Calculate the pH of a 0.0065 M Solution of NaOH
Use this interactive chemistry calculator to determine hydroxide concentration, pOH, and pH for sodium hydroxide solutions. The default setup solves the exact problem: the pH of a 0.0065 M NaOH solution.
Calculator
This calculator uses the strong-base assumption: NaOH dissociates completely, so [OH-] equals the NaOH concentration for a dilute solution.
How to Calculate the pH of a 0.0065 M Solution of NaOH
To calculate the pH of a 0.0065 M solution of sodium hydroxide, you use the fact that NaOH is a strong base. In introductory and most general chemistry contexts, strong bases are treated as substances that dissociate completely in water. That means every dissolved formula unit of NaOH separates into sodium ions, Na+, and hydroxide ions, OH-. Since sodium hydroxide contributes one hydroxide ion per formula unit, the hydroxide concentration is equal to the stated concentration of the base.
For a 0.0065 M NaOH solution, the first conclusion is straightforward: [OH-] = 0.0065 M. Once you know the hydroxide concentration, you calculate pOH using the common logarithm relationship pOH = -log10[OH-]. Then, assuming standard conditions at 25 C where pH + pOH = 14.00, you can determine pH from pH = 14.00 – pOH.
Final answer: For 0.0065 M NaOH, pOH = 2.19 and pH = 11.81 at 25 C.
Step-by-Step Solution
1. Write the dissociation equation
Sodium hydroxide dissociates in water as follows:
NaOH(aq) → Na+(aq) + OH-(aq)
Because NaOH is a strong base, this reaction is treated as going essentially to completion in a dilute aqueous solution.
2. Determine hydroxide concentration
The molarity of the NaOH solution is 0.0065 M. Since each mole of NaOH releases one mole of OH-, the hydroxide ion concentration is the same:
[OH-] = 0.0065 M
3. Compute pOH
Use the definition:
pOH = -log10[OH-]
Substitute 0.0065:
pOH = -log10(0.0065)
pOH ≈ 2.19
4. Compute pH
At 25 C, the relationship between pH and pOH is:
pH + pOH = 14.00
So:
pH = 14.00 – 2.19 = 11.81
Why NaOH Is Treated as a Strong Base
Sodium hydroxide belongs to the category of strong Arrhenius bases. In water, it dissociates nearly completely, unlike weak bases such as ammonia, which establish an equilibrium and require a base dissociation constant, Kb, for accurate calculation. The strong-base assumption makes NaOH problems much simpler because you do not need to solve an ICE table or equilibrium expression in normal dilute concentration ranges.
That matters because many students confuse the process for strong and weak bases. If the solute were NH3 instead of NaOH, you would not simply say [OH-] equals the formal concentration. But for NaOH, KOH, and other strong metal hydroxides commonly encountered in chemistry classes, complete dissociation is usually the correct starting point.
What the Result Means
A pH of 11.81 indicates a distinctly basic solution. Neutral water at 25 C has a pH of 7.00. Every one-unit change in pH corresponds to a tenfold change in hydrogen ion concentration, so a solution with pH 11.81 is far more basic than neutral water and strongly favors hydroxide ions over hydronium ions. In practical terms, a 0.0065 M NaOH solution is dilute compared with many stock laboratory bases, but it is still corrosive enough to require appropriate handling and eye protection in educational or laboratory settings.
Because pH is logarithmic, students sometimes expect a concentration under 0.01 M to produce a pH only slightly above 7. In reality, even millimolar concentrations of a strong base can push pH well into the basic region. Here, 0.0065 M corresponds to 6.5 × 10^-3 moles of hydroxide per liter, which is more than enough to create a pH above 11.
Common Mistakes When Solving This Problem
- Using pH = -log[OH-] instead of pOH = -log[OH-]. The negative log of hydroxide concentration gives pOH, not pH.
- Forgetting the final subtraction from 14 at 25 C. After finding pOH, you must use pH = 14 – pOH.
- Assuming [H+] = 0.0065 M. That would apply only if the solute were a strong acid with one acidic proton, not a strong base.
- Confusing molarity with a decimal power. 0.0065 M is not 6.5 M; it is 6.5 × 10^-3 M.
- Rounding too early. If you round intermediate values aggressively, your final pH can shift by a few hundredths.
