Acids and Bases – Summary
Types of problems:
1. Identifying conjugate acid-base pairs
2. The ion-product constant (Kw) of water
3. pH calculations
- pH from [H+]
- [H+] from pH
4. Calculating pH of strong acids and bases
5. Weak acids
6. Calculating pH of polyprotic acids
7. Calculating pH of weak bases
8. Acid-base properties of salt solutions
Problem Type 1: Identifying conjugate acid-base pairs
A conjugate base is the species remaining when one proton (H+) has been removed from a Brønsted acid. A conjugate acid is the species remaining when one proton (H+) is added to a Brønsted base. Consider the following reaction:
HCl + H2O ↔ H3O+ + Cl-
The conjugate base to the Brønsted acid, HCl, is Cl-. The conjugate acid to the Brønsted base, H2O is H3O+. Thus, HCl and Cl- are defined as a conjugate acid-base pair, and H2O and H3O+ are similarly defined as a conjugate acid-base pair.
Another way to view this relationship is:
HCl + H2O ↔ H3O+ + Cl-
acid1 base2 acid2 base1
Problem Type 2: The ion-product constant (Kw) of water
Although water is a weak conductor of electricity, it does ionize to a small extent:
H2O(l) = H+(aq) + OH-(aq)
The equilibrium expression for this equilibrium reaction is written as:
However, since the equilibrium is shifted very far to the left, [H2O] is essentially constant and we can rewrite the expression as:
Example 1: What is the [H+] of a solution if the [OH-] is 5.0 x 10-5M?
[OH]- = 5.0 x 10-5 M
Problem Type 3: pH calculations
H+ and OH- concentrations are very small and inconvenient to express and work with. Possible concentrations have a very wide range (like from 1 M to 100,000,000,000,000 M!!!), so that the placement of the decimal point in the concentration is more informative than the actual digits. For example it seems far more informative to know that the H+ concentration of a solution is “in the millions” rather than “in the hundreds” than to know whether it is 6 million or 6.5 million.
Thus, [H+] is commonly given by expressing it as the exponent (power) of 10 that calculates to the actual concentration, and is referred to as the “pH” or “power of H.” That is:
pH from [H+]
Example 2: What is the pH of a 0.015 M [H+] solution?
Where [H+] = 0.015 M
Note that the 2 significant figures in the concentration result in 2 decimal place
precision in the pH value.
[H+] from pH
Example 3: What is the [H+] of a solution with a pH = 3.8?
Note that 1 decimal place in the pH results in 1 significant figure in the concentration.
Problem Type 4: Calculating pH of strong acids and bases
Strong acids and strong bases are electrolytes that are assumed to ionize completely in water. Thus, there really is no such thing as 0.5 M HCl, because this strong acid completely ionizes to 0.5 M H+ and 0.5 M Cl-. A more accurate way of writing the reaction for the dissolving of HCl is:
HCl + H2O = H3O+ + Cl-
You need to memorize the following strong acids and bases:
Acids Bases
HClO4 perchloric acid LiOH lithium hydroxide
HI hydroiodic acid NaOH sodium hydroxide
HBr hydrobromic acid KOH potassium hydroxide
HCl hydrochloric acid RbOH rubidium hydroxide
H2SO4 sulfuric acid (1st H only) CsOH cesium hydroxide
HNO3 nitric acid Ba(OH)2 barium hydroxide
(all alkalai metals except barium)
Example 4: What is the pH of a 0.10 M Ba(OH)2 solution?
First write the balanced equation, then make an “accounting” spread sheet detailing all species in solutions:
Ba(OH)2 = Ba2+ + 2OH- _
Initial concentration 0.10 M 0 0
Change -0.10 +0.10 +0.20 _
Equilibrium 0 0.10 M 0.20 M
There are now two valid pathways to solve this problem:
Problem Type 5: Weak acids
Most acids are weak acids in that they ionize in water to a limited extent, so that the parent acid, HA is in equilibrium with the resulting ions, H+ and A-. Since this is an equilibrium system, an equilibrium expression can be written for the reaction:
HA + H2O = H3O+ + A-
But the concentration of water is presumed to be near 1 and nearly constant, so
Calculations involving weak acids are just like the equilibrium calculations performed when studying equilibria, probably your previous topic. The main difference is the extra pH calculation. Often, you will be instructed to calculate the concentrations of all species in solution. You may even be given the pH and asked to calculate the Ka.
Example 5: What is the pH of a 0.20 M acetic acid (CH3COOH) solution?
First write the balanced equation, then make an “accounting” spread sheet detailing all species in solutions:
CH3COOH = H+ + CH3COO- _
Initial concentration 0.20 M 0 0
Change -x +x +x _
Equilibrium 0.20-x x x
Since Ka is quite small, x must be quite small, and .20-x ≈ .20, so that
CH3COOH = H+ + CH3COO- _
Initial concentration 0.20 M 0 0
Change -x +x +x _
Equilibrium 0.20-x x x
Equilibrium 0.20 M
The assumption that .20-x ≈ .20 is valid only if x<5% of .20 which it is (.0019/.2=1%.)
