Weak Acid HX: Settling a Dispute About the Approximation
A 0.150 mol/L solution of an unknown monoprotic weak acid, HX, has a measured pH of 2.62 at 25 °C. One student calculates Ka by substituting [HX] ≈ 0.150 mol/L directly into Ka = [H⁺]²/[HX] and obtains 3.8 × 10⁻⁵. A second student insists the calculation is invalid because the equilibrium concentration of HX must be lower than 0.150 mol/L, so the denominator is too large and the answer is wrong. The teacher asks the class to settle the dispute with a percent-ionization check and then to predict what happens to the pH when the same acid is diluted tenfold to 0.0150 mol/L. A 0.100 mol/L solution of the sodium salt NaX is also available for testing.
1. Calculate Ka both with and without the approximation, use percent ionization to judge whether the approximation was justified, and state the best value of Ka.
Read the worked answer
Find [H⁺]
[H⁺] = 10⁻²·⁶² = 2.40 × 10⁻³ mol/L
Without the approximation (exact)
[HX] = 0.150 − 0.00240 = 0.1476 mol/L
Ka = (2.40 × 10⁻³)² / 0.1476 = 5.76 × 10⁻⁶ / 0.1476 = 3.9 × 10⁻⁵
With the approximation
Ka = (2.40 × 10⁻³)² / 0.150 = 3.8 × 10⁻⁵
Percent ionization
% ionization = (2.40 × 10⁻³ / 0.150) × 100% = 1.6%
Verdict
The 5% rule is satisfied (1.6% ≪ 5%), so the approximation is justified: only 1.6% of the acid ionized, and ignoring that in the denominator changes Ka by about 1.6%, comparable to the rounding uncertainty in a pH quoted to two decimals. The second student is conceptually correct that [HX] is lower than 0.150 mol/L, but the effect is negligible here, so the first student's answer is not 'wrong' - the exact treatment agrees within rounding.
Best value: Ka ≈ 3.9 × 10⁻⁵ (the exact calculation is preferred; the difference is not chemically significant).
2. Predict qualitatively how the pH changes when the solution is diluted tenfold to 0.0150 mol/L, then calculate the new pH and compare it with the prediction that dilution raises the pH by exactly 1.00 unit.
Read the worked answer
Qualitative prediction
Dilution lowers [HX], which by Le Châtelier's principle shifts HX ⇌ H⁺ + X⁻ to the right, increasing the fraction ionized. So [H⁺] falls by less than a factor of 10, and the pH rises by less than 1.00 unit.
Calculation (0.0150 mol/L)
Ka = x²/(0.0150 − x) = 3.9 × 10⁻⁵
Approximation first: x ≈ √((3.9 × 10⁻⁵)(0.0150)) = √(5.85 × 10⁻⁷) = 7.6 × 10⁻⁴ mol/L
Check: 7.6 × 10⁻⁴/0.0150 = 5.1% - right at the 5% limit, so refine with the quadratic:
x² + (3.9 × 10⁻⁵)x − 5.85 × 10⁻⁷ = 0 → x = 7.5 × 10⁻⁴ mol/L
pH = −log(7.5 × 10⁻⁴) = 3.12
Comparison
If the acid behaved ideally (constant percent ionization), a tenfold dilution would give pH = 2.62 + 1.00 = 3.62. The actual rise is only 3.12 − 2.62 = 0.50 units, because percent ionization rose from 1.6% to about 5%. Weak acids do not follow the tenfold-dilution rule; only strong acids do.
3. For the 0.100 mol/L NaX solution, state whether the solution is acidic, basic, or neutral, calculate its pH, and justify the answer in terms of conjugate acid–base behaviour.
Read the worked answer
Prediction: basic
X⁻ is the conjugate base of a weak acid (Ka = 3.9 × 10⁻⁵), so it is a weak base and hydrolyses water:
X⁻ + H₂O ⇌ HX + OH⁻
Na⁺ is a spectator ion (conjugate of a strong base, no hydrolysis).
Calculate Kb
Kb = Kw/Ka = (1.0 × 10⁻¹⁴)/(3.9 × 10⁻⁵) = 2.6 × 10⁻¹⁰
Calculate [OH⁻]
Kb ≈ [OH⁻]²/[X⁻] → [OH⁻] = √((2.6 × 10⁻¹⁰)(0.100)) = √(2.6 × 10⁻¹¹) = 5.1 × 10⁻⁶ mol/L
(Check: 5.1 × 10⁻⁶/0.100 = 0.005% - approximation is excellent.)
pOH = −log(5.1 × 10⁻⁶) = 5.29
pH = 14.00 − 5.29 = 8.71
Justification
A salt of a weak acid and a strong base gives a basic solution. The value is fully consistent: Kb is small (2.6 × 10⁻¹⁰), so hydrolysis is slight and the pH is only mildly basic. Note the symmetry worth flagging to students - Ka of the acid (3.9 × 10⁻⁵) and Kb of its conjugate (2.6 × 10⁻¹⁰) multiply to Kw, so the weaker the acid, the stronger its conjugate base.