Calculate pH for OH 1.0 × 10-5 M
Use this premium calculator to convert hydroxide ion concentration into pOH and pH. The default example is [OH–] = 1.0 × 10-5 M, which gives a basic solution at standard room temperature.
Enter or keep the default value 1.0 × 10-5 M and click the button to see the pOH, pH, interpretation, and chart.
How to calculate pH for OH 1.0 × 10-5 M
When you are asked to calculate pH for OH 1.0 × 10-5 M, you are starting from the hydroxide ion concentration, written as [OH–] = 1.0 × 10-5 mol/L. This is a classic introductory chemistry problem because it shows how pOH and pH are related through logarithms and through the ion product of water. At 25 C, the water equilibrium constant is commonly expressed as Kw = 1.0 × 10-14, which leads to the familiar relation pH + pOH = 14.
The calculation is short, but understanding why it works is valuable. Hydroxide concentration tells you how basic a solution is. A larger [OH–] means a more basic solution, a smaller pOH, and a larger pH. Because pH and pOH are logarithmic quantities, every factor of 10 change in hydroxide concentration changes pOH by 1 unit. That logarithmic structure is the reason scientific notation is so common in acid-base chemistry.
Step by step solution
- Write the given hydroxide concentration: [OH–] = 1.0 × 10-5 M.
- Use the pOH formula: pOH = -log10[OH–].
- Substitute the value: pOH = -log10(1.0 × 10-5).
- Evaluate the logarithm. Since log10(1.0 × 10-5) = -5, pOH = 5.
- At 25 C, use pH = 14 – pOH.
- Therefore pH = 14 – 5 = 9.
Why the answer is basic
A neutral aqueous solution at 25 C has [H+] = 1.0 × 10-7 M and [OH–] = 1.0 × 10-7 M, corresponding to pH 7 and pOH 7. In this problem, the hydroxide concentration is 100 times larger than neutral water because 10-5 is two powers of ten larger than 10-7. That means the solution is definitely basic. The pOH drops from 7 to 5, and the pH rises from 7 to 9. This is a useful mental check before you even complete the arithmetic.
Important formula relationships
- pOH = -log10[OH–]
- pH = -log10[H+]
- Kw = [H+][OH–]
- pKw = pH + pOH
- At 25 C, pKw = 14.00, so pH + pOH = 14.00
Students often memorize pH + pOH = 14, but the more complete statement is pH + pOH = pKw. The value 14 applies at 25 C. As temperature changes, Kw changes, so pKw changes too. That is why the calculator above gives you a temperature choice.
Comparison table: hydroxide concentration and resulting pH at 25 C
| [OH–] in M | pOH | pH at 25 C | Interpretation |
|---|---|---|---|
| 1.0 × 10-7 | 7.00 | 7.00 | Neutral water benchmark |
| 1.0 × 10-6 | 6.00 | 8.00 | Mildly basic |
| 1.0 × 10-5 | 5.00 | 9.00 | Clearly basic |
| 1.0 × 10-4 | 4.00 | 10.00 | More strongly basic |
| 1.0 × 10-3 | 3.00 | 11.00 | Strongly basic region |
Temperature matters more than many learners expect
One subtle point in pH calculations is the role of temperature. In many classroom examples, 25 C is assumed automatically, so pH + pOH = 14 is used without discussion. In more rigorous chemistry, however, the autoionization of water changes with temperature. As temperature rises, Kw increases and pKw decreases. That means the neutral point shifts numerically even though the solution is still neutral in the sense that [H+] equals [OH–].
For your specific problem, if the same [OH–] = 1.0 × 10-5 M were considered at 50 C rather than 25 C, pOH would still be 5 because that part depends only on the hydroxide concentration. But pH would be pKw – 5, not 14 – 5. If pKw is about 13.26 at 50 C, then the pH would be about 8.26 instead of 9.00. The solution is still basic relative to neutrality at that temperature, but the numerical pH is lower than the room-temperature result.
