Acid-Base Reactions and pH Calculations: Solved Problems
1. Identifying Conjugate Acid-Base Pairs
Identify the conjugate acid-base pairs in the following reaction:
CH3COO– + HCN ⇄ CH3COOH + CN–
Solution:
According to the Bronsted-Lowry theory, an acid is a chemical species that can donate an H+ ion. A base is a species that can accept an H+ ion. Therefore, by reacting an acid with a base, a new acid and a new base will always be formed.
Looking at the acid-base reaction of this problem:
CH3COO– + HCN ⇄ CH3COOH + CN–
You can see that the species CH3COO– (acetate ion) is a proton acceptor because it has electron pairs to share on the O atom bearing the negative charge and is therefore a base, while HCN is an acid (hydrocyanic acid) because it yields the H+ ion to the acetate ion:
CH3COO– + HCN ⇄ CH3COOH + CN–
base1 acid2 acid1 base2
Upon transfer of a proton, the base CH3COO– is transformed into an acid, CH3COOH, which is therefore its conjugate acid. Moreover, the acid HCN becomes the cyanide ion, CN–, which is its conjugate base.
In summary, the conjugate acid-base pairs found in this reaction are:
HCN/CN– and CH3COOH/CH3COO–
2. Conjugate Acids of Bases
Write the formula for the conjugate acids of the following bases: (a) HS– (b) H2PO4–
Solution:
(a) A base corresponds to a chemical species that accepts a proton, therefore when the HS– accepts a proton:
HS– + H+ it is transformed into H2S.
(b) Proceeding similarly, the ion H2PO4– accepts a proton and becomes H3PO4.
3. Conjugate Bases of Acids
Write the formula for the conjugate bases of the following acids: (a) HSO3– (b) H2PO4–
Solution:
(a) By definition, when an acid loses a proton it is transformed into its conjugate base. To determine the formula of the conjugate base of HSO3–, we remove a proton. Therefore, we should remove an H atom from the formula, and the total charge of the species becomes more negative since the H atom leaves its electron. The formula of the conjugate base of the ion HSO3– is then SO32-.
(b) Proceeding similarly, we remove an H from the formula H2PO4– and increase the negative charge by one unit, resulting in the conjugate base of this species as HPO42-.
4. Calculating pH from Hydronium Ion Concentration
Calculate the pH of a solution that has the following hydronium ion concentration (H3O+ or H+): (a) 4.75 × 10-4 M, (b) 0.0188 M, (c) 5.79 × 10-10 M.
Solution:
(a) If the hydronium ion concentration of the solution is known, to calculate the pH, simply apply the definition of pH:
pH = -log[H+]
pH = -(log 4.75 × 10-4)
pH = -(-3.323)
pH = 3.32
(b) pH = -(log 0.0188) = -(-1.726)
pH = 1.73
(c) pH = -(log 5.79 × 10-10) = -(-9.237)
pH = 9.24
5. Calculating pH from Hydroxyl Ion Concentration
Calculate the pH of a solution whose concentration of hydroxyl ions (OH–) is (a) 4.5 × 10-12 M, (b) 0.00316 M, (c) 2.3 × 10-4 M.
Solution:
(a) To calculate the pH of a basic solution, you can use any of the following methods:
First, calculate the pOH, applying:
pOH = -log[OH–]
pOH = -(log 4.5 × 10-12) = -(-11.35)
pOH = 11.4
To calculate the pH, apply the relationship:
pH + pOH = 14
pH = 14 – pOH
pH = 14 – 11.4 = 2.6
(b) [OH–] = 0.00316 M
pOH = -(log 0.00316) = -(-2.50) = 2.50
pH = 14 – pOH = 14 – 2.50 = 11.5
(c) [OH–] = 2.3 × 10-4 M
pOH = -(log 2.3 × 10-4) = -(-3.6) = 3.6
pH = 14 – 3.6 = 10.4
6. Calculating pH and pOH of Strong Acid/Base Solutions
Calculate pH and pOH of (a) a solution of 0.15 M HCl, and (b) a solution of 0.2 M NaOH.
Solution:
(a) HCl is a strong acid and therefore is considered completely dissociated:
HCl → H+ + Cl–
[H+] = 0.15 M
pH = -log[H+] = -(log 0.15) = -(-0.824)
pH = 0.82
From the relationship pH + pOH = 14, we can solve for pOH:
pOH = 14 – pH = 14 – 0.82
pOH = 13.18
(b) NaOH, being a strong base, is also considered completely dissociated:
NaOH → Na+(aq) + OH–
pOH = -(log 0.2) = -(-0.70)
pOH = 0.70
pH = 14 – pOH = 14 – 0.70 = 13.30
7. Calculating Ion Concentrations from pH
Calculate the concentration of hydronium ions and hydroxyl ions from a solution whose pH is 11.5.
Solution:
To calculate the concentration of H+ from pH, the operations should be performed in reverse:
-log[H+] = pH / (-1)
log[H+] = -pH / applying antilog
[H+] = antilog(-pH)
[H+] = antilog(-11.5)
[H+] = 3.16 × 10-12 M
8. pH of Diluted Nitric Acid Solution
Determine the pH of a solution prepared by diluting 150 mL of a 2.5 M nitric acid (HNO3) solution to a total volume of 1 L (MW = 63 g/mol).
Solution:
Nitric acid is a strong acid; therefore, it dissociates 100%. We need to calculate the concentration of the HNO3 solution after dilution to determine the acid concentration in the dilute solution and, from it, the concentration of H+:
The concentration of the diluted solution can be determined using the expression:
Vf = Vi × Ni × Nf
150 mL × 2.5 M = 1000 mL × X
X = 0.375 M
Once the concentration of the final solution is known, we can determine the hydrogen ion concentration. If this is 0.375 M and it dissociates 100%, then the concentration of H+ ions in the solution is also 0.375 M because it is a monoprotic acid.
pH = -(log 0.375) = -(-0.43) = 0.43
9. pH of Calcium Hydroxide Solution
Calculate the pH of a solution made by dissolving 0.4 g of calcium hydroxide in water to make 1.5 L (MW = 74 amu).
Solution:
Calcium is an alkaline-earth element, and therefore Ca(OH)2 is a strong electrolyte (100% dissociation):
Ca(OH)2 → Ca2+ + 2OH–
First, we need to calculate the concentration of Ca(OH)2 in the solution to calculate the concentration of hydroxyl ions in the solution from it. The molarity of this solution is:
M = moles / (molar mass × volume (liter))
M = 0.4 g / (74 g/mol × 1.5 L)
M = 0.0036
According to the equation, 1 mole of Ca(OH)2 produces 2 moles of OH–, and as it is a strong electrolyte, a 0.0036 M concentration of the base produces twice the concentration of OH–, that is, 0.0072 M.
pOH = -log(0.0072) = -(-2.14) = 2.1
pH = 14 – pOH = 14 – 2.1 = 11.9
10. Concentration of Perchloric Acid Solution
Calculate the concentration of a HClO4 solution that has pH = 2.4.
Solution:
First, consider that HClO4 is a strong acid, therefore 100% dissociated:
HClO4 → 100% H+ + ClO4–
Therefore, the concentration of H+ is equal to the initial concentration of HClO4:
[H+] = antilog(-2.4) = 4 × 10-3 M
The HClO4 solution is then 4 × 10-3 M.