Calculate the pH of a 0.800 m NaC₂H₃O₂ Solution
Use this premium chemistry calculator to estimate the pH of sodium acetate in water by weak-base hydrolysis. Enter the acetate concentration, acetic acid Ka, and temperature assumptions to generate pH, pOH, hydroxide concentration, and a visual chart.
How to calculate the pH of a 0.800 m NaC₂H₃O₂ solution
When you are asked to calculate the pH of a 0.800 m NaC₂H₃O₂ solution, you are dealing with a salt of a weak acid and a strong base. Sodium acetate, NaC₂H₃O₂, dissolves completely in water to give sodium ions and acetate ions. The sodium ion is essentially a spectator ion in this context, but the acetate ion behaves as a weak base because it is the conjugate base of acetic acid. That means the pH of the solution will be greater than 7 at standard conditions.
The important chemical idea is hydrolysis. Acetate reacts with water according to the equilibrium:
This equilibrium produces hydroxide ions, which raise the pH. To solve the problem correctly, you generally start with the acid dissociation constant of acetic acid, then convert it to the base dissociation constant for acetate using the relationship Kb = Kw / Ka. At 25°C, a common textbook value for the acid dissociation constant of acetic acid is 1.8 × 10-5, and the ion-product constant of water is 1.0 × 10-14.
Step 1: Write the dissociation and hydrolysis reactions
Sodium acetate is a soluble ionic compound, so it dissociates essentially completely:
- NaC₂H₃O₂ → Na+ + C₂H₃O₂–
- Only C₂H₃O₂– significantly affects pH
- Acetate acts as a weak base in water
Since the solution concentration is given as 0.800 m, many introductory problems treat that as approximately 0.800 mol per liter in dilute aqueous work unless density corrections are specifically required. This calculator follows that standard educational assumption. In more advanced physical chemistry, molality and molarity are distinct, but for a typical pH problem like this, the approximation is usually accepted unless your instructor specifies otherwise.
Step 2: Convert Ka to Kb
Acetate is the conjugate base of acetic acid, so its basicity is determined from acetic acid’s Ka:
- Take Ka for acetic acid = 1.8 × 10-5
- Take Kw = 1.0 × 10-14
- Compute Kb = Kw / Ka
Using those values:
This is a very small Kb, which tells us acetate is only a weak base. Even so, at 0.800 concentration there is enough acetate present to generate a measurable hydroxide concentration.
Step 3: Set up the ICE framework
For the hydrolysis reaction:
Let the initial acetate concentration be 0.800. Initially, the hydroxide formed by hydrolysis is negligible compared with the amount produced by the equilibrium shift, so we write the change as x:
- Initial: [C₂H₃O₂–] = 0.800, [HC₂H₃O₂] = 0, [OH–] = 0
- Change: -x, +x, +x
- Equilibrium: 0.800 – x, x, x
Substitute into the Kb expression:
Because Kb is very small, x is much smaller than 0.800, so 0.800 – x is approximately 0.800. That gives the standard approximation:
Evaluating this gives:
Step 4: Convert hydroxide concentration to pOH and pH
Once hydroxide concentration is known, use:
- pOH = -log[OH–]
- pH = 14.00 – pOH
Substituting [OH–] ≈ 2.11 × 10-5:
pH ≈ 9.32
Therefore, the pH of a 0.800 m sodium acetate solution is approximately 9.32 at 25°C when using the usual textbook constants and the weak-base approximation.
Why sodium acetate makes the solution basic
Sodium acetate comes from the neutralization of acetic acid, a weak acid, with sodium hydroxide, a strong base. Salts formed from a weak acid and a strong base generally yield basic solutions because the anion can react with water to produce hydroxide. In contrast, salts of strong acids and strong bases, such as sodium chloride, are usually neutral because neither ion hydrolyzes appreciably.
Acetate has enough affinity for protons that it partially pulls a proton from water, making acetic acid and hydroxide. The reaction does not go to completion because acetate is still only a weak base, but even partial hydrolysis shifts the pH upward.