Comparison table: approximate pKw and neutral pH by temperature
| Temperature | Approximate pKw | Neutral pH | pH for [OH–] = 1.0 × 10-5 M |
|---|---|---|---|
| 0 C | 14.94 | 7.47 | 9.94 |
| 25 C | 14.00 | 7.00 | 9.00 |
| 40 C | 13.60 | 6.80 | 8.60 |
| 50 C | 13.26 | 6.63 | 8.26 |
How to do the logarithm quickly
For values written as a coefficient times a power of ten, you can often estimate pOH mentally. Use this identity:
log(a × 10b) = log(a) + b
So if [OH–] = 1.0 × 10-5, then log(1.0) = 0 and the exponent contributes -5, giving log = -5. Taking the negative of that gives pOH = 5. If the coefficient were not exactly 1.0, such as 3.2 × 10-5, then pOH would be -[log(3.2) – 5] ≈ 4.49. This is why the coefficient matters whenever it is not 1.
Common mistakes to avoid
- Confusing pH with pOH. If the problem gives [OH–], calculate pOH first, then convert to pH.
- Dropping the negative sign. Since concentrations less than 1 have negative logarithms, the leading negative in the definition is essential.
- Assuming 14 at all temperatures. Use pKw when the temperature differs significantly from 25 C.
- Misreading scientific notation. 1.0 × 10-5 is not the same as 105; the sign of the exponent changes everything.
- Ignoring reasonableness. A hydroxide concentration larger than neutral water should produce a pH above 7 at 25 C.
What this means chemically
A pH of 9 at 25 C indicates a basic solution, but not an extremely strong base. In environmental and biological contexts, pH changes of even 1 unit can be significant because the pH scale is logarithmic. A shift from pH 7 to pH 9 means a 100-fold change in hydrogen ion concentration relative to neutral conditions. This is why pH is such a powerful summary metric in chemistry, water quality analysis, and laboratory work.
For example, natural waters often occupy a much narrower pH range than the full 0 to 14 scale. Regulatory and environmental discussions commonly focus on whether water lies within a range compatible with aquatic life, corrosion control, treatment efficiency, or analytical stability. Even if your current problem is purely academic, it directly connects to practical measurements used in municipal water treatment, environmental monitoring, and industrial process control.
Worked example in words
Suppose your instructor asks, “Calculate the pH of a solution for which [OH–] = 1.0 × 10-5 M.” You would respond as follows. First, compute pOH from the hydroxide concentration. Since pOH = -log[OH–] and log(1.0 × 10-5) = -5, the pOH is 5. Then, assuming the standard 25 C condition used in general chemistry unless otherwise stated, use pH = 14 – 5 = 9. The solution is therefore basic.
Useful mental checks for exam settings
- Neutral water at 25 C has [OH–] = 10-7 M.
- Your value, 10-5 M, is 100 times larger than neutral water.
- That means the solution must be basic.
- If [OH–] is 10-5, then pOH should be around 5.
- If pOH is 5 at 25 C, pH must be around 9.
These checks make it easier to catch calculator errors. If you ever obtain a pH less than 7 from this starting value at 25 C, something has gone wrong.
Authority sources for deeper reading
If you want official or university-backed context on pH, water chemistry, and acid-base fundamentals, these sources are helpful:
- USGS explanation of pH in water systems
- U.S. EPA discussion of pH and aquatic systems
- Purdue University review of pH calculations and acid-base concepts
Final takeaway
To calculate pH for OH 1.0 × 10-5 M, first compute pOH using the negative logarithm of hydroxide concentration. That gives pOH = 5. Then convert pOH to pH using pH = 14 – pOH at 25 C, giving pH = 9. The solution is basic. If temperature is not 25 C, replace 14 with the appropriate pKw. The calculator on this page automates all of that while also showing the concentration, pOH, pH, and a visual comparison chart.