Calculate Ph Of 1.0 10 2 M Solution Of Naoh

Use this premium calculator to find hydroxide concentration, pOH, and pH for a sodium hydroxide solution written in scientific notation. By default, the setup matches the classic chemistry problem: 1.0 × 10^-2 M NaOH at 25 degrees C.

NaOH pH Calculator

Results

Quick answer for the default case: 1.0 × 10^-2 M NaOH gives pOH = 2.00 and pH = 12.00 at 25 degrees C.

How to calculate the pH of a 1.0 × 10^-2 M solution of NaOH

If you need to calculate pH of 1.0 10 2 m solution of NaOH, the key idea is that sodium hydroxide is a strong base. In introductory and general chemistry, NaOH is treated as dissociating completely in water:

NaOH(aq) → Na⁺(aq) + OH^–(aq)

Because the dissociation is essentially complete for a dilute solution like 1.0 × 10^-2 M, the hydroxide concentration is taken to be the same as the formal concentration of NaOH. That means:

[OH^–] = 1.0 × 10^-2 M

Next, calculate pOH using the base ten logarithm:

pOH = -log[OH^–] = -log(1.0 × 10^-2) = 2.00

At 25 degrees C, pH + pOH = 14.00, so:

pH = 14.00 – 2.00 = 12.00

Therefore, the pH of a 1.0 × 10^-2 M NaOH solution is 12.00 at 25 degrees C.

Why this result is straightforward for NaOH

This problem is easier than weak acid or weak base calculations because NaOH is a strong electrolyte. It separates almost completely into ions in water, so you do not usually need an ICE table, an equilibrium constant expression, or a quadratic equation. In most classroom settings, the whole process is just three steps:

1. Write the dissociation equation for NaOH.
2. Set [OH^–] equal to the NaOH molarity.
3. Find pOH and then convert to pH.

That is exactly why this type of question often appears early in acid-base units. It checks whether you know the difference between strong bases and weak bases, and whether you can move correctly between concentration, pOH, and pH.

Step by step solution for 1.0 × 10^-2 M NaOH

1. Interpret the concentration. The expression 1.0 × 10^-2 M means 0.010 M.
2. Use complete dissociation. Since NaOH is a strong base, 0.010 M NaOH gives 0.010 M OH^–.
3. Find pOH. pOH = -log(0.010) = 2.00.
4. Convert to pH. At 25 degrees C, pH = 14.00 – 2.00 = 12.00.

Notice the logarithm pattern here. Any concentration written as 1.0 × 10^-n produces a pOH of n if the coefficient is exactly 1.0. So if [OH^–] = 1.0 × 10^-2, the pOH is 2. If [OH^–] = 1.0 × 10^-3, the pOH is 3, and so on.

Common student mistakes

• Using the NaOH concentration directly as pH instead of first finding pOH.
• Forgetting that NaOH is a base, so the direct ion produced is OH^–, not H⁺.
• Subtracting from 7 instead of 14 at 25 degrees C.
• Misreading 1.0 × 10^-2 as 10².
• Ignoring temperature when a problem specifically gives a non-25 degrees C condition.

The most important check is conceptual: a sodium hydroxide solution should be basic, so its pH must be greater than 7 under ordinary classroom assumptions. If you obtain 2.00 or 0.010 as the final pH, you likely stopped too early or used the wrong formula.

Strong base behavior compared with weak bases

One reason chemistry instructors like this example is that it highlights how strong bases differ from weak bases such as ammonia. With NaOH, dissociation is effectively complete. With NH₃, only a fraction reacts with water to generate OH^–. That means NaOH problems are concentration driven, while weak base problems are equilibrium driven.

Solution	Formal Concentration	Main Model Used	Approximate [OH-]	Approximate pH at 25 degrees C
NaOH	1.0 × 10^-2 M	Complete dissociation	1.0 × 10^-2 M	12.00
KOH	1.0 × 10^-2 M	Complete dissociation	1.0 × 10^-2 M	12.00
Ba(OH)2	1.0 × 10^-2 M	Complete dissociation with 2 OH- per formula unit	2.0 × 10^-2 M	12.30
NH3	1.0 × 10^-2 M	Weak base equilibrium	Much less than 1.0 × 10^-2 M	About 10.6

The table shows why stoichiometry matters. For monohydroxide strong bases such as NaOH and KOH, one mole of solute gives one mole of hydroxide. For barium hydroxide, each formula unit contributes two hydroxides, so the hydroxide concentration doubles. That directly affects pOH and pH.

