-(aq) Ksp = [Ag+][Cl-]
Ksp is called the solubility product constant or solubility product.
The solubility product of a compound is the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium equation.
The following cases are more complex than AgCl:
· MgF2(s) D Mg2+(aq) + 2F-(aq) Ksp = [Mg2+][F-]2
· Ag2CO3(s) D 2Ag+(aq) + CO32-(aq) Ksp = [Ag+]2[CO32-]
· Ca3(PO4)2(s) D 3Ca2+(aq) + 2PO43-(aq) Ksp = [Ca2+]3[PO43-]2
Ion Product (Q): is the product of the molar concentrations of ions each raised to the power of their stoichiometric coefficients.
Q = [Ag+]o[Cl-]o
The subscript o reminds us that these are initial concentration and don’t necessarily correspond to those at equilibrium.
For the dissolution of an ionic solid in aqueous solution, any one of the following conditions may exist:
· Q < Ksp unsaturated solution no precipitate.
· Q = Ksp saturated solution.
· Q > Ksp Supersaturated solution Precipitate will form
Molar Solubility and Solubility:
The value of Ksp indicates the solubility of an ionic compound, the smaller the value of Ksp , the less soluble the compound in water.
Molar Solubility: is the number of moles of solute in 1 L of a saturated solution (mol/L).
Solubility: is the number of grams of solute in 1 L of a saturated solution (grams/L).
All these expressions refer to the concentration of saturated solutions at some given temperature (usually 25oC).
Examples 17.7, 17.8.
Predicting Precipitation Reactions:
From a knowledge of the solubility rules and the solubility products , we can predict whether a precipitate will form when we mix two solutions or add a soluble compound to a solution.
Example 17.9.
The Common Ion Effect and Solubility:
Suppose we study a solution containing two dissolved substances that share a common ion , say, AgCl and AgNO3 .
AgCl(s) D Ag+(aq) + Cl-(aq)
If we add AgNO3 to the solution, Ag+(aq) will increase, and will make Q > Ksp ,this will drive the equilibrium to the left and some AgCl precipitate out of the solution, as Le Chatelier’s principle would predict.
The effect of adding a common ion, then is a decrease in the solubility of the AgCl in solution. Note that this [Ag+] is no longer equal to [Cl-] at equilibrium; rather, [Ag+] > [Cl-].
Example 17.10.
Complex Ion Equilibria and Solubility:
Complexes are compounds formed from combination of metal ions (Lewis acids = electron-pair acceptor) with ligands (Lewis base = electron-pair donor).
Ag+(aq) + 2NH3(aq) D Ag(NH3)2+(aq)
Acid base
A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions.
Transition metals have a particular tendency to form complex ions.
A ligand increases the solubility of a slightly soluble ionic compound if it forms a complex ion with the cation.
If an excess of NH3 is added to a light-blue precipitate of Cu(OH)2, the blue precipitate redissolves to produce a dark-blue color solution as a result of the formation of the complex ion Cu(NH3)42+.
Cu(OH)2(s) + 4NH3(aq) D Cu(NH3)42+ + 2OH-(aq)
Thus the formation of the complex ion Cu(NH3)42+ increases the solubility of Cu(OH)2.
The formation constant or stability constant (Kf) is the equilibrium constant for the complex ion formation.
The larger Kf is, the more stable the complex ion is.
The formation of the Cu(NH3)42+ ion can be expressed as:
Cu2+(aq) + 4NH3(aq) D Cu(NH3)42+ Kf = [Cu(NH3)42+]/[Cu2+][NH3]4 = 5.0*1013
The very large value of Kf in this case indicates the great stability of the complex ion in solution and accounts for very low concentration of Cu2+ ions at equilibrium.
Example 17.11.
Amphoteric hydroxide like Al(OH)3 precipitate which can be formed when NaOH is added to Al(NO3)3 solution. With further addition of NaOH solution, the precipitate dissolves due to the formation of the complex ion Al(OH)4-.
Al(OH)3(s) + OH-(aq) D Al(OH)4-
Application and Solubility Product Principle to Qualitative Analysis:
Selected Problems: 5, 8-10, 12, 14, 15, 19, 20, 25, 26, 33-41, 45-47, 51-53, 60, 61.
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