Quick Reference Table for Similar NaOH Concentrations
The following table shows how pOH and pH change for several representative NaOH concentrations at 25 C. These values illustrate the logarithmic nature of the pH scale and help you compare 0.0065 M with nearby concentrations often used in homework and laboratory examples.
| NaOH Concentration (M) | [OH-] (M) | pOH | pH at 25 C |
|---|---|---|---|
| 0.0010 | 0.0010 | 3.00 | 11.00 |
| 0.0050 | 0.0050 | 2.30 | 11.70 |
| 0.0065 | 0.0065 | 2.19 | 11.81 |
| 0.0100 | 0.0100 | 2.00 | 12.00 |
| 0.1000 | 0.1000 | 1.00 | 13.00 |
Molarity Versus Molality in This Problem
You may see the concentration written as either 0.0065 M or 0.0065 m. These units are not identical. Molarity, M, means moles of solute per liter of solution. Molality, m, means moles of solute per kilogram of solvent. In strict physical chemistry, these are different concentration measures, and exact conversion depends on solution density. However, for very dilute aqueous solutions, the numerical values of molarity and molality are often close enough that introductory chemistry problems may use them almost interchangeably as an approximation.
If your assignment specifically says 0.0065 m NaOH, and no density data are given, many instructors expect you to treat it like a dilute aqueous solution and approximate the hydroxide concentration as 0.0065. Under that assumption, the calculated pH remains about 11.81. In advanced work, especially outside the dilute limit, you would need activity corrections and possibly a molality-to-molarity conversion for more precise results.
Relationship Between pH, pOH, and Kw
The familiar identity pH + pOH = 14.00 comes from the ion product of water, Kw = 1.0 × 10^-14 at 25 C. Since Kw = [H+][OH-], taking the negative logarithm of both sides leads to the common pH plus pOH expression. This identity is one of the most useful tools in acid-base chemistry because it lets you work from either hydrogen ion or hydroxide ion concentration.
It is worth noting that Kw changes with temperature. That means the sum pH + pOH is exactly 14.00 only at 25 C. In many educational settings, unless the problem explicitly specifies a different temperature and asks for temperature-dependent water ionization, 25 C is assumed.
| Chemistry Quantity | Symbol | Value Used Here | Purpose in the Calculation |
|---|---|---|---|
| Sodium hydroxide concentration | C or [NaOH] | 0.0065 M | Equals [OH-] for a strong base with one OH- per formula unit |
| Hydroxide ion concentration | [OH-] | 0.0065 M | Input to the pOH formula |
| Hydroxide potential | pOH | 2.19 | Found from -log10[OH-] |
| Hydrogen potential | pH | 11.81 | Found from 14.00 – pOH at 25 C |
| Ion product of water | Kw | 1.0 × 10^-14 | Basis for pH + pOH = 14.00 at 25 C |
How This Compares with Strong Acids and Weak Bases
If you compare 0.0065 M NaOH with a 0.0065 M strong acid such as HCl, the math is nearly symmetrical. For HCl, complete dissociation gives [H+] = 0.0065 M, so pH = -log10(0.0065) = 2.19. For NaOH, complete dissociation gives [OH-] = 0.0065 M, so pOH = 2.19 and pH = 11.81. The values are reflections across neutral pH 7 when the solutions are calculated at 25 C.
By contrast, if the solute were a weak base such as NH3 at 0.0065 M, the hydroxide concentration would be much lower than 0.0065 M because only a fraction of dissolved ammonia reacts with water to form OH-. That would result in a lower pH than the NaOH case.
Practical Laboratory Context
In laboratory work, sodium hydroxide solutions are widely used for titrations, pH adjustment, cleaning procedures, and synthesis. Even dilute NaOH should be handled with care because strong bases can damage tissue and certain materials. Typical safe practice includes goggles, gloves, and immediate rinsing of affected surfaces or skin with water according to institutional safety procedures.
At 0.0065 M, the concentration is much lower than common 0.1 M or 1.0 M standard base solutions, but it is still chemically significant. The pH of 11.81 confirms that it is far from neutral and should never be treated as harmless water.
Authoritative Resources for Further Reading
- U.S. Environmental Protection Agency: pH Overview
- Chemistry LibreTexts Educational Resources
- CDC NIOSH Sodium Hydroxide Safety Information
Final Takeaway
To calculate the pH of a 0.0065 M solution of NaOH, treat sodium hydroxide as a strong base that dissociates completely. Set the hydroxide concentration equal to the sodium hydroxide concentration, calculate pOH with the negative logarithm, and then subtract from 14.00 at 25 C. The result is:
- [OH-] = 0.0065 M
- pOH = -log10(0.0065) = 2.19
- pH = 14.00 – 2.19 = 11.81
If you remember one rule from this problem, let it be this: for a strong base like NaOH, the hydroxide concentration comes directly from the stated concentration. That shortcut makes pH calculations fast, reliable, and easy to check.