Scientific notation and why 1.0 × 10^-2 matters

Scientific notation is common in chemistry because concentrations often span many powers of ten. The notation 1.0 × 10^-2 M is equivalent to 0.010 M. When calculating pOH, the exponent largely drives the answer:

• 1.0 × 10^-1 M OH- gives pOH = 1
• 1.0 × 10^-2 M OH- gives pOH = 2
• 1.0 × 10^-3 M OH- gives pOH = 3

If the coefficient is not exactly 1.0, then the logarithm will shift the answer slightly. For example, [OH^–] = 3.2 × 10^-2 M gives a pOH a little below 2 because the coefficient 3.2 is greater than 1.

Temperature and the value of pK_w

In many basic chemistry courses, you use pH + pOH = 14.00 because the temperature is assumed to be 25 degrees C. However, the ionic product of water changes with temperature. That means pK_w is not always exactly 14.00. If your instructor includes temperature, you should use:

pH + pOH = pK_w

Here is a practical summary of commonly cited pK_w values used in educational references and chemistry data tables.

Temperature	Approximate pKw	Neutral pH	pH of 1.0 × 10^-2 M NaOH
0 degrees C	14.94	7.47	12.94
20 degrees C	14.17	7.09	12.17
25 degrees C	14.00	7.00	12.00
40 degrees C	13.68	6.84	11.68

These values explain an important point: neutral pH is not always 7.00. At temperatures above 25 degrees C, neutral water has a pH below 7, even though it is still neutral because [H⁺] = [OH^–]. For classroom problems that do not mention temperature, assume 25 degrees C unless your instructor says otherwise.

Why the answer is 12.00 and not 2.00

Many learners find acid-base problems easier when they check what the number represents. In this case, 2.00 is the pOH, not the pH. Since NaOH is a base, a low pOH corresponds to a high pH. Converting from pOH to pH flips the result using pH = 14.00 – pOH at 25 degrees C.

Another way to think about it is relative acidity and basicity. A pH of 12 is strongly basic compared with ordinary water, while a pH of 2 would describe a strongly acidic solution, which would be incompatible with dissolved sodium hydroxide.

Worked comparison examples

1. 0.0010 M NaOH: [OH-] = 1.0 × 10^-3 M, pOH = 3.00, pH = 11.00 at 25 degrees C.
2. 0.10 M NaOH: [OH-] = 1.0 × 10^-1 M, pOH = 1.00, pH = 13.00 at 25 degrees C.
3. 0.010 M Ba(OH)2: [OH-] = 2.0 × 10^-2 M, pOH = 1.70, pH = 12.30 at 25 degrees C.

These examples show two useful patterns. First, each tenfold change in hydroxide concentration changes pOH by one unit. Second, bases that release more than one OH^– per formula unit must be adjusted for stoichiometry before calculating pOH.

Authoritative chemistry references

If you want to verify acid-base definitions, pH fundamentals, and water chemistry data, consult reliable educational and government sources. The following references are useful:

• Chemistry LibreTexts educational chemistry reference
• U.S. EPA overview of pH and water chemistry
• U.S. Geological Survey explanation of pH and water
• MIT Chemistry department resources

Exam strategy for this exact question

When you see a prompt like calculate pH of 1.0 10 2 m solution of NaOH, immediately rewrite it mentally as 1.0 × 10^-2 M NaOH. Then move through a short checklist:

1. Identify NaOH as a strong base.
2. Set [OH-] equal to 1.0 × 10^-2 M.
3. Compute pOH = 2.00.
4. Convert to pH = 12.00 at 25 degrees C.

This approach is quick, reliable, and exactly what most instructors expect. If the question is multiple choice, you can often eliminate wrong answers fast by checking whether they are acidic values or whether they confuse pH with pOH.

Final answer summary

For a 1.0 × 10^-2 M aqueous solution of sodium hydroxide:

• NaOH dissociates completely.
• [OH-] = 1.0 × 10^-2 M.
• pOH = 2.00.
• pH = 12.00 at 25 degrees C.

That is the standard textbook answer and the one returned by the calculator above when using the default values.

Quick memory tip: strong base first gives OH-, then pOH, then pH. For 1.0 × 10^-2 M NaOH at 25 degrees C, the answer is pH 12.00.