Calculate The Ph Of A 0.15 M Solution Of Ch3Coona

This premium calculator estimates the pH of a sodium acetate solution by treating CH3COONa as a salt of a weak acid and a strong base. Because acetate ion hydrolyzes in water to form OH-, the solution is basic. Enter or confirm the concentration and acid dissociation constant of acetic acid, then calculate the resulting Kb, hydroxide concentration, pOH, and pH.

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How to calculate the pH of a 0.15 M solution of CH3COONa

To calculate the pH of a 0.15 M solution of CH3COONa, you need to recognize what the compound is doing in water. CH3COONa is sodium acetate, a salt formed from acetic acid, which is a weak acid, and sodium hydroxide, which is a strong base. When sodium acetate dissolves, it dissociates essentially completely into sodium ions and acetate ions:

CH3COONa -> Na+ + CH3COO-

The sodium ion does not significantly affect pH in water, but the acetate ion does. Acetate is the conjugate base of acetic acid, so it can react with water and produce hydroxide ions:

CH3COO- + H2O ⇌ CH3COOH + OH-

Because hydroxide ions are formed, the solution becomes basic and its pH rises above 7. This is the central idea behind the entire calculation. You are not dealing with a neutral salt such as sodium chloride. You are dealing with a weak-base hydrolysis equilibrium.

Step 1: Identify the needed equilibrium constant

Since acetate behaves as a weak base in water, the most useful equilibrium constant is Kb for acetate. In many chemistry problems, the value provided is actually the acid dissociation constant Ka for acetic acid. That is common because acetic acid is a standard weak acid with a well-known Ka around 1.8 x 10^-5 at 25 C.

The relationship between Ka and Kb is:

Ka x Kb = Kw

At 25 C, Kw = 1.0 x 10^-14. Therefore:

Kb = Kw / Ka = (1.0 x 10^-14) / (1.8 x 10^-5) = 5.56 x 10^-10

This tells you that acetate is a weak base, but it is still strong enough at 0.15 M to raise the pH noticeably above neutrality.

Step 2: Set up the hydrolysis equilibrium

Start with the base hydrolysis reaction:

CH3COO- + H2O ⇌ CH3COOH + OH-

If the initial concentration of acetate is 0.15 M, then an ICE setup looks like this:

• Initial: [CH3COO-] = 0.15, [CH3COOH] = 0, [OH-] = 0
• Change: [-x, +x, +x]
• Equilibrium: [CH3COO-] = 0.15 – x, [CH3COOH] = x, [OH-] = x

Now substitute into the base dissociation expression:

Kb = [CH3COOH][OH-] / [CH3COO-] = x^2 / (0.15 – x)

Plug in Kb = 5.56 x 10^-10:

5.56 x 10^-10 = x^2 / (0.15 – x)

Step 3: Solve for hydroxide concentration

Because Kb is very small compared with the initial concentration, many textbooks use the weak-base approximation and assume that x is much smaller than 0.15. If you do that, the equation becomes:

x^2 / 0.15 = 5.56 x 10^-10

x^2 = 8.33 x 10^-11

x = 9.13 x 10^-6 M

Since x represents [OH-], the hydroxide concentration is about 9.13 x 10^-6 M.

If you solve with the quadratic equation rather than the approximation, the answer changes only negligibly. That is why the approximation is accepted here. The percent ionization is extremely small relative to the starting concentration.

Step 4: Convert [OH-] to pOH and pH

Once [OH-] is known, the rest is straightforward:

1. pOH = -log[OH-]
2. pH = 14 – pOH

Substituting the value:

pOH = -log(9.13 x 10^-6) ≈ 5.04

pH = 14.00 – 5.04 = 8.96

So, the pH of a 0.15 M solution of CH3COONa is approximately 8.96 at 25 C when Ka for acetic acid is taken as 1.8 x 10^-5.

Final answer: A 0.15 M sodium acetate solution has a pH of about 8.96 under standard 25 C conditions.

Why sodium acetate gives a basic pH

Students often wonder why a salt can change pH at all. The key is the strength of the parent acid and parent base. Salts of a strong acid and a strong base, such as NaCl, are typically neutral because neither ion hydrolyzes significantly. Salts of a weak acid and a strong base, like sodium acetate, are basic because the anion is the conjugate base of a weak acid and therefore reacts with water to generate OH-. Conversely, salts of a strong acid and a weak base, such as ammonium chloride, are acidic because the cation reacts with water to produce H3O+.

