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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Weak acids and weak bases: finding [H₃O⁺] and [OH⁻]
Chemistry Sec 3 Term 1, Chapter 3: pp. 71–77
GoalFind [H₃O⁺] of a weak acid with √(Ka × C) and [OH⁻] of a weak base with √(Kb × C), and check that the approximation is allowed.
New words
weak base · a base that ionises only partly in water, such as ammoniaKb · the ionisation constant of a weak base, such as [NH₄⁺][OH⁻] ÷ [NH₃] for ammoniaapproximation · treating C − x as C because x is very small compared with C
Explanation
Take a weak acid HA of concentration C mol/L in water: HA + H₂O ⇌ H₃O⁺ + A⁻. Each HA that ionises gives one H₃O⁺ and one A⁻, so the two ions are equal in number: let each be x. Then [HA] = C − x and Ka = x² ÷ (C − x). A weak acid ionises only a little, so x is much smaller than C and C − x is almost C. This gives Ka ≈ x² ÷ C, so x = √(Ka × C): [H₃O⁺] = √(Ka × C). A weak base works the same way. For ammonia, NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, with Kb = [NH₄⁺][OH⁻] ÷ [NH₃] = 1.8 × 10⁻⁵, the equal ions give [OH⁻] = √(Kb × C). The rule is an approximation: after the answer, check that x is small beside C, for example under 5 % of it. Method: write the equilibrium, multiply Ka (or Kb) by C, then take the square root: halve the exponent, and take the square root of the front number. If the exponent is odd, change it first, for example 2.5 × 10⁻⁷ = 25 × 10⁻⁸.
ƒWeak acid and weak base
[H₃O⁺] = √(Ka × C) [OH⁻] = √(Kb × C)
an approximation: check that the answer is small beside C
Worked examples
ƒInvented acid HA
[H₃O⁺] = √(1.0 × 10⁻⁶ × 0.010) = 1.0 × 10⁻⁴ mol/L
A weak acid. An invented acid HA has Ka = 1.0 × 10⁻⁶ and C = 0.010 mol/L. [H₃O⁺] = √(Ka × C) = √(1.0 × 10⁻⁶ × 0.010) = √(1.0 × 10⁻⁸) = 1.0 × 10⁻⁴ mol/L. Check: 1.0 × 10⁻⁴ ÷ 0.010 = 0.010, which is 1 % of C, so the approximation is fine.
A weak base. An invented base has Kb = 2.0 × 10⁻⁵ and C = 0.050 mol/L. [OH⁻] = √(Kb × C) = √(2.0 × 10⁻⁵ × 0.050) = √(1.0 × 10⁻⁶) = 1.0 × 10⁻³ mol/L. Check: 1.0 × 10⁻³ ÷ 0.050 = 0.020, which is 2 % of C, so the approximation is fine.
Ammonia. For 0.020 mol/L ammonia, Kb = 1.8 × 10⁻⁵: [OH⁻] = √(1.8 × 10⁻⁵ × 0.020) = √(3.6 × 10⁻⁷) = √(36 × 10⁻⁸) = 6.0 × 10⁻⁴ mol/L. This is also [NH₄⁺]. The ammonia left is 0.020 − 0.00060 = 0.0194 mol/L, nearly the original, and 6.0 × 10⁻⁴ ÷ 0.020 = 3 % of C, so the rule works. The odd exponent was changed first: 3.6 × 10⁻⁷ = 36 × 10⁻⁸.
Class activity
Square-root cards, by hand, no calculator, with a grown-up. Make four cards with these products Ka × C on the front: 1.0 × 10⁻⁸, 4.0 × 10⁻⁶, 9.0 × 10⁻¹⁰ and 2.5 × 10⁻⁷. Work out the square root of each and write it on the back. Hint: halve the exponent and take the root of the front number; for the last card write 25 × 10⁻⁸ first. You should find 1.0 × 10⁻⁴, 2.0 × 10⁻³, 3.0 × 10⁻⁵ and 5.0 × 10⁻⁴. Swap the cards with a friend to check each other. Paper only.
Practice
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An invented acid has Ka = 4.0 × 10⁻⁶ and C = 0.010 mol/L. Find [H₃O⁺] and check the approximation.
[H₃O⁺] = √(4.0 × 10⁻⁶ × 0.010) = √(4.0 × 10⁻⁸) = 2.0 × 10⁻⁴ mol/L. Check: 2.0 × 10⁻⁴ ÷ 0.010 = 0.020 = 2 % of C, so it is acceptable.
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An invented weak base has Kb = 9.0 × 10⁻⁶ and C = 0.010 mol/L. Find [OH⁻].
