Calculate The Ph Of 0.100 M Sodium Phenolate

Use this premium calculator to determine the pH of sodium phenolate solution from phenol acidity data. Sodium phenolate, also called sodium phenoxide, is the conjugate base of phenol and hydrolyzes in water to produce hydroxide ions.

Quick chemistry summary

• Sodium phenolate dissociates to Na⁺ and phenolate ion, C₆H₅O^–.
• Phenolate is the conjugate base of phenol, a weak acid.
• Hydrolysis reaction: C₆H₅O^– + H₂O ⇌ C₆H₅OH + OH^–.
• K_b = K_w / K_a.
• For phenol with pK_a = 9.95, K_a ≈ 1.12 × 10^-10.
• That gives K_b ≈ 8.91 × 10^-5 for phenolate.
• At 0.100 concentration, pH is expected to be strongly basic, near 11.5.

How to calculate the pH of 0.100 m sodium phenolate

To calculate the pH of 0.100 m sodium phenolate, you treat the solute as a salt that contains the conjugate base of a weak acid. Sodium phenolate is the sodium salt of phenol. In water, sodium ions are spectator ions, while the phenolate ion reacts with water and generates hydroxide. That means the solution becomes basic. The entire problem is a classic weak-base hydrolysis calculation.

The key idea is that sodium phenolate does not behave like a strong base in the way sodium hydroxide does. Instead, its basicity comes from the equilibrium between phenolate and water. The better you understand that equilibrium, the easier the pH calculation becomes. For a standard 0.100 m or approximately 0.100 M aqueous solution at 25 degrees C, the answer is around pH 11.48 when you use the commonly cited phenol pK_a of 9.95.

Step 1: Write the hydrolysis reaction

Sodium phenolate dissociates essentially completely in water:

NaC₆H₅O → Na⁺ + C₆H₅O^–

The phenolate ion then acts as a weak base:

C₆H₅O^– + H₂O ⇌ C₆H₅OH + OH^–

This reaction produces hydroxide ion, so the pH rises above 7. The equilibrium constant that governs this step is K_b, the base dissociation constant of phenolate.

Step 2: Convert phenol pKa into Ka and then into Kb

Most reference data list the acidity of phenol, not the basicity of phenolate. So the first move is to convert phenol pK_a to K_a:

1. Use pK_a = 9.95
2. Calculate K_a = 10^-9.95 ≈ 1.12 × 10^-10
3. Use K_b = K_w / K_a
4. At 25 degrees C, K_w = 1.0 × 10^-14
5. Therefore K_b ≈ 8.91 × 10^-5

This K_b value tells you phenolate is a weak base, but not an extremely weak one. That is why a 0.100 concentration gives a noticeably basic solution.

Quantity	Symbol	Typical value at 25 degrees C	Meaning
Phenol acidity constant	Ka	1.12 × 10^-10	Acid strength of phenol
Phenol acid dissociation index	pKa	9.95	Logarithmic expression of phenol acidity
Water ion product	Kw	1.0 × 10^-14	Links pH and pOH at 25 degrees C
Phenolate basicity constant	Kb	8.91 × 10^-5	Base strength of phenolate ion

Step 3: Set up the equilibrium expression

Start with an initial phenolate concentration of 0.100. Let x be the amount that reacts with water.

• Initial: [C₆H₅O^–] = 0.100, [C₆H₅OH] = 0, [OH^–] = 0
• Change: -x, +x, +x
• Equilibrium: 0.100 – x, x, x

Now write the base equilibrium expression:

K_b = [C₆H₅OH][OH^–] / [C₆H₅O^–] = x² / (0.100 – x)

Insert K_b = 8.91 × 10^-5:

8.91 × 10^-5 = x² / (0.100 – x)

Step 4: Solve for hydroxide concentration

You can solve this either by approximation or with the exact quadratic formula. Since K_b is much smaller than the initial concentration, the approximation works well:

x ≈ √(K_bC) = √[(8.91 × 10^-5)(0.100)] ≈ 2.99 × 10^-3 M

So the hydroxide concentration is approximately:

[OH^–] ≈ 2.99 × 10^-3 M

If you solve exactly using the quadratic equation, you get essentially the same result:

x = [-K_b + √(K_b² + 4K_bC)] / 2 ≈ 2.94 × 10^-3 M

Because x is only about 2.9 percent of the starting concentration, the weak-base approximation is acceptable for most classroom and lab calculations.

Step 5: Convert hydroxide concentration to pOH and pH

Once [OH^–] is known, finding pOH is straightforward:

pOH = -log(2.94 × 10^-3) ≈ 2.53

At 25 degrees C:

pH = 14.00 – 2.53 = 11.47

Depending on the exact pK_a value you adopt for phenol and whether you use the exact or approximate method, you will usually report the pH of 0.100 m sodium phenolate as about 11.48.