Salt Type	Parent Acid	Parent Base	Typical Water Behavior	Example
Strong acid + strong base	Strong	Strong	Approximately neutral, pH near 7	NaCl
Weak acid + strong base	Weak	Strong	Basic due to anion hydrolysis	CH3COONa
Strong acid + weak base	Strong	Weak	Acidic due to cation hydrolysis	NH4Cl
Weak acid + weak base	Weak	Weak	Depends on relative Ka and Kb	NH4CH3COO

Approximation versus quadratic solution

In weak acid and weak base calculations, the approximation method is popular because it simplifies the algebra dramatically. For sodium acetate, the approximation is excellent because the change in concentration caused by hydrolysis is tiny compared with the initial 0.15 M. A quick check confirms this:

• Calculated x = 9.13 x 10^-6 M
• Initial concentration = 0.15 M
• Percent change = (9.13 x 10^-6 / 0.15) x 100 ≈ 0.0061%

Since the change is much less than 5%, the approximation is fully justified. Still, a premium calculator should allow the quadratic route because it teaches good equilibrium practice and lets users compare methods directly.

How concentration affects pH in sodium acetate solutions

The pH of a sodium acetate solution depends on concentration. More acetate generally means more hydroxide is generated, so the pH increases gradually with concentration. The relationship is not linear because the hydrolysis equilibrium depends on the square root of concentration when the approximation is used:

[OH-] ≈ sqrt(Kb x C)

This means if concentration increases by a factor of 100, hydroxide concentration increases by a factor of 10. That is why pH changes steadily but not explosively as the solution becomes more concentrated.

CH3COONa Concentration (M)	Approx. [OH-] (M)	Approx. pOH	Approx. pH
0.010	2.36 x 10^-6	5.63	8.37
0.050	5.27 x 10^-6	5.28	8.72
0.100	7.46 x 10^-6	5.13	8.87
0.150	9.13 x 10^-6	5.04	8.96
0.500	1.67 x 10^-5	4.78	9.22

Common mistakes to avoid

• Assuming CH3COONa is neutral just because it is a salt. It is not neutral in water.
• Using Ka directly in the hydrolysis expression for acetate instead of converting to Kb.
• Forgetting that the solution becomes basic, so pH must be greater than 7.
• Mixing up pOH and pH after calculating hydroxide concentration.
• Ignoring temperature effects when very high precision is required, because Kw changes with temperature.

Real chemistry context: where sodium acetate pH matters

Sodium acetate appears in buffer preparation, food chemistry, textile processing, laboratory standardization, and biochemistry. In many practical settings, sodium acetate is mixed with acetic acid to form an acetate buffer. The pH of the pure salt solution is useful because it gives a baseline for what acetate does in water before acid is added. In laboratory work, understanding hydrolysis also helps explain why salts of weak acids can shift measured pH in analytical samples.

In environmental and biological systems, acetate is a familiar anion. However, the exact pH effect always depends on concentration, ionic strength, and temperature. The textbook result of 8.96 is appropriate for a straightforward general chemistry calculation at 25 C with idealized behavior.

Authoritative references for acid-base constants and water chemistry

For deeper verification and reference data, consult: NIST, LibreTexts Chemistry, U.S. Environmental Protection Agency, U.S. Geological Survey, and MIT Chemistry.

If you want only .gov or .edu style authority sources especially relevant to water chemistry and acid-base equilibrium concepts, good starting points include the EPA, the USGS, and chemistry departments from major universities such as MIT.

Summary of the calculation

1. Recognize CH3COONa as a salt of a weak acid and strong base.
2. Write the hydrolysis reaction: CH3COO- + H2O ⇌ CH3COOH + OH-.
3. Convert Ka of acetic acid to Kb for acetate using Kb = Kw / Ka.
4. Set up the equilibrium expression Kb = x^2 / (C – x).
5. Solve for x = [OH-].
6. Find pOH = -log[OH-].
7. Find pH = 14 – pOH.

Using C = 0.15 M, Ka = 1.8 x 10^-5, and Kw = 1.0 x 10^-14, the result is:

pH ≈ 8.96

This guide is intended for educational use and standard chemistry problem solving. For advanced analytical work, corrections for ionic strength, activity coefficients, and temperature dependence may be necessary.