Calculate the pH of 6.4 × 10-8
Use this premium pH calculator to solve hydrogen ion or hydroxide ion concentration problems in scientific notation. The default example is set to 6.4 × 10-8 M, which is the exact expression most students search for.
How to calculate the pH of 6.4 × 10-8
If you want to calculate the pH of 6.4 × 10-8, the key question is what this number represents. In standard chemistry notation, pH is calculated from the hydrogen ion concentration, written as [H+]. If the value 6.4 × 10-8 is the hydrogen ion concentration in moles per liter, then the textbook formula is straightforward:
pH = -log10[H+]
Substituting the value gives:
pH = -log10(6.4 × 10-8)
Using logarithm rules:
- log10(6.4 × 10-8) = log10(6.4) + log10(10-8)
- log10(6.4) ≈ 0.80618
- log10(10-8) = -8
- So log10(6.4 × 10-8) ≈ 0.80618 – 8 = -7.19382
- Apply the negative sign: pH ≈ 7.19382
Rounded to two decimal places, the pH is 7.19.
Why this answer surprises many students
A lot of learners assume that any number written with a negative exponent must produce an acidic pH, but that is not how the scale works. The pH scale is logarithmic. Pure water at 25 C has [H+] = 1.0 × 10-7 M and a pH of 7.00. Since 6.4 × 10-8 is smaller than 1.0 × 10-7, it means the solution has slightly less hydrogen ion than neutral water. Less hydrogen ion means a slightly higher pH. That is why the answer comes out to 7.19, which is slightly basic.
Quick intuition check
- If [H+] = 1 × 10-7, then pH = 7.00
- If [H+] is larger than 1 × 10-7, pH drops below 7 and becomes acidic
- If [H+] is smaller than 1 × 10-7, pH rises above 7 and becomes basic
- Since 6.4 × 10-8 is less than 1 × 10-7, the pH must be greater than 7
Step by step method you can use on any similar problem
Method 1: Direct logarithm method
- Identify whether the concentration is [H+] or [OH-]
- If it is [H+], use pH = -log10[H+]
- Rewrite the number in scientific notation if needed
- Take the base-10 logarithm
- Apply the negative sign and round properly
Method 2: Break apart the scientific notation
This method is especially useful when doing chemistry homework by hand.
- Write the concentration as a coefficient times a power of ten
- Use log(ab) = log(a) + log(b)
- Use log(10n) = n
- Combine the terms
- Take the negative of the final logarithm
For 6.4 × 10-8:
- log(6.4) ≈ 0.80618
- log(10-8) = -8
- Total log = -7.19382
- pH = 7.19382
What if 6.4 × 10-8 were [OH-] instead?
If the given concentration is hydroxide ion concentration instead of hydrogen ion concentration, the workflow changes:
- Calculate pOH = -log10[OH-]
- At 25 C, use pH + pOH = 14.00
- Solve for pH = 14.00 – pOH
So if [OH-] = 6.4 × 10-8 M:
- pOH = 7.19
- pH = 14.00 – 7.19 = 6.81
This is why it is essential to identify what the concentration refers to before you calculate anything. The same numerical concentration gives a different final pH depending on whether it is [H+] or [OH-].
Comparison table: common concentrations and their pH values
| Hydrogen ion concentration [H+] in M | Calculated pH | Interpretation |
|---|---|---|
| 1.0 × 10-1 | 1.00 | Strongly acidic |
| 1.0 × 10-3 | 3.00 | Acidic |
| 1.0 × 10-7 | 7.00 | Neutral at 25 C |
| 6.4 × 10-8 | 7.19 | Slightly basic |
| 1.0 × 10-8 | 8.00 | Basic |
| 1.0 × 10-10 | 10.00 | Moderately basic |
Important advanced note: water autoionization
For introductory chemistry, pH is usually found by directly applying pH = -log[H+]. That gives the expected answer of 7.19 for 6.4 × 10-8 M H+. However, advanced chemistry points out an important subtlety. Pure water already contributes hydrogen ions and hydroxide ions through autoionization. At 25 C, the ion product of water is:
Kw = [H+][OH-] = 1.0 × 10-14
When an acid concentration is extremely small, especially on the order of 10-8 M, the natural hydrogen ion concentration from water itself can no longer be ignored. In that case, a more rigorous equilibrium treatment may be needed. This is why some advanced solutions do not simply stop at 7.19 when discussing very dilute acid systems.
