Calculate the pH of 0.14 M NaF Solution
Use this interactive chemistry calculator to determine the pH, pOH, hydroxide concentration, and hydrolysis behavior of a sodium fluoride solution. The calculator applies weak base hydrolysis of fluoride ion derived from the weak acid HF.
How to calculate the pH of 0.14 M NaF solution
To calculate the pH of 0.14 M NaF solution, you treat sodium fluoride as a salt that fully dissociates in water into Na+ and F–. Sodium ion is essentially a spectator ion because it comes from the strong base NaOH. The fluoride ion is the important species because it is the conjugate base of hydrofluoric acid, HF, which is a weak acid. Since F– can react with water to generate OH–, the solution becomes basic, meaning the pH is greater than 7.
The core equilibrium is:
F– + H2O ⇌ HF + OH–
That reaction tells you immediately that the problem is not a strong base pH problem and not a neutral salt problem. It is a weak base hydrolysis problem. The basicity of fluoride depends on the acid dissociation constant of HF. Once you know the Ka of HF, you can calculate the base dissociation constant Kb of fluoride using:
Kb = Kw / Ka
Using common 25 degrees C values:
- Ka of HF = 6.8 × 10-4
- Kw = 1.0 × 10-14
- Kb of F– = 1.0 × 10-14 / 6.8 × 10-4 ≈ 1.47 × 10-11
Now set up the base hydrolysis equilibrium for a 0.14 M fluoride concentration. If x is the amount of OH– formed, then:
- [F–] initial = 0.14
- [HF] initial = 0
- [OH–] initial ≈ 0
- [F–] change = -x
- [HF] change = +x
- [OH–] change = +x
So the equilibrium expression becomes:
Kb = x2 / (0.14 – x)
Because Kb is very small, x is much smaller than 0.14, so many chemistry students first use the approximation:
x ≈ √(KbC)
Substituting the numbers:
x ≈ √[(1.47 × 10-11)(0.14)] ≈ 1.44 × 10-6 M
That x value is the hydroxide concentration:
[OH–] ≈ 1.44 × 10-6 M
Then calculate pOH:
pOH = -log(1.44 × 10-6) ≈ 5.84
Finally:
pH = 14.00 – 5.84 = 8.16
Why NaF gives a basic solution
Students often wonder why sodium fluoride is basic while salts like NaCl are neutral. The answer lies in the parent acid and base. NaCl is made from a strong acid, HCl, and a strong base, NaOH. Neither ion significantly hydrolyzes in water, so the solution stays near neutral. NaF is different because HF is a weak acid. Its conjugate base, F–, is therefore basic enough to remove a proton from water and make OH–. That small but measurable OH– production shifts the pH above 7.
Quick classification rule
- Strong acid + strong base salt: usually neutral
- Weak acid + strong base salt: basic
- Strong acid + weak base salt: acidic
- Weak acid + weak base salt: depends on relative Ka and Kb
Because NaF comes from weak acid HF and strong base NaOH, it falls into the second category. That is why the pH must be basic, not acidic.
Step by step solution using the quadratic method
If you want the more exact solution, start from:
Kb = x2 / (C – x)
Rearrange:
x2 + Kb x – Kb C = 0
For this problem:
- C = 0.14
- Kb ≈ 1.4706 × 10-11
Then:
x = [-Kb + √(Kb2 + 4KbC)] / 2
That gives a hydroxide concentration essentially identical to the approximation, around 1.43 × 10-6 M. The pH remains about 8.16. In this case, the approximation is excellent because x is tiny relative to 0.14 M.
