Calculate the pH of 0.005 M NaOH Solution
Use this premium calculator to find pOH, pH, hydroxide concentration, and temperature-adjusted values for a sodium hydroxide solution. The default example is 0.005 M NaOH, a common strong-base pH calculation used in chemistry classes, lab prep, and exam practice.
NaOH pH Calculator
For sodium hydroxide, we assume complete dissociation in dilute aqueous solution. That means the hydroxide concentration is approximately equal to the NaOH molarity.
How to calculate the pH of 0.005 M NaOH solution
To calculate the pH of a 0.005 M sodium hydroxide solution, you use one of the most direct strong-base methods in general chemistry. Sodium hydroxide, written as NaOH, dissociates essentially completely in water into sodium ions and hydroxide ions. Because of that full dissociation, the hydroxide ion concentration is approximately equal to the original NaOH concentration. In this case, a 0.005 M NaOH solution gives an OH concentration of 0.005 M under standard classroom assumptions.
Once you know the hydroxide concentration, the next step is to calculate pOH using the formula pOH = -log[OH]. Then, at 25 C, you use the relationship pH + pOH = 14. For 0.005 M NaOH, the arithmetic is straightforward:
- Write the concentration of hydroxide ions: [OH] = 0.005 M
- Calculate pOH = -log(0.005)
- pOH = 2.301
- Calculate pH = 14.000 – 2.301
- pH = 11.699
So, the pH of 0.005 M NaOH solution at 25 C is approximately 11.70. This is strongly basic, as expected for a sodium hydroxide solution. The answer is above 7 because NaOH increases hydroxide ion concentration in water, shifting the acid-base balance toward basicity.
Why NaOH is treated as a strong base
In introductory chemistry and many practical lab contexts, sodium hydroxide is categorized as a strong base. That classification means it dissociates nearly 100 percent in dilute water solution:
NaOH(aq) → Na+(aq) + OH-(aq)
This is important because not every base behaves this way. Weak bases such as ammonia do not fully ionize, so you would need an equilibrium expression and a base dissociation constant. For NaOH, however, you usually skip equilibrium setup and move directly to hydroxide concentration. That saves time and reduces the chance of error.
Detailed explanation of the formula
The pOH formula is based on the common logarithm of hydroxide concentration:
- pOH = -log[OH]
- pH + pOH = 14 at 25 C
- pH = 14 – pOH
For 0.005 M NaOH, note that 0.005 can also be written in scientific notation as 5.0 × 10-3. This makes the logarithm easier to understand conceptually:
-log(5.0 × 10-3) = 3 – log(5.0) = 3 – 0.699 = 2.301
Then:
pH = 14.000 – 2.301 = 11.699
Rounded to two decimal places, the pH is 11.70. Rounded to three decimal places, the pH is 11.699.
Common student mistakes when solving this problem
Even though this is a classic chemistry problem, students often make a few predictable errors. If you avoid these, you will almost always get the correct answer.
- Using pH directly from NaOH concentration. The concentration gives you hydroxide ion concentration, so you must calculate pOH first, not pH.
- Forgetting that NaOH is a base. Bases produce OH, not H.
- Dropping a zero incorrectly. 0.005 is not the same as 0.05. A factor of ten changes pOH by 1 unit and changes pH by 1 unit.
- Assuming pH + pOH = 14 at all temperatures. That rule is exact at 25 C in standard textbook treatment, but pKw changes with temperature.
- Mixing M and mM. 0.005 M is 5 mM. If you enter 0.005 mM by mistake, the result will be very different.
Comparison table for common NaOH concentrations at 25 C
The table below shows how pOH and pH change for several common sodium hydroxide concentrations. This helps you compare the target problem, 0.005 M NaOH, to nearby examples used in homework and laboratory calculations.
| NaOH Concentration | [OH-] Assumed | pOH at 25 C | pH at 25 C | Basicity Level |
|---|---|---|---|---|
| 0.0001 M | 0.0001 M | 4.000 | 10.000 | Moderately basic |
| 0.001 M | 0.001 M | 3.000 | 11.000 | Clearly basic |
| 0.005 M | 0.005 M | 2.301 | 11.699 | Strongly basic |
| 0.01 M | 0.01 M | 2.000 | 12.000 | Strongly basic |
| 0.1 M | 0.1 M | 1.000 | 13.000 | Very strongly basic |
How temperature affects the result
Most classroom problems silently assume 25 C, where the ionic product of water leads to pKw = 14.000. In more advanced chemistry, this value changes with temperature. As temperature rises, the ionization of water changes, which means pH + pOH is not always exactly 14. For practical work, this can matter if you are doing analytical chemistry, environmental sampling, or process chemistry where temperature differs significantly from room temperature.