Final answer for 0.100 m sodium phenolate

For a 0.100 m sodium phenolate solution at 25 degrees C, using phenol pK_a = 9.95 and K_w = 1.0 × 10^-14, the calculated pH is:

pH ≈ 11.48

This result confirms that sodium phenolate solutions are distinctly basic, though not nearly as basic as a 0.100 M solution of sodium hydroxide. That difference is important. Sodium hydroxide dissociates fully and directly supplies 0.100 M hydroxide, while sodium phenolate only partially generates hydroxide through equilibrium hydrolysis.

Solution	Input concentration	Main source of OH^–	Approximate [OH^–]	Approximate pH at 25 degrees C
Sodium phenolate	0.100	Weak-base hydrolysis	2.94 × 10^-3 M	11.47 to 11.48
Ammonia	0.100	Weak-base ionization	1.33 × 10^-3 M	11.12
Sodium hydroxide	0.100	Strong-base dissociation	1.00 × 10^-1 M	13.00

Why sodium phenolate is basic

The logic comes from conjugate acid-base relationships. Phenol is a weak acid because it can donate a proton, but it does so only to a limited extent in water. Its conjugate base, phenolate, therefore has measurable affinity for protons. When phenolate is dissolved in water, it takes a proton from water molecules and forms phenol plus hydroxide. The hydroxide generated is what raises the pH.

This behavior is typical for salts formed from a strong base and a weak acid. Sodium phenolate originates from the weak acid phenol and the strong base sodium hydroxide. The sodium ion itself does not affect pH significantly, but the phenolate ion does. This is exactly why the solution must be analyzed with hydrolysis equilibrium rather than with simple complete dissociation assumptions.

Common mistakes students make

• Using the pK_a of phenol directly as if it were the pK_b of phenolate.
• Forgetting to convert pK_a to K_a before finding K_b.
• Treating sodium phenolate like a strong base such as NaOH.
• Using pH = -log[OH^–] instead of pOH = -log[OH^–].
• Ignoring the temperature dependence of K_w if the problem is not at 25 degrees C.
• Mixing up molality and molarity without stating the approximation used.

Does 0.100 m mean the same as 0.100 M?

Strictly speaking, no. A molal concentration is moles of solute per kilogram of solvent, while molarity is moles of solute per liter of solution. However, for many dilute aqueous chemistry problems, especially at introductory level, 0.100 m is often treated as approximately equal to 0.100 M because the density of the solution is close to that of water. The calculator above lets you keep the unit label visible while still performing the standard equilibrium calculation.

In high-precision analytical work, the distinction matters more. If you were performing a rigorous thermodynamic calculation, you might also need activity coefficients, ionic strength corrections, and a more exact relation between concentration scales. But for a standard acid-base equilibrium problem, the expected answer remains very close to pH 11.48.

Reference data and authoritative resources

If you want to verify the constants or review pH fundamentals, consult authoritative sources such as the NIST Chemistry WebBook, the USGS Water Science School page on pH and water, and MIT OpenCourseWare for acid-base equilibrium instruction. These resources are especially useful when you need trustworthy physical constants and a stronger conceptual foundation for conjugate acid-base systems.

How the exact quadratic method improves accuracy

In many weak-acid and weak-base calculations, the shortcut x = √(KC) is taught first because it is fast and often gives a very close answer. That approximation assumes x is small compared with the initial concentration. For 0.100 sodium phenolate, the approximation works quite well because the degree of hydrolysis is only a few percent. Even so, the exact quadratic formula is better practice when calculators and software are readily available.

Using the exact approach, the hydroxide concentration is found from:

1. x² + K_bx – K_bC = 0
2. x = [-K_b + √(K_b² + 4K_bC)] / 2
3. Then pOH = -log x
4. Finally pH = pK_w – pOH

This exact method becomes more important when the concentration is lower, the equilibrium constant is larger, or the percent ionization rises enough that the small-x assumption begins to fail. The calculator on this page supports both methods so you can compare them directly.

Practical interpretation of the result

A pH of roughly 11.48 means the solution is substantially basic. In a laboratory, that level of basicity can influence indicators, reaction mechanisms, and the solubility or stability of acid-sensitive compounds. It also explains why sodium phenolate is often used as a reactive nucleophilic or basic intermediate in organic chemistry.

At the same time, the pH is much lower than a strong base of equal formal concentration. That distinction matters when selecting reagents, estimating corrosion risk, or anticipating side reactions. Sodium phenolate creates a basic environment because of equilibrium hydrolysis, not because it instantly delivers the full stoichiometric concentration of hydroxide.

Bottom line

To calculate the pH of 0.100 m sodium phenolate, convert the pK_a of phenol to K_a, determine K_b from K_w/K_a, solve the hydrolysis equilibrium for [OH^–], and then convert to pOH and pH. Using phenol pK_a = 9.95 at 25 degrees C gives a pH of about 11.48. That is the standard, chemically correct answer for this equilibrium problem.

Educational note: actual measured pH may differ slightly from ideal calculations because of ionic strength, concentration scale differences, temperature shifts, and instrument calibration.