Calculate The Ph Of A 0.200 M Nach3Co2 Solution

Use this interactive sodium acetate hydrolysis calculator to find pH, pOH, Kb, and equilibrium hydroxide concentration for an aqueous NaCH3CO2 solution. The default values match the classic chemistry problem: 0.200 M sodium acetate at 25 degrees Celsius.

How to calculate the pH of a 0.200 M NaCH3CO2 solution

To calculate the pH of a 0.200 M NaCH3CO2 solution, you identify the solute as sodium acetate, the salt of a strong base and a weak acid. Sodium acetate dissociates essentially completely in water into sodium ions and acetate ions. The sodium ion, Na⁺, is a spectator ion under normal introductory chemistry conditions. The acetate ion, CH₃CO₂^–, is the conjugate base of acetic acid and reacts with water to produce hydroxide ions. That hydrolysis reaction is why the solution is basic and why the pH comes out above 7.

The key equilibrium is:

CH3CO2- + H2O ⇌ CH3CO2H + OH-

Because hydroxide ions are produced, the quantity you usually solve for first is [OH-]. Once you know hydroxide concentration, you calculate pOH from pOH = -log[OH-] and then calculate pH from pH = 14.00 – pOH at 25 degrees C.

Step 1: Start with the acid dissociation constant of acetic acid

Most general chemistry textbooks use a Ka for acetic acid close to 1.8 × 10^-5 at 25 degrees C. Since acetate is the conjugate base of acetic acid, its base dissociation constant is found from:

Kb = Kw / Ka

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

Kb = (1.0 × 10^-14) / (1.8 × 10^-5) = 5.56 × 10^-10

Step 2: Set up the hydrolysis equilibrium table

For a 0.200 M sodium acetate solution, the initial acetate concentration is 0.200 M. Let x equal the amount of acetate that reacts with water:

• Initial: [CH3CO2-] = 0.200, [OH-] ≈ 0, [CH3CO2H] = 0
• Change: -x, +x, +x
• Equilibrium: 0.200 – x, x, x

Insert these into the Kb expression:

Kb = [CH3CO2H][OH-] / [CH3CO2-] = x^2 / (0.200 – x)

Step 3: Solve for hydroxide concentration

Because Kb is small, many classroom solutions use the approximation 0.200 – x ≈ 0.200. That gives:

x = square root of (Kb × 0.200)

x = square root of (5.56 × 10^-10 × 0.200) = square root of 1.11 × 10^-10

x ≈ 1.05 × 10^-5 M

So the hydroxide concentration is approximately 1.05 × 10^-5 M. If you solve the quadratic exactly, you get essentially the same answer at this concentration because the degree of hydrolysis is tiny relative to the 0.200 M starting concentration.

Step 4: Convert hydroxide concentration into pOH and pH

1. pOH = -log(1.05 × 10^-5) ≈ 4.98
2. pH = 14.00 – 4.98 = 9.02

This means the pH of a 0.200 M NaCH3CO2 solution is about 9.02 at 25 degrees C when you use Ka = 1.8 × 10^-5 for acetic acid.

Bottom line: For the standard textbook problem, the pH is approximately 9.02. Small variations in the chosen Ka value or temperature can shift the result slightly, often by only a few hundredths of a pH unit.

Why sodium acetate makes water basic

This problem is a classic example of salt hydrolysis. Salts can be neutral, acidic, or basic in water depending on the strength of the parent acid and base. Sodium acetate comes from sodium hydroxide, a strong base, and acetic acid, a weak acid. The strong-base cation does not significantly affect pH, but the weak-acid conjugate base does. The acetate ion can accept a proton from water, generating hydroxide. That is the central chemical reason the solution is basic.

Salt	Parent acid	Parent base	Expected aqueous behavior	Typical pH trend at moderate concentration
NaCl	HCl, strong acid	NaOH, strong base	Essentially neutral	Near 7.00
NH4Cl	HCl, strong acid	NH3, weak base	Acidic	Below 7
NaCH3CO2	CH3CO2H, weak acid	NaOH, strong base	Basic	Above 7
Na2CO3	H2CO3, weak acid	NaOH, strong base	More strongly basic	Often well above 10

Exact method vs approximation

Students often ask whether the approximation is acceptable. In this problem, yes, it is. A good checkpoint is the 5 percent rule. If the change x is less than 5 percent of the initial concentration, then replacing 0.200 – x with 0.200 is usually justified.

For this solution, x ≈ 1.05 × 10^-5. Compared with 0.200 M, that is a tiny fraction:

(1.05 × 10^-5 / 0.200) × 100 ≈ 0.0053%

That is vastly below 5 percent, so the approximation is excellent. Still, using the exact quadratic solution is a nice way to confirm the answer and avoid approximation limits in more concentrated or more weakly buffered systems.