So,
Problem Type 6: Calculating pH of polyprotic acids
Polyprotic (diprotic, triprotic, etc) acids may yield more than one H+ per molecule. Usually, each succeeding proton is less likely to dissociate from the parent species, so that each ionization equilibrium has a distinctly smaller acid dissociation constant, Ka. For the diprotic acid H2A:
H2A = H+ + HA- and
HA- = H+ + A2- usually Ka1Ka2
Where HA- is both the conjugate base to the acid H2A and the acid in the second ionization.
Example 6: Calculate the concentrations of H2A, HA-, A2-, and H+ in a 1.0 M solution of H2A, where Ka1 = 1.3 x 10-2 and Ka2 = 6.3 x 10-8.
Start by considering only the first ionization.
H2A = H+ + HA-
Initial 1.0 0 0
Change -x +x +x
Equilibrium 1.0-x x x
Equilibrium 0.89M 0.11M 0.11M
Now consider the second ionization.
HA- = H+ + A2-
Initial .11 .11 0
Change -y +y +y
Equilibrium .11-y .11+ y y
Equilibrium 0.11M 0.11M M
Concluding:
Problem Type 7: Calculating pH of weak bases
Don’t let these calculations for weak bases confuse you. They are the same as for acids, except you need Kb instead of Ka and the balanced equation needs to show OH- as a product.
Example 7: What is the pH of a 0.10 M solution of pyridine (C5H5N)? Kb = 1.7 x 10-9
First write the balanced equation, then make an “accounting” spread sheet detailing all species in solutions:
C5H5N + H2O = C5H5NH+ + OH- _
Initial concentration 0.10 M N/A 0 0
Change -x +x +x _
Equilibrium 0.10-x x x
Since Kb is quite small, x must be quite small, and .10-x ≈ .10, so that
(1.3 x 10-5/.1 = .013% < 5%)
C5H5N + H2O = C5H5NH+ + OH- _
Initial concentration 0.10 M N/A 0 0
Change -x +x +x _
Equilibrium 0.10-x x x
Equilibrium 0.10 M
Problem Type 8: Acid-base properties of salt solutions
When salts are dissolved in water they can produce acid, base, or neutral solutions. There are four possibilities:
Salts that produce neutral solutions: Any salt that yields an anion that is the conjugate base of a strong acid or a cation that is the conjugate acid of a strong base will produce a neutral solution.
Example 8: Will NaNO3 produce an acid, basic, or neutral solution when dissolved in water?
When dissolved: NaNO3 = Na+ + NO3-
However, NO3- can react with H2O as follows: NO3- + H2O = HNO3 + OH-
Clearly, this would result in a basic solution since it produces excess OH- ions.
However, HNO3 is a strong acid, so the equilibrium would be shifted completely to the left and no excess OH- would be produced. Conclusion: NaNO3 will produce a neutral solution.
In general, NO3- is the conjugate base to the strong acid HNO3, so any salt that produces NO3- will yield a neutral solution.
Salts that produce basic solutions: Salts that produce the conjugate base of a weak acid will yield a solution that is basic.
Example 9: Will NaF produce an acid, basic, or neutral solution when dissolved in water?
When dissolved: NaF = Na+ + F-
However, F- can react with H2O as follows: F- + H2O = HF + OH-
Clearly, this would result in a basic solution since it produces excess OH- ions. HF is a weak acid, so any HF produced would tend to remain dissolved in its molecular form rather than dissociate back into H+ and F-. Conclusion: NaF will produce a basic solution.
In general, F- is the conjugate base to the weak acid HF, so any salt that produces F- will yield a basic solution.
Salts that produce acidic solutions: Salts that produce the conjugate acid of a weak base will yield a solution that is acidic.
Example 10: Will NH4Cl produce an acid, basic, or neutral solution when dissolved in water?
When dissolved: NH4Cl = NH4+ + Cl-
However, NH4+ can react with H2O as follows: NH4+ + H2O = NH3+ H3O+
Clearly, this would result in an acidic solution since it produces excess H3O+ ions. NH3 is a weak base, so any NH3 produced would tend to remain dissolved in its molecular form rather than re-protonate back into NH4+ and H2O. Conclusion: NH4Cl will produce an acidic solution.
In general, NH4+ is the conjugate acid to the weak base NH3, so any salt that produces NH4+ will yield a basic solution.
Salts in which both the cation and anion react with H2O (both hydrolyze): The result depends on the relative strengths of the cation and anion. Here, there are three possibilities:
If Kb > Ka: If Kb for the anion is greater than Ka for the cation, then the solution is basic, because the anion will hydrolyze to a greater extent than the cation producing more OH- than H+.
If Kb Ka: If Kb for the anion is smaller than Ka for the cation, then the solution is acidic, because the cation will hydrolyze to a greater extent than the anion producing more H+ than OH-.
If Kb = Ka: If Kb for the anion is approximately equal to Ka for the cation, then the solution is neutral, because the cation will hydrolyze to the same extent as the anion producing the same amounts of H+ and OH-.