A brief note on Acids, Bases and pH . Read it before a test. Try each question at the end before you open its answer.
Three ideas Arrhenius: acids give H + \text{H}^+ H + , bases give OH − \text{OH}^- OH − , in water. Bronsted-Lowry: acids donate protons, bases accept them. A conjugate pair differs by one proton. Lewis: acids accept an electron pair, bases donate one. BF 3 \text{BF}_3 BF 3 is a Lewis acid. Strong acids and bases ionise fully. A strong acid has a very weak conjugate base. Water and pH K w = [ H 3 O + ] [ OH − ] = 1.0 × 10 − 14 K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1.0 \times 10^{-14} K w = [ H 3 O + ] [ OH − ] = 1.0 × 1 0 − 14 at 298 K 298\ \text{K} 298 K . It grows with temperature. pH = − log 10 [ H 3 O + ] \text{pH} = -\log_{10}[\text{H}_3\text{O}^+] pH = − log 10 [ H 3 O + ] , and pH + pOH = 14 \text{pH} + \text{pOH} = 14 pH + pOH = 14 at 298 K 298\ \text{K} 298 K . One pH unit is a tenfold change. Neutral means equal ions. Weak acids and bases K a = [ H 3 O + ] [ A − ] [ HA ] K_a = \dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]} K a = [ HA ] [ H 3 O + ] [ A − ] and p K a = − log 10 K a \text{p}K_a = -\log_{10} K_a p K a = − log 10 K a . Smaller pKa, stronger acid. α ≈ K a / C \alpha \approx \sqrt{K_a/C} α ≈ K a / C and [ H 3 O + ] ≈ K a C [\text{H}_3\text{O}^+] \approx \sqrt{K_a C} [ H 3 O + ] ≈ K a C when α \alpha α is small. Weak base: [ OH − ] ≈ K b C [\text{OH}^-] \approx \sqrt{K_b C} [ OH − ] ≈ K b C . Conjugate pair: K a K b = K w K_a K_b = K_w K a K b = K w , and p K a + p K b = 14 \text{p}K_a + \text{p}K_b = 14 p K a + p K b = 14 . A worked example Compare 0.050 mol/L 0.050\ \text{mol/L} 0.050 mol/L of a strong acid with 0.050 mol/L 0.050\ \text{mol/L} 0.050 mol/L of acetic acid.
Strong acid. [ H 3 O + ] = 0.050 mol/L [\text{H}_3\text{O}^+] = 0.050\ \text{mol/L} [ H 3 O + ] = 0.050 mol/L , so the pH is 1.30 1.30 1.30 . Acetic acid. [ H 3 O + ] ≈ 1.8 × 10 − 5 × 0.050 = 9.5 × 10 − 4 mol/L [\text{H}_3\text{O}^+] \approx \sqrt{1.8 \times 10^{-5} \times 0.050} = 9.5 \times 10^{-4}\ \text{mol/L} [ H 3 O + ] ≈ 1.8 × 1 0 − 5 × 0.050 = 9.5 × 1 0 − 4 mol/L , so the pH is about 3.0 3.0 3.0 . Meaning. Same concentration, but the weak acid has about fifty times fewer hydronium ions. Practice with new numbers
A program to try The program finds the pH of the strong acid and of acetic acid, solving the weak one exactly.
Run it. How far apart are the two pH values?
1. Change C to 0.0050. Does the gap grow or shrink?
Mistakes to avoid The minus sign in pH. pH is minus the logarithm. The square root. K a C \sqrt{K_a C} K a C , not K a C K_a C K a C . The pOH of a base. Take it away from 14 for the pH. Two hydroxides. Ba(OH) 2 \text{Ba(OH)}_2 Ba(OH) 2 gives two OH − \text{OH}^- OH − per unit. Weak and dilute. They mean different things. Check yourself What is the conjugate acid of NH 3 \text{NH}_3 NH 3 ? Stuck? Show a hint Show the answerWhat is the pH of a solution with [ H 3 O + ] = 1.0 × 10 − 6 mol/L [\text{H}_3\text{O}^+] = 1.0 \times 10^{-6}\ \text{mol/L} [ H 3 O + ] = 1.0 × 1 0 − 6 mol/L ? Stuck? Show a hint Show the answerWhich is the stronger acid: pKa 2 or pKa 5? Stuck? Show a hint Show the answerWhy is BF 3 \text{BF}_3 BF 3 an acid though it has no hydrogen? Stuck? Show a hint Show the answerA weak acid is diluted. Does its pH rise or fall? Stuck? Show a hint Show the answer