Parameter	Approximation method	Exact quadratic method	Interpretation
Kb used	5.56 × 10^-10	5.56 × 10^-10	Same chemistry input
[OH-]	About 1.05 × 10^-5 M	About 1.05 × 10^-5 M	Difference is negligible here
pOH	About 4.98	About 4.98	Both routes agree
pH	About 9.02	About 9.02	Standard final answer

Important constants and data behind the calculation

Chemistry calculations depend on accepted equilibrium constants and the reference temperature. For the sodium acetate problem, the most important values are the ion-product constant of water and the acid dissociation constant for acetic acid. In many educational settings, Kw = 1.0 × 10^-14 and Ka = 1.8 × 10^-5 are the standard values at 25 degrees C. These values are representative and are common in introductory and analytical chemistry materials.

For authoritative reference material, consult university and government sources such as:

• LibreTexts Chemistry for equilibrium tutorials used by many colleges
• U.S. Environmental Protection Agency for pH fundamentals and water chemistry context
• NIST Chemistry WebBook for chemical reference data

Common mistakes when solving sodium acetate pH problems

1. Using Ka directly instead of converting to Kb

The acetate ion is a base in water, not an acid. Students sometimes plug acetic acid Ka directly into an expression for pH without converting through Kb = Kw / Ka. That leads to the wrong type of equilibrium and an incorrect pH.

2. Forgetting that sodium is a spectator ion

In this problem, Na⁺ does not meaningfully hydrolyze water. The chemistry comes from acetate. If you focus on the spectator ion, you will miss the real source of basicity.

3. Solving for pH before finding pOH

Because the hydrolysis reaction creates hydroxide, the direct quantity obtained from the equilibrium calculation is [OH-]. The proper route is usually hydroxide concentration, then pOH, then pH.

4. Misreading the formula NaCH3CO2

This formula is another way of writing sodium acetate, often also expressed as CH₃COONa. Both formulas represent the same compound. The acetate ion is CH₃COO^–, which is equivalent to CH₃CO₂^–.

5. Ignoring temperature assumptions

The familiar relation pH + pOH = 14.00 is exact only at 25 degrees C under standard assumptions. Many teaching problems imply 25 degrees C unless told otherwise, but in advanced settings this detail matters.

Short worked example for the exact problem

1. Given: C = 0.200 M, Ka = 1.8 × 10^-5, Kw = 1.0 × 10^-14
2. Find base constant: Kb = 1.0 × 10^-14 / 1.8 × 10^-5 = 5.56 × 10^-10
3. Set equilibrium: Kb = x^2 / (0.200 – x)
4. Approximate: x = square root of (5.56 × 10^-10 × 0.200) = 1.05 × 10^-5
5. Compute pOH: pOH = 4.98
6. Compute pH: pH = 14.00 – 4.98 = 9.02

How concentration changes the pH of sodium acetate solutions

As sodium acetate concentration increases, the equilibrium hydroxide concentration generally increases too, so pH rises modestly. However, because hydroxide depends on the square root of concentration in the approximation, pH does not increase linearly with molarity. For example, moving from 0.020 M to 0.200 M increases concentration by a factor of 10, but hydroxide rises by only about a factor of square root of 10, and pH rises by roughly 0.5 units. This is a useful conceptual point when interpreting weak-base hydrolysis.

That trend helps explain why even a fairly concentrated 0.200 M sodium acetate solution is not extremely basic. Sodium acetate is a weak base because its conjugate acid, acetic acid, is only weakly acidic. In practical terms, that means the pH lands in the mildly basic range near 9 rather than the strongly basic range above 12 associated with strong bases like sodium hydroxide.

Why this result matters in lab and industry

Sodium acetate appears in buffer preparation, analytical chemistry, biochemistry workflows, textile processing, food applications, and teaching laboratories. Knowing how to calculate its pH is important because it affects indicator choice, reaction conditions, enzyme stability, and metal-ion behavior. In aqueous systems, a pH near 9 can alter solubility, reaction rate, and the charge state of molecules. Even if the textbook problem seems simple, the underlying method is foundational for understanding real chemical systems.

Final answer

If you are asked to calculate the pH of a 0.200 M NaCH3CO2 solution at 25 degrees C using Ka = 1.8 × 10^-5 for acetic acid, the expected answer is:

pH ≈ 9.02

This result comes from acetate hydrolysis in water, with Kb = 5.56 × 10^-10, [OH-] ≈ 1.05 × 10^-5 M, and pOH ≈ 4.98.