This calculator estimates pKw from 0 C to 60 C and adjusts the pH result accordingly. For the specific case of 0.005 M NaOH, the hydroxide concentration remains the same, so pOH from concentration does not change much in the idealized model, but pH shifts because pKw shifts.
| Temperature | Approximate pKw | pOH for 0.005 M NaOH | Calculated pH | Interpretation |
|---|---|---|---|---|
| 0 C | 14.94 | 2.301 | 12.639 | Higher pH because pKw is larger |
| 25 C | 14.00 | 2.301 | 11.699 | Standard textbook answer |
| 40 C | 13.53 | 2.301 | 11.229 | Lower pH at higher temperature |
| 60 C | 13.02 | 2.301 | 10.719 | Still basic, but lower numerical pH |
Step by step worked example for 0.005 M NaOH
Here is the full worked method in the exact order many instructors prefer:
- Identify the solute as sodium hydroxide, a strong base.
- Write the dissociation reaction: NaOH → Na+ + OH-.
- Set hydroxide concentration equal to NaOH concentration: [OH-] = 0.005 M.
- Use the pOH formula: pOH = -log(0.005).
- Evaluate the logarithm to get pOH = 2.301.
- Apply pH + pOH = 14 at 25 C.
- Compute pH = 14 – 2.301 = 11.699.
- Round based on the requested precision, usually 11.70.
If your assignment asks for significant figures, think carefully about the concentration. The number 0.005 often implies one significant figure, but in many textbook contexts instructors still accept pH = 11.70 because logarithmic reporting follows decimal-place conventions. Always match your course rule.
Why 0.005 M NaOH is a useful teaching example
This concentration is useful because it sits between simple powers of ten. If the problem were 0.001 M NaOH, the pOH would be exactly 3 and the pH would be exactly 11 at 25 C. If the problem were 0.01 M NaOH, the pOH would be exactly 2 and the pH would be exactly 12. By choosing 0.005 M, the problem forces you to actually use logarithms and understand the relationship between concentration and pH rather than memorizing a shortcut.
It also illustrates an important fact about logarithmic scales: concentration changes are not linear in pH. Doubling the concentration does not increase pH by a fixed simple amount. Instead, each tenfold concentration change shifts pOH by 1 unit. That logarithmic behavior is essential in acid-base chemistry, analytical chemistry, biochemistry, and environmental chemistry.
Real-world context for sodium hydroxide solutions
Sodium hydroxide is one of the most important industrial and laboratory bases. It is used in titrations, soap production, cleaning formulations, paper processing, and chemical manufacturing. In the lab, even relatively dilute NaOH solutions can be hazardous because they are caustic and can damage skin or eyes. A 0.005 M solution is much less concentrated than stock laboratory base, but it is still basic enough to affect indicators, biological materials, and pH-sensitive systems.
Because sodium hydroxide is so common, understanding how to compute its pH is foundational. Once you know how to handle 0.005 M NaOH, you can usually solve any standard strong-base pH problem by following the same method.
Authoritative references for water chemistry and pH concepts
If you want to verify the chemistry concepts behind this calculator, these authoritative educational and government sources are useful:
- U.S. Environmental Protection Agency: pH basics and water chemistry
- LibreTexts Chemistry educational resources used by universities
- U.S. Geological Survey: pH and water science
Fast answer summary
If you only need the final result, here it is:
- Given: 0.005 M NaOH
- Since NaOH is a strong base: [OH-] = 0.005 M
- pOH = -log(0.005) = 2.301
- pH = 14.000 – 2.301 = 11.699
- Final pH at 25 C: 11.70