Weak acids and bases, Kw, pH and the solubility product
الأسبوع 14 · الباب الثالث (٥): تركيز أيوني الهيدرونيوم والهيدروكسيل للأحماض والقواعد الضعيفة وتأين الماء وحاصل الضرب الأيوني Kw وpH وpOH والذوبانية وحاصل الإذابة Ksp (أثقل أسبوع: ثلاثة دروس حسابية) · 3 دروس
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The ionisation of water, Kw, pH and pOH
Chemistry Sec 3 Term 1, Chapter 3: pp. 71–77
GoalUse Kw = [H⁺][OH⁻] = 1 × 10⁻¹⁴ to find [OH⁻], and convert between concentration, pH and pOH with exact powers of ten and the given logarithms.
New words
ionic product of water Kw · [H⁺][OH⁻], equal to 1 × 10⁻¹⁴ at 25 °C in water and in every dilute solutionpH · −log[H⁺]: the smaller it is, the more acidic the solutionpOH · −log[OH⁻]; pH + pOH = 14
Explanation
Pure water ionises a very little: H₂O ⇌ H⁺ + OH⁻. Its constant, with the solvent left out, is the ionic product of water, Kw = [H⁺][OH⁻] = 1 × 10⁻¹⁴ at 25 °C (extra: Kw changes with temperature). In pure water [H⁺] = [OH⁻] = 10⁻⁷ mol/L, and the water is neutral. In an acid [H⁺] is above 10⁻⁷; if one ion rises the other falls so that the product stays 10⁻¹⁴. The pH scale takes away the powers of ten: pH = −log[H⁺] and pOH = −log[OH⁻], so pH + pOH = 14. A neutral solution has pH 7, an acidic one below 7 and a basic one above 7. One pH unit is a factor of 10 in [H⁺]. In this course we avoid a calculator: either [H⁺] is an exact power of ten, such as 10⁻³ (pH 3), or we are given log 2 = 0.30, log 3 = 0.48 and log 5 = 0.70. For [H⁺] = a × 10⁻ⁿ, pH = n − log a. A strong acid is fully ionised, so [H⁺] equals its concentration; the same is true of [OH⁻] for a strong base such as NaOH. For a base, find pOH first and then subtract it from 14.
Worked examples
Solution
[H⁺]
[OH⁻]
pH
pOH
0.0010 mol/L HCl
10⁻³
10⁻¹¹
3
11
0.010 mol/L NaOH
10⁻¹²
10⁻²
12
2
Strong acid and strong base. 0.0010 mol/L HCl is fully ionised, so [H⁺] = 1.0 × 10⁻³ and pH = 3. Then [OH⁻] = 10⁻¹⁴ ÷ 10⁻³ = 1.0 × 10⁻¹¹ and pOH = 11; check 3 + 11 = 14. For 0.010 mol/L NaOH, [OH⁻] = 10⁻², so pOH = 2 and pH = 14 − 2 = 12.
Weak acid and weak base from lesson 1. For the invented acid with Ka = 4.0 × 10⁻⁸ and C = 0.010 mol/L, [H⁺] = √(4.0 × 10⁻⁸ × 0.010) = √(4.0 × 10⁻¹⁰) = 2.0 × 10⁻⁵ mol/L, and pH = 5 − log 2 = 5 − 0.30 = 4.70. For the weak base with Kb = 2.0 × 10⁻⁵ and C = 0.050 mol/L, [OH⁻] = 1.0 × 10⁻³, so pOH = 3 and pH = 14 − 3 = 11.
!Careful!
pH = −log[H⁺], not log[H⁺]. Then pH + pOH = 14.
Back and forth, with log 2 = 0.30. [H⁺] = 2 × 10⁻⁴ gives pH = 4 − 0.30 = 3.70. From pH 9: [H⁺] = 1 × 10⁻⁹ and pOH = 5, so [OH⁻] = 1 × 10⁻⁵. A dilution: 10 mL of 0.010 mol/L HCl contains 0.010 × 0.010 = 1.0 × 10⁻⁴ mol of acid; in 1.0 L this is 1.0 × 10⁻⁴ mol/L, so pH = 4. A thousandfold dilution would only change the pH by 3 units. Careful: pH = −log[H⁺]; leaving out the minus sign gives a negative pH for an acid.
Class activity
Cabbage-juice ranking, with a grown-up. Squeeze a few red cabbage leaves by hand in a cup of cold water to make a purple juice. Pour equal amounts into four clear cups. Add one teaspoon of kitchen vinegar to the first, one teaspoon of lemon juice to the second, one teaspoon of baking soda stirred in water to the third, and nothing but a teaspoon of tap water to the fourth. See what happens and rank the cups from most acidic to most basic by colour. This is a ranking only, not a pH value. No tasting, no cleaners, no bleach, no ammonia: wash up after. Then mark the four cups as points on a drawn pH strip without numbers.
