Calculate the pH of 0.10 m NaC2H3O2 Solution
Use this interactive sodium acetate calculator to estimate the pH of an aqueous NaC2H3O2 solution from concentration and acetic acid dissociation data. For dilute solutions, this is the standard weak-base hydrolysis approach used in general chemistry.
- Default concentration is set to 0.10 for the common textbook problem.
- Default Ka is 1.8 × 10-5 for acetic acid at 25 degrees C.
- The calculator assumes a dilute aqueous solution and treats the entered concentration as the formal acetate concentration.
Hydrolysis overview chart
How to calculate the pH of 0.10 m NaC2H3O2 solution
Sodium acetate, written as NaC2H3O2 or CH3COONa, is the sodium salt of acetic acid. When this salt dissolves in water, it dissociates essentially completely into sodium ions and acetate ions. The sodium ion is a spectator ion in acid-base chemistry, but the acetate ion is the conjugate base of a weak acid. That means acetate reacts with water to produce a small amount of hydroxide, making the solution basic. This is why a 0.10 sodium acetate solution has a pH greater than 7.
In a typical textbook problem, the notation may appear as 0.10 m or 0.10 M. Strictly speaking, lowercase m means molality, while uppercase M means molarity. For dilute aqueous chemistry problems, these are often treated similarly as an approximation because the density is close to that of water. This calculator follows that classroom convention unless you are doing high-precision physical chemistry work. The key chemistry remains the same: acetate hydrolyzes in water because it is the conjugate base of a weak acid.
Step 1: Write the hydrolysis reaction
The chemically important equilibrium is:
This tells us that acetate accepts a proton from water and forms acetic acid plus hydroxide ion. Because hydroxide is produced, the pH rises above neutral.
Step 2: Find Kb from Ka
Acetate is the conjugate base of acetic acid, so its base dissociation constant is related to the acid dissociation constant by:
At 25 degrees C, a commonly used value for acetic acid is Ka = 1.8 × 10-5, and Kw = 1.0 × 10-14. Therefore:
Step 3: Set up the equilibrium expression
If the initial concentration of acetate is 0.10 and x is the amount that reacts with water, then at equilibrium:
- [C2H3O2-] = 0.10 – x
- [HC2H3O2] = x
- [OH-] = x
The equilibrium expression for Kb is:
Because Kb is very small, x is much smaller than 0.10, so many instructors allow the approximation 0.10 – x ≈ 0.10. That gives:
Since x equals the hydroxide concentration:
Step 4: Convert hydroxide concentration to pOH and pH
Now calculate pOH:
Then use the relationship:
So the pH of a 0.10 sodium acetate solution at 25 degrees C is approximately 8.87.
Why sodium acetate is basic
This is one of the most important patterns in introductory acid-base chemistry. Salts can be neutral, acidic, or basic depending on the acid and base from which they are formed. Sodium acetate comes from NaOH, a strong base, and HC2H3O2, a weak acid. The cation from a strong base, Na+, does not significantly hydrolyze. The anion from a weak acid, acetate, does hydrolyze. Therefore the solution becomes basic.
By contrast, a salt such as sodium chloride is formed from a strong acid and strong base, so neither ion appreciably hydrolyzes and the pH stays near 7. A salt like ammonium chloride behaves differently because NH4+ is the conjugate acid of a weak base, making the solution acidic. Understanding this pattern helps you predict pH before doing any calculations.
Worked example for 0.10 NaC2H3O2
- Write the base hydrolysis reaction for acetate.
- Calculate Kb using Kb = Kw / Ka.
- Substitute the formal concentration into the ICE setup.
- Find [OