Calculate The Ph Of Ch3Coona The Ka Is 1.8X10-54

Use this premium sodium acetate pH calculator to estimate hydrolysis, pOH, pH, and related concentrations. It also explains why many textbook problems actually mean 1.8 x 10^-5, the familiar Ka of acetic acid at 25°C.

If you truly use Ka = 1.8 x 10^-54, the conjugate base becomes extraordinarily strong, and the usual weak-base assumptions become unrealistic. This calculator still returns the mathematically constrained result by limiting hydrolysis to the available acetate concentration.

How to Calculate the pH of CH3COONa When Ka Is 1.8 x 10^-54

Sodium acetate, written as CH3COONa, is the salt of a weak acid and a strong base. In water it dissociates almost completely into sodium ions and acetate ions:

CH3COONa → Na⁺ + CH3COO^–

The sodium ion is essentially a spectator ion in ordinary acid-base calculations. The acetate ion is the important species because it acts as a base and reacts with water:

CH3COO^– + H2O ⇌ CH3COOH + OH^–

This reaction produces hydroxide ions, so a sodium acetate solution is basic. The pH therefore depends on the base hydrolysis constant of acetate, Kb, which is related to the acid dissociation constant of acetic acid, Ka, by the fundamental equilibrium relation:

Ka x Kb = Kw

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

Kb = Kw / Ka

Step 1: Convert Ka into Kb

If the problem literally states that the Ka is 1.8 x 10^-54, then:

Kb = (1.0 x 10^-14) / (1.8 x 10^-54) = 5.56 x 10^39

That is an astronomically large base dissociation constant. It implies that acetate would behave as an extremely strong base, which is not chemically consistent with real acetic acid chemistry. In real general chemistry, acetic acid is usually given as Ka = 1.8 x 10^-5. That means many students encounter this exact problem statement because of a missing caret or formatting issue. The difference between 10^-54 and 10^-5 is enormous, so checking the intended exponent matters.

Step 2: Set up the hydrolysis equilibrium

Suppose the sodium acetate concentration is C mol/L. Let x be the amount of acetate that reacts with water to produce hydroxide. Then:

• Initial [CH3COO^–] = C
• Change = -x
• Equilibrium [CH3COO^–] = C – x
• Equilibrium [CH3COOH] = x
• Equilibrium [OH^–] = x

So the equilibrium expression is:

Kb = x² / (C – x)

Rearranging gives the quadratic equation:

x² + Kb x – Kb C = 0

The physically meaningful root is:

x = [-Kb + √(Kb² + 4KbC)] / 2

Step 3: Find pOH and pH

Once x = [OH^–] is known:

1. pOH = -log[OH^–]
2. pH = 14 – pOH

For realistic weak-base systems, this works cleanly. However, if Kb is extremely large, the math suggests hydrolysis is essentially complete. Since one acetate can generate at most one hydroxide ion through this stoichiometric step, the hydroxide concentration cannot exceed the original acetate concentration in this simplified model. That is why a practical calculator should limit [OH^–] ≤ C under such extreme inputs.

Worked example with the exact prompt: Ka = 1.8 x 10^-54

Assume a 0.10 M sodium acetate solution. Then:

1. Ka = 1.8 x 10^-54
2. Kb = 1.0 x 10^-14 / 1.8 x 10^-54 = 5.56 x 10^39
3. This Kb is so huge that acetate hydrolysis goes essentially to completion
4. Thus the hydroxide concentration approaches the initial salt concentration, about 0.10 M
5. pOH ≈ -log(0.10) = 1.00
6. pH ≈ 14.00 – 1.00 = 13.00

So if you literally interpret the input as written, a 0.10 M CH3COONa solution gives an approximate pH of 13.00. Again, that is a consequence of the extremely tiny Ka value and does not reflect actual sodium acetate behavior in ordinary chemistry.

Worked example with the standard acetic acid value: Ka = 1.8 x 10^-5

Now use the usual acetic acid Ka value for comparison:

1. Ka = 1.8 x 10^-5
2. Kb = 1.0 x 10^-14 / 1.8 x 10^-5 = 5.56 x 10^-10
3. For C = 0.10 M, the weak-base approximation gives x ≈ √(KbC)
4. x ≈ √(5.56 x 10^-11) ≈ 7.46 x 10^-6 M
5. pOH ≈ 5.13
6. pH ≈ 8.87

This is the familiar answer students usually expect for a sodium acetate solution. It is mildly basic, not strongly caustic.

Scenario	Ka of CH3COOH	Kb of CH3COO-	Assumed [CH3COONa]	Approximate [OH-]	Approximate pH
Literal prompt	1.8 x 10^-54	5.56 x 10^39	0.10 M	0.10 M	13.00
Standard textbook acetic acid	1.8 x 10^-5	5.56 x 10^-10	0.10 M	7.46 x 10^-6 M	8.87

Why sodium acetate is basic

The reason CH3COONa has a pH above 7 is rooted in conjugate acid-base behavior. Acetate is the conjugate base of acetic acid. Since acetic acid is weak, its conjugate base is strong enough to react measurably with water. By contrast, sodium comes from sodium hydroxide, a strong base, and does not appreciably affect pH in dilute aqueous solution. This is one of the standard rules for salt hydrolysis:

• Strong acid + strong base salt: roughly neutral
• Weak acid + strong base salt: basic
• Strong acid + weak base salt: acidic
• Weak acid + weak base salt: depends on both Ka and Kb

When the square-root shortcut is valid

Students often use the approximation:

[OH^–] ≈ √(KbC)

This comes from assuming x is much smaller than C, so C – x ≈ C. The shortcut is appropriate when Kb is small and the degree of hydrolysis is limited. For normal sodium acetate calculations, it works very well. For the extreme value Ka = 1.8 x 10^-54, the approximation completely fails because hydrolysis is not small relative to concentration. That is why the calculator above lets you choose either the full quadratic method or the weak-base approximation.

Comparison table: expected pH at several sodium acetate concentrations

The next table uses the standard acetic acid value Ka = 1.8 x 10^-5, which corresponds to Kb = 5.56 x 10^-10 at 25°C. These are typical instructional values for sodium acetate solutions.

[CH3COONa] (M)	Approximate [OH-] (M)	pOH	Approximate pH	Interpretation
0.001	7.46 x 10^-7	6.13	7.87	Slightly basic
0.010	2.36 x 10^-6	5.63	8.37	Mildly basic
0.100	7.46 x 10^-6	5.13	8.87	Typical classroom result
1.000	2.36 x 10^-5	4.63	9.37	More strongly basic, still weak-base behavior

Common mistakes in this problem type

• Using Ka directly to find pH. For a salt of a weak acid, you normally need Kb of the conjugate base.
• Ignoring the concentration of the salt. pH depends strongly on the initial CH3COONa molarity.
• Assuming the prompt must be correct. A Ka of 1.8 x 10^-54 for acetic acid is almost certainly a typo or formatting error in most educational settings.
• Using the weak-base shortcut when Kb is huge. If Kb is large, solve the exact expression or apply a physically meaningful limit.
• Forgetting that pH + pOH = 14 only at 25°C. This calculator uses that standard condition.

Best final answer to report

If your teacher or source explicitly states Ka = 1.8 x 10^-54 and the sodium acetate concentration is 0.10 M, then the chemically constrained result is:

pH ≈ 13.00

If the intended value was the standard acetic acid constant Ka = 1.8 x 10^-5, then for a 0.10 M sodium acetate solution the expected answer is:

pH ≈ 8.87

So the key to solving this question correctly is not only the equilibrium math but also recognizing whether the exponent in Ka was entered properly.

Authoritative references for pH, water equilibrium, and salt hydrolysis

• USGS: pH and Water
• University of Wisconsin: Salt Hydrolysis Concepts
• Purdue University: pH and Acid-Base Equilibria Review

This page is designed for educational calculation at 25°C with idealized aqueous behavior. Real concentrated solutions can deviate from ideal activity-based assumptions.