If 6.4 × 10-8 M represented a strong acid dissolved in water, a more complete treatment would include water equilibrium. That rigorous value comes out closer to about pH 6.86, not 7.19. Both values appear in chemistry discussions, but they answer slightly different questions:
- 7.19 is the direct textbook result when 6.4 × 10-8 M is taken as the actual [H+]
- 6.86 is the corrected result for a very dilute strong acid solution when water autoionization is included
In most homework and exam contexts that literally ask for the pH of 6.4 × 10-8, the intended answer is usually 7.19 unless the instructor explicitly says to consider water autoionization.
Comparison table: standard classroom result vs rigorous equilibrium treatment
| Scenario | Starting value | Method used | Result |
|---|---|---|---|
| Given [H+] directly | 6.4 × 10-8 M | pH = -log[H+] | 7.19 |
| Very dilute strong acid in water | C = 6.4 × 10-8 M | Include Kw = 1.0 × 10-14 | Approximately 6.86 |
| Given [OH-] directly | 6.4 × 10-8 M | pOH first, then pH = 14 – pOH | 6.81 |
Why logarithms matter so much in pH problems
The pH scale compresses an enormous concentration range into manageable numbers. Hydrogen ion concentrations in chemistry often vary from about 1 M to 1 × 10-14 M or even more extreme values in some contexts. A logarithmic scale lets chemists compare these concentrations efficiently. Every decrease of one pH unit corresponds to a tenfold increase in hydrogen ion concentration. Likewise, every increase of one pH unit means the hydrogen ion concentration is ten times smaller.
For example, a solution at pH 6 has ten times the [H+] of a solution at pH 7. A solution at pH 5 has one hundred times the [H+] of a solution at pH 7. This is one reason students must be careful not to think of pH as a linear scale. A change from 7.19 to 6.19 is not a tiny shift. It means hydrogen ion concentration has increased by a factor of 10.
Common mistakes when calculating the pH of 6.4 × 10-8
- Dropping the negative sign. Since pH = -log[H+], you must apply the negative sign after evaluating the logarithm.
- Confusing [H+] with [OH-]. The formulas are different, so always read the problem carefully.
- Misreading scientific notation. 6.4 × 10-8 is not the same as 6.4 × 108.
- Rounding too early. Keep extra digits through the intermediate log calculation, then round the final pH.
- Ignoring advanced context. In highly dilute acid solutions, water autoionization may matter.
Best practice for homework, exams, and lab work
The safest way to solve this type of question is to start by writing what is given and what is being asked. If your problem literally states that [H+] = 6.4 × 10-8 M, then pH = 7.19 is the clean answer. If the problem says a strong acid has concentration 6.4 × 10-8 M in water and asks for a rigorous pH, you should consider water equilibrium. If it gives [OH-] instead, compute pOH first. The wording matters.
A simple exam-ready framework
- Write the formula that matches the species given
- Substitute using scientific notation correctly
- Evaluate the logarithm carefully
- Round only at the end
- Add a one-line interpretation such as acidic, neutral, or basic
Authoritative chemistry references
Final answer summary
If the quantity 6.4 × 10-8 is the hydrogen ion concentration, then the standard pH calculation is:
pH = -log(6.4 × 10-8) = 7.19
That means the solution is slightly basic. If your class or textbook specifically asks for a rigorous treatment of a very dilute acid solution, then you may need the water autoionization correction, which gives a value closer to 6.86. For most direct pH exercises, though, the expected answer is 7.19.