Calculate the pH of a 0.100 m NaCH3CO2 Solution
Use this premium acetate hydrolysis calculator to estimate the pH of aqueous sodium acetate, NaCH3CO2. The calculator defaults to a 0.100 m solution and standard acetic acid dissociation data, then shows the hydrolysis constant, hydroxide concentration, pOH, and final pH with a responsive chart.
Calculator
Results
pH = 8.87
- Kb = Kw / Ka
- [OH-] approximately sqrt(Kb multiplied by C)
- Default 0.100 m sodium acetate is basic
pH Trend vs Sodium Acetate Concentration
This chart shows how the estimated pH changes as acetate concentration varies around your selected value while keeping Ka and Kw fixed.
Expert Guide: How to Calculate the pH of a 0.100 m NaCH3CO2 Solution
To calculate the pH of a 0.100 m NaCH3CO2 solution, you are solving a classic weak-base hydrolysis problem. Sodium acetate, written as NaCH3CO2 or more commonly NaC2H3O2, is the sodium salt of acetic acid. In water it dissociates essentially completely into Na+ and CH3CO2-. The sodium ion does not affect pH in any meaningful way under ordinary classroom assumptions, but the acetate ion does. Because acetate is the conjugate base of a weak acid, it reacts with water to produce a small amount of hydroxide. That hydroxide makes the solution basic, so the pH is greater than 7.
The key idea is simple: sodium acetate is not itself a strong base, but its anion hydrolyzes. Once dissolved, the relevant equilibrium is CH3CO2- + H2O ⇌ CH3CO2H + OH-. This tells you that hydroxide is produced, so the route to pH is to determine Kb for acetate, estimate the equilibrium hydroxide concentration, convert to pOH, and finally calculate pH. For a 0.100 m solution, using common 25 degrees C values for acetic acid Ka of 1.8 × 10^-5 and water Kw of 1.0 × 10^-14, the pH comes out to about 8.87. That is the standard textbook answer when molality is treated approximately like molarity in dilute solution.
Step 1: Identify the Acid Base Role of the Salt
Whenever you are given a salt and asked for pH, start by identifying the parent acid and base. Sodium acetate comes from sodium hydroxide, a strong base, and acetic acid, a weak acid. Salts formed from a strong base and a weak acid produce basic solutions. That quick classification already tells you the pH must be above neutral.
- Na+ is the cation from a strong base and is essentially pH neutral.
- CH3CO2- is the conjugate base of acetic acid.
- Conjugate bases of weak acids react with water to make OH-.
- Therefore, the solution is basic.
Step 2: Write the Hydrolysis Reaction
After sodium acetate dissolves, the equilibrium that matters is the acetate hydrolysis reaction:
CH3CO2- + H2O ⇌ CH3CO2H + OH-
This is the reaction for a weak base in water. The equilibrium constant for this process is Kb, not Ka. However, Kb is easy to obtain from the acid dissociation constant of acetic acid because conjugate acid base pairs are linked by the water ion product.
Step 3: Convert Ka to Kb
The relationship between Ka and Kb is:
Kb = Kw / Ka
Using typical 25 degrees C constants:
- Kw = 1.0 × 10^-14
- Ka for acetic acid = 1.8 × 10^-5
So:
Kb = (1.0 × 10^-14) / (1.8 × 10^-5) = 5.56 × 10^-10
This very small Kb shows that acetate is only a weak base, which means it generates only a small amount of hydroxide even when the initial concentration is fairly substantial.
| Quantity | Symbol | Typical 25 degrees C Value | Meaning |
|---|---|---|---|
| Acetic acid dissociation constant | Ka | 1.8 × 10^-5 | Strength of CH3CO2H as a weak acid |
| Water ion product | Kw | 1.0 × 10^-14 | Links Ka and Kb at 25 degrees C |
| Acetate base dissociation constant | Kb | 5.56 × 10^-10 | Strength of CH3CO2- as a weak base |
| Acetic acid pKa | pKa | 4.76 | Widely used for buffer and equilibrium work |
Step 4: Set Up the Equilibrium Expression
If the sodium acetate concentration is 0.100 m and the solution is dilute enough that you treat molality as approximately equal to molarity, then the initial acetate concentration is taken as 0.100 M. Let x represent the hydroxide generated by hydrolysis.
- Initial: [CH3CO2-] = 0.100, [CH3CO2H] = 0, [OH-] = 0
- Change: [CH3CO2-] decreases by x, [CH3CO2H] increases by x, [OH-] increases by x
- Equilibrium: [CH3CO2-] = 0.100 – x, [CH3CO2H] = x, [OH-] = x
Substitute into the Kb expression:
Kb = x² / (0.100 – x)
Because Kb is very small, x will be much smaller than 0.100, so the standard weak base approximation is valid:
0.100 – x ≈ 0.100
Then:
x² / 0.100 = 5.56 × 10^-10
x² = 5.56 × 10^-11
x = 7.45 × 10^-6
This x value is the hydroxide concentration:
[OH-] = 7.45 × 10^-6 M
Step 5: Convert Hydroxide Concentration to pH
Once [OH-] is known, calculate pOH:
pOH = -log(7.45 × 10^-6) = 5.13
At 25 degrees C:
pH = 14.00 – 5.13 = 8.87
So the calculated pH of a 0.100 m NaCH3CO2 solution is:
pH ≈ 8.87
Why the Approximation Works So Well
A useful self-check is the 5 percent rule. Compare x with the initial concentration:
(7.45 × 10^-6 / 0.100) × 100 = 0.00745 percent
That is far below 5 percent, so ignoring x in the denominator is completely justified. In practical teaching and homework contexts, the square root approximation is more than accurate enough.
Molality vs Molarity: Does 0.100 m Matter?
The problem statement uses molality, written with a lowercase m, not molarity, written with an uppercase M. Strictly speaking, molality is moles of solute per kilogram of solvent, while molarity is moles of solute per liter of solution. In concentrated or high precision work, this distinction matters. However, for dilute aqueous solutions near room temperature, a 0.100 m sodium acetate solution is often treated as approximately 0.100 M for introductory calculations. That is why many textbook solutions use the same weak-base setup shown above.
If you wanted a more rigorous answer, you would need solution density and potentially activity coefficients. Those refinements are important in analytical chemistry, geochemistry, and process calculations, but they are usually beyond the scope of the standard general chemistry question.
| NaCH3CO2 Concentration | Estimated [OH-] | Estimated pOH | Estimated pH |
|---|---|---|---|
| 0.010 | 2.36 × 10^-6 | 5.63 | 8.37 |
| 0.050 | 5.27 × 10^-6 | 5.28 | 8.72 |
| 0.100 | 7.45 × 10^-6 | 5.13 | 8.87 |
| 0.500 | 1.67 × 10^-5 | 4.78 | 9.22 |
| 1.000 | 2.36 × 10^-5 | 4.63 | 9.37 |
Common Mistakes Students Make
- Using Ka directly instead of converting to Kb for the acetate ion.
- Assuming sodium acetate is neutral because it contains sodium, which is incorrect.
- Forgetting that pH must be greater than 7 because acetate generates OH-.
- Mixing up pOH and pH after calculating hydroxide concentration.
- Treating acetic acid as a strong acid or acetate as a strong base, which it is not.
Shortcut Formula for Weak Base Salts
For a salt containing the conjugate base of a weak acid, you can often use a shortcut. Once you have Kb and the initial concentration C, estimate hydroxide concentration as:
[OH-] ≈ √(Kb × C)
Then compute pOH and pH. For sodium acetate at 0.100 concentration units:
[OH-] ≈ √((5.56 × 10^-10)(0.100)) = 7.45 × 10^-6
This immediately leads to the same final pH of about 8.87.
Real Chemistry Context
Sodium acetate is important far beyond homework problems. It is used in buffering systems, textile processing, food applications, and laboratory preparations. In acetic acid acetate buffer systems, the pKa of acetic acid near 4.76 makes the pair especially useful around mildly acidic pH values. By contrast, a pure sodium acetate solution without added acetic acid is mildly basic because the conjugate base hydrolyzes in water. This distinction is worth remembering: a buffer and a simple salt solution are not the same thing, even though they involve the same conjugate pair.
Quick Final Answer
If your instructor expects the standard weak-base salt method at 25 degrees C, then the solution is straightforward:
- Find Kb = Kw / Ka = 1.0 × 10^-14 / 1.8 × 10^-5 = 5.56 × 10^-10
- Use [OH-] ≈ √(KbC) = √((5.56 × 10^-10)(0.100)) = 7.45 × 10^-6
- pOH = 5.13
- pH = 14.00 – 5.13 = 8.87
Therefore, the pH of a 0.100 m NaCH3CO2 solution is approximately 8.87.
Authoritative Reference Sources
For readers who want to verify equilibrium constants, acid base theory, or water properties from authoritative sources, these references are excellent starting points: