Calculate The Ph Of A 0.0150 M Solution Of Naoh.

Use this premium chemistry calculator to find hydroxide concentration, pOH, and pH for a sodium hydroxide solution. The form is prefilled for 0.0150 M NaOH at 25 degrees Celsius, which is the standard classroom assumption for pH calculations.

NaOH pH Calculator

For sodium hydroxide, assume complete dissociation in dilute aqueous solution: NaOH → Na⁺ + OH^–

Expert Guide: How to Calculate the pH of a 0.0150 M Solution of NaOH

If you need to calculate the pH of a 0.0150 M solution of sodium hydroxide, the good news is that this is one of the most straightforward acid-base problems in general chemistry. Sodium hydroxide, written as NaOH, is a strong base. In dilute aqueous solution, it dissociates essentially completely into sodium ions and hydroxide ions. That single fact makes the calculation simple, predictable, and highly accurate for typical classroom and laboratory exercises.

The target result for a 0.0150 M NaOH solution at 25 degrees Celsius is a pH of approximately 12.176, usually rounded to 12.18. To understand why, you only need to move through a short sequence: determine hydroxide concentration, calculate pOH, and convert pOH to pH.

Step 1: Recognize that NaOH is a strong base

NaOH belongs to the family of strong hydroxide bases. In water, the dissociation is treated as complete:

NaOH(aq) → Na⁺(aq) + OH^–(aq)

Because one mole of NaOH produces one mole of hydroxide ions, the hydroxide concentration is equal to the initial molar concentration of NaOH, assuming the solution is not so concentrated that activity corrections become significant and not so dilute that water autoionization dominates.

• Given concentration of NaOH = 0.0150 M
• Stoichiometric OH^– released per formula unit = 1
• Therefore, [OH^–] = 0.0150 M

Step 2: Calculate pOH

The definition of pOH is:

pOH = -log[OH^–]

Substitute the hydroxide concentration:

pOH = -log(0.0150)

Using base-10 logarithms:

pOH = 1.8239

With appropriate rounding, pOH is usually reported as 1.824 or 1.82, depending on the number of decimal places required.

Step 3: Convert pOH to pH

At 25 degrees Celsius, the standard relationship is:

pH + pOH = 14.00

So:

pH = 14.00 – 1.8239 = 12.1761

Rounded properly, the final answer is:

pH = 12.176 or 12.18

Final answer for 0.0150 M NaOH

1. NaOH dissociates completely.
2. [OH^–] = 0.0150 M
3. pOH = -log(0.0150) = 1.8239
4. pH = 14.00 – 1.8239 = 12.1761

Therefore, the pH of a 0.0150 M solution of NaOH is 12.18 at 25 degrees Celsius.

Why this problem is easier than weak base problems

Strong base calculations are direct because you do not need an equilibrium table, a base dissociation constant, or an approximation method. Weak bases such as ammonia require the use of K_b, and the hydroxide concentration must be solved from equilibrium rather than assigned directly from the initial concentration. Sodium hydroxide does not create that extra layer of work in introductory calculations.

Type of base	Example	How [OH-] is found	Main formula used	Difficulty level
Strong base	NaOH, KOH	Equal to formal concentration times OH stoichiometry	pOH = -log[OH-]	Low
Strong dibasic base	Ba(OH)2	Twice the formal concentration in ideal dilution	pOH = -log(2C)	Low to moderate
Weak base	NH3	Found from equilibrium expression	Kb = [BH+][OH-]/[B]	Moderate

M versus m: an important notation detail

The prompt often appears as “calculate the pH of a 0.0150 m solution of NaOH.” In strict chemistry notation, lowercase m usually means molality, which is moles of solute per kilogram of solvent, while uppercase M means molarity, which is moles of solute per liter of solution. Most classroom pH problems involving NaOH actually intend 0.0150 M. For a dilute aqueous solution, the numerical difference between 0.0150 m and 0.0150 M is usually small, but the units are not identical and should not be casually interchanged in rigorous work.

In practical educational settings, if your instructor or source gives 0.0150 “m” in a simple pH problem without density information, the expectation is almost always to treat it as a molar concentration value and solve using the strong base method shown above. If density or mass of solvent is provided, then a true molality-based treatment may be needed.

What assumptions are built into the calculation?

• Complete dissociation: NaOH is treated as fully ionized in water.
• Dilute solution behavior: Ion activities are approximated by concentrations.
• Standard temperature: The equation pH + pOH = 14.00 is used at 25 degrees Celsius.
• No side reactions: We assume no significant reaction with atmospheric carbon dioxide or other dissolved species.
• Negligible autoionization of water: Since 0.0150 M is much larger than 1.0 × 10^-7 M, hydroxide from water itself is negligible.

How concentration changes pH for strong bases

The pH scale is logarithmic, not linear. That means a tenfold change in hydroxide concentration shifts pOH by 1 unit, and correspondingly shifts pH by 1 unit in the opposite direction at 25 degrees Celsius. This is why even apparently small changes in concentration can produce a visible shift in pH.

NaOH concentration (M)	[OH-] (M)	pOH	pH at 25 degrees Celsius	Interpretation
0.00100	0.00100	3.000	11.000	Basic
0.0100	0.0100	2.000	12.000	Strongly basic
0.0150	0.0150	1.824	12.176	Target problem
0.100	0.100	1.000	13.000	Very strongly basic

How to check your answer quickly

A strong base with concentration a little above 0.01 M should have a pOH a little below 2. Since pH = 14 – pOH, the pH should be a little above 12. That quick estimate already tells you that 12.18 is reasonable. If you get an answer below 7, or even near 7, something went wrong, because NaOH is definitely basic.

Common mistakes students make

1. Using pH = -log[OH-]
   That formula gives pOH, not pH.
2. Forgetting complete dissociation
   NaOH is a strong base, so [OH-] is not solved by equilibrium in basic introductory problems.
3. Confusing M with m
   Molarity and molality are different concentration units.
4. Dropping the stoichiometric factor for polyhydroxide bases
   For Ba(OH)2, [OH-] is roughly 2C, not C.
5. Rounding too early
   Keep extra digits through the logarithm step, then round the final result.

Real-world context for pH and strong bases

Strongly basic solutions matter in water treatment, industrial cleaning, chemical manufacturing, titration work, and lab safety. Sodium hydroxide is widely used because it is a powerful base, inexpensive, and reliable. However, it is also caustic, so pH calculations are not just theoretical. They help predict corrosion risk, neutralization requirements, and safe handling conditions.

Environmental agencies and water science programs monitor pH because it affects aquatic life, metal solubility, chemical reactivity, and treatment efficiency. Most natural waters lie far closer to neutral than a 0.0150 M NaOH solution. A pH around 12.18 is far above the range tolerated by most natural systems without serious chemical impact.

Relevant reference ranges and comparison data

Authoritative sources such as the U.S. Geological Survey and the U.S. Environmental Protection Agency explain that pH is a core water-quality parameter. Natural waters commonly fall near pH 6.5 to 8.5, while a sodium hydroxide solution like the one calculated here is dramatically more basic. That difference highlights how concentrated laboratory bases depart from environmental norms.

• Typical drinking water guideline discussions often reference pH ranges around 6.5 to 8.5.
• A 0.0150 M NaOH solution has a pH near 12.18, which is far more alkaline than ordinary natural waters.
• This means the solution is highly reactive compared with typical environmental samples.

Authority sources for further reading

Explore these reliable resources for background on pH, water chemistry, and acid-base concepts:
USGS: pH and Water
U.S. EPA: pH Overview
University-supported chemistry reference on water autoionization and pH

Summary of the method

To calculate the pH of a 0.0150 M NaOH solution, identify NaOH as a strong base, set the hydroxide concentration equal to 0.0150 M, compute pOH using the negative logarithm, and then subtract from 14.00 at 25 degrees Celsius. The result is 12.176, which rounds to 12.18. This exact workflow is one of the foundational skills in acid-base chemistry, and once you understand it for NaOH, you can apply the same logic to many other strong hydroxide bases.

Quick answer: For 0.0150 M NaOH at 25 degrees Celsius, [OH-] = 0.0150 M, pOH = 1.824, and pH = 12.176, usually reported as 12.18.