Comparison table: NaF concentration versus pH
The pH of sodium fluoride solution increases gradually with concentration because more fluoride ion is present to hydrolyze. The change is not linear. Because pH depends logarithmically on hydroxide concentration, the solution becomes more basic as concentration rises, but not dramatically.
| NaF Concentration (M) | Approximate [OH-] (M) | Approximate pOH | Approximate pH |
|---|---|---|---|
| 0.010 | 3.83 × 10-7 | 6.42 | 7.58 |
| 0.050 | 8.57 × 10-7 | 6.07 | 7.93 |
| 0.100 | 1.21 × 10-6 | 5.92 | 8.08 |
| 0.140 | 1.44 × 10-6 | 5.84 | 8.16 |
| 0.200 | 1.71 × 10-6 | 5.77 | 8.23 |
| 0.500 | 2.71 × 10-6 | 5.57 | 8.43 |
Important chemistry ideas behind this calculation
1. Complete dissociation of NaF
Sodium fluoride is a soluble ionic compound, so in introductory and general chemistry it is treated as dissociating completely in water:
NaF → Na+ + F–
Therefore, the initial fluoride concentration is equal to the formal NaF concentration, here 0.14 M.
2. Fluoride is a weak base
The fluoride ion is not a strong base like OH–. It reacts only slightly with water. That is why Kb is very small and the resulting pH is only mildly basic, around 8.16 rather than something extreme like 12 or 13.
3. Ka and Kb are linked
Conjugate acid-base pairs are related by:
Ka × Kb = Kw
That means if HF is stronger, F– is weaker, and vice versa. Since HF is weak but not extremely weak, F– is a weak base but not a completely negligible one.
4. The 5 percent rule
Approximation methods are often justified if x is less than 5 percent of the initial concentration. Here:
(1.44 × 10-6 / 0.14) × 100 ≈ 0.0010%
That is vastly below 5 percent, so the approximation is fully valid.
Common mistakes when solving the pH of 0.14 M NaF solution
- Treating NaF as a neutral salt. It is not neutral because F– hydrolyzes.
- Using Ka directly as if HF were present initially. The solution contains fluoride ion, so you should use Kb for F– or convert from Ka using Kb = Kw / Ka.
- Forgetting to convert pOH to pH. Once [OH–] is found, compute pOH first, then use pH = 14 – pOH at 25 degrees C.
- Assuming the solution is strongly basic. Fluoride is a weak base, so the pH is only slightly above 7.
- Mixing up initial concentration and equilibrium concentration. Initial [F–] is 0.14 M, but equilibrium [OH–] is much smaller.
Comparison table: salt type and expected solution pH
| Salt Example | Parent Acid | Parent Base | Expected Solution Type | Typical pH Trend |
|---|---|---|---|---|
| NaCl | HCl, strong | NaOH, strong | Neutral | Near 7 |
| NaF | HF, weak | NaOH, strong | Basic | Above 7 |
| NH4Cl | HCl, strong | NH3, weak | Acidic | Below 7 |
| NH4CN | HCN, weak | NH3, weak | Depends on Ka vs Kb | Variable |
Practical significance of fluoride solution chemistry
Fluoride chemistry matters in environmental science, analytical chemistry, dentistry, and industrial process control. While the concentration in this worked example is much higher than typical fluoride concentrations encountered in drinking water, the same acid-base principles apply. The behavior of fluoride in water influences speciation, buffering interactions, and compatibility with other dissolved substances. Understanding the hydrolysis of fluoride can help students make sense of why some salts shift pH and how conjugate acid-base relationships predict real solution behavior.
In more advanced systems, ionic strength, activity corrections, and temperature dependence can slightly change the numerical answer. However, for standard general chemistry calculations, using ideal solution assumptions and 25 degrees C constants gives a reliable result. For classroom work, laboratory pre-labs, homework, and exam settings, a pH of approximately 8.16 is the accepted answer for 0.14 M NaF when Ka(HF) = 6.8 × 10-4.
Short answer summary
- NaF dissociates completely to give F–.
- F– is the conjugate base of weak acid HF.
- Compute Kb using Kb = Kw / Ka.
- Use weak base hydrolysis to find [OH–].
- For 0.14 M NaF, pH ≈ 8.16 at 25 degrees C.
Authoritative chemistry references
National Institute of Standards and Technology (NIST)
Chemistry LibreTexts educational resource
U.S. Environmental Protection Agency (EPA)
For foundational chemical constants and data methods, you may also consult university and government resources such as NIST Chemistry WebBook, educational chemistry libraries from universities and colleges, and public science references from the EPA Office of Ground Water and Drinking Water.