Approximation versus quadratic solution
In weak acid and weak base chemistry, the square-root shortcut is widely used because it is fast and usually very accurate when x is much smaller than the initial concentration. For 0.800 sodium acetate, the approximation works extremely well. A quadratic solution gives nearly the same answer because the degree of hydrolysis is tiny compared with the starting concentration.
| Method | Formula Used | Calculated [OH-] | Calculated pH | Comment |
|---|---|---|---|---|
| Approximation | x = √(KbC) | 2.11 × 10-5 | 9.32 | Fast and standard for textbook work |
| Quadratic | x² + Kbx – KbC = 0 | 2.11 × 10-5 | 9.32 | More rigorous, almost identical result here |
Comparison with other acetate concentrations
The pH increases as acetate concentration increases, but not in a simple one-to-one way. Because hydroxide concentration depends on the square root of concentration in the weak-base approximation, a 100-fold increase in concentration increases [OH-] by about 10 times, changing pH by roughly 1 unit. That pattern is helpful when doing quick estimation.
| Sodium acetate concentration | Assumed Ka of acetic acid | Kb of acetate | Estimated pH at 25°C | Interpretation |
|---|---|---|---|---|
| 0.0800 | 1.8 × 10-5 | 5.56 × 10-10 | 8.82 | Mildly basic |
| 0.800 | 1.8 × 10-5 | 5.56 × 10-10 | 9.32 | Clearly basic, typical answer for this problem |
| 1.60 | 1.8 × 10-5 | 5.56 × 10-10 | 9.47 | Higher pH, but still weak-base behavior |
Common student mistakes when calculating the pH of sodium acetate
- Using Ka directly instead of converting to Kb.
- Treating sodium acetate as if it were a strong base.
- Forgetting that the question asks for pH, not pOH.
- Writing the wrong hydrolysis reaction.
- Confusing molality with molarity without checking the problem’s assumptions.
- Rounding too early during logarithm calculations.
The biggest error is often conceptual. Sodium acetate itself does not contain free OH– ions the way sodium hydroxide does. Its basicity is indirect, arising from equilibrium hydrolysis of acetate in water. Once you identify that, the calculation becomes straightforward.
Does 0.800 m mean the same thing as 0.800 M?
Strictly speaking, no. Molality, symbol m, means moles of solute per kilogram of solvent, while molarity, symbol M, means moles of solute per liter of solution. They are not identical because solution volume changes with temperature and composition. However, many general chemistry pH problems use m and M interchangeably in practice when discussing dilute aqueous solutions or when density information is not supplied. If you are in an analytical chemistry or physical chemistry course, check whether your instructor expects a density correction.
For the current problem, using 0.800 as the working concentration gives the accepted classroom result near pH 9.32. The difference introduced by a density correction is usually much smaller than the uncertainty coming from rounded equilibrium constants in a typical textbook exercise.
Real chemistry context: acetate in laboratory and biological systems
Acetate salts are common in labs because they are inexpensive, water-soluble, and useful in buffer preparation. Sodium acetate is often paired with acetic acid to create acetate buffers in the mildly acidic range. By itself, however, sodium acetate gives a basic solution because only the conjugate base is present. This distinction is important in biochemistry, food science, and pharmaceutical formulation, where pH controls reaction rates, stability, enzyme activity, and solubility.
In industrial and educational settings, sodium acetate is also valued for its relatively low hazard profile compared with strong bases. But low hazard does not mean no effect on pH. At moderate concentration, the solution remains measurably basic and can influence indicator color, precipitation equilibria, and buffer capacity calculations.
Quick expert summary
- Identify sodium acetate as the salt of a weak acid and strong base.
- Recognize acetate as a weak base.
- Calculate Kb from Ka using Kb = Kw / Ka.
- Use x = √(KbC) to estimate [OH-].
- Find pOH from -log[OH-].
- Find pH from 14.00 – pOH.
With Ka = 1.8 × 10-5 and C = 0.800, the result is: pH ≈ 9.32.
Authoritative references for equilibrium constants and aqueous chemistry
For additional verification and deeper reading, review these authoritative sources:
- LibreTexts Chemistry educational resources
- U.S. Environmental Protection Agency
- NIST Chemistry WebBook
Note: NIST and EPA are authoritative U.S. government sources, while university chemistry departments and educational repositories can provide supporting explanations of acid-base equilibria. If your course gives a different Ka value for acetic acid, use your instructor’s value because the final pH will shift slightly.