Calculate pH of 0.68 M HF with Ka = 6.8 × 10^-4
Use this premium hydrofluoric acid calculator to find the equilibrium hydrogen ion concentration, pH, percent ionization, and compare the exact quadratic method with the weak acid approximation.
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Click Calculate pH to solve the equilibrium for 0.68 M HF.
Expert guide: how to calculate the pH of 0.68 M HF when Ka = 6.8 × 10^-4
To calculate the pH of a 0.68 M hydrofluoric acid solution, you use a weak acid equilibrium model rather than the complete dissociation approach used for strong acids. HF is unusual because it is highly reactive and dangerous, yet in water it behaves as a weak acid compared with hydrochloric acid, nitric acid, or perchloric acid. That means the concentration of H+ produced in solution is much lower than the starting analytical concentration of HF, and the equilibrium constant Ka determines how far the dissociation proceeds.
For this problem, the given data are straightforward: the initial concentration is 0.68 M and the acid dissociation constant is 6.8 × 10^-4. The target is the pH, but along the way you can also determine the equilibrium concentration of H+, the concentration of fluoride ion, the amount of undissociated HF remaining, and the percent ionization. These are the quantities that matter in general chemistry, analytical chemistry, and introductory equilibrium calculations.
Step 1: Write the balanced dissociation reaction
Hydrofluoric acid dissociates in water according to the equilibrium:
HF ⇌ H+ + F-
Because this is a weak acid, you should not assume that all 0.68 moles per liter of HF become H+. Instead, only a fraction ionizes. The Ka value tells you how extensive that ionization is at equilibrium.
Step 2: Build the ICE framework
The standard setup uses an ICE table, meaning Initial, Change, and Equilibrium. If the initial HF concentration is 0.68 M and there is effectively no added H+ or F- from other sources, then:
- Initial [HF] = 0.68
- Initial [H+] = 0
- Initial [F-] = 0
Let x be the amount of HF that dissociates.
- Change in [HF] = -x
- Change in [H+] = +x
- Change in [F-] = +x
At equilibrium:
- [HF] = 0.68 – x
- [H+] = x
- [F-] = x
Step 3: Substitute into the Ka expression
The acid dissociation expression is:
Ka = [H+][F-] / [HF]
Substitute the equilibrium terms:
6.8 × 10^-4 = x^2 / (0.68 – x)
This is the heart of the calculation. At this point, you can either solve the equation exactly using the quadratic formula or use the weak acid approximation if x is small compared with 0.68.
Step 4: Solve exactly using the quadratic formula
Rearrange the equation:
x^2 = (6.8 × 10^-4)(0.68 – x)
x^2 = 4.624 × 10^-4 – 6.8 × 10^-4 x
x^2 + 6.8 × 10^-4 x – 4.624 × 10^-4 = 0
Using the quadratic formula, the physically meaningful solution is:
x = (-6.8 × 10^-4 + √((6.8 × 10^-4)^2 + 4(6.8 × 10^-4)(0.68))) / 2
This gives:
- [H+] = x ≈ 0.02117 M
- [F-] ≈ 0.02117 M
- [HF]eq ≈ 0.65883 M
Then compute pH:
pH = -log10(0.02117) ≈ 1.67
Rounded appropriately, the pH of 0.68 M HF with Ka = 6.8 × 10^-4 is 1.67.
Step 5: Check the approximation method
For a weak acid, many students use the shortcut:
x ≈ √(KaC)
Substituting the numbers:
x ≈ √((6.8 × 10^-4)(0.68)) = √(4.624 × 10^-4) ≈ 0.02150 M
This leads to:
pH ≈ -log10(0.02150) ≈ 1.67
The approximation is very close to the exact answer. The reason it works well is that the dissociated amount is small relative to the initial concentration. Specifically, x / 0.68 is only a few percent, so subtracting x from 0.68 only changes the denominator slightly.
| Method | [H+] (M) | Calculated pH | Difference from exact |
|---|---|---|---|
| Exact quadratic | 0.02117 | 1.674 | 0.000 |
| Weak acid approximation | 0.02150 | 1.667 | 0.007 pH units |
Percent ionization of 0.68 M HF
Percent ionization is a useful diagnostic because it shows how much of the acid actually dissociates:
Percent ionization = ([H+]eq / initial concentration) × 100
Percent ionization = (0.02117 / 0.68) × 100 ≈ 3.11%
That means only about 3.11% of the HF molecules ionize under these conditions. This is exactly why HF must be treated as a weak acid in equilibrium calculations even though it is very hazardous in practical handling and toxicology.
Why HF is called weak even though it is dangerous
In chemistry, the terms weak and strong describe the degree of ionization in water, not the real-world hazard of the substance. Hydrofluoric acid is a classic example where chemical hazard and acid strength are not the same thing. HF has a lower Ka than strong mineral acids, so it does not dissociate completely in water. However, it is extremely dangerous because fluoride ions penetrate tissue and can bind calcium and magnesium in the body. So from an equilibrium standpoint HF is weak, but from a safety standpoint it is one of the most serious acids encountered in laboratory and industrial settings.
Common mistakes when solving this exact problem
- Treating HF as a strong acid. If you assume complete dissociation, you would wrongly set [H+] = 0.68 M and get pH around 0.17, which is far too low.
- Using pKa incorrectly. A pKa of about 3.17 corresponds to Ka = 6.8 × 10^-4. If you confuse pKa and pH, the final answer will be incorrect.
- Dropping the x without checking. The approximation works here, but you should still compare the exact and approximate values when accuracy matters.
- Forgetting scientific notation. Ka = 6.8 × 10^-4 is not 6.8 or 0.68. A small exponent error changes the answer dramatically.
Comparison with other common weak acids
It helps to see hydrofluoric acid in context. The table below compares accepted approximate acid dissociation constants for several common weak acids at room temperature. Lower pKa means a stronger acid among weak acids.
| Acid | Formula | Ka | pKa | Relative strength note |
|---|---|---|---|---|
| Hydrofluoric acid | HF | 6.8 × 10^-4 | 3.17 | Stronger than acetic acid, still weak vs strong acids |
| Formic acid | HCOOH | 1.8 × 10^-4 | 3.75 | Weaker than HF |
| Acetic acid | CH3COOH | 1.8 × 10^-5 | 4.76 | Much weaker than HF |
| Hypochlorous acid | HOCl | 3.0 × 10^-8 | 7.52 | Far weaker than HF |
How concentration affects the pH of HF
For weak acids, the pH does not decrease linearly with concentration because equilibrium shifts as the solution becomes more or less concentrated. As concentration rises, the hydrogen ion concentration increases, but percent ionization usually decreases. That is why a 0.68 M HF solution is substantially acidic while still only a few percent ionized. If you dilute the solution heavily, the percent ionization goes up even though the absolute hydrogen ion concentration goes down.
This behavior is important in laboratory calculations, industrial cleaning formulations, and environmental chemistry. Researchers and students often compare the exact equilibrium approach with the square root shortcut to decide whether the simplification is acceptable. In this problem, the shortcut gives a pH within about 0.007 units of the exact answer, which is quite good for many classroom settings.
When should you use the exact quadratic method?
- When your instructor specifically asks for the exact pH
- When precision matters in analytical work
- When the 5% rule is borderline or questionable
- When comparing multiple weak acid systems and you need consistent treatment
In professional work, exact equilibrium calculations are often preferred because computers can solve them instantly and remove uncertainty about whether the approximation is acceptable.
Safety and authoritative references
If you are studying HF beyond the math, review high-quality safety and pH resources. The following sources are especially useful:
- CDC NIOSH hydrofluoric acid safety guidance
- USGS explanation of pH and water chemistry
- MIT OpenCourseWare chemistry resources
Final answer
For a solution of 0.68 M HF with Ka = 6.8 × 10^-4, the equilibrium hydrogen ion concentration is about 0.02117 M, and the resulting pH is 1.67. The percent ionization is approximately 3.11%. If you use the weak acid approximation, you get nearly the same result, but the exact quadratic method is the best way to present the final answer when accuracy is important.
If you want to explore related problems, you can use this calculator to change the concentration or Ka value and instantly compare how the equilibrium concentrations and pH shift. That makes it ideal for homework practice, exam review, and quick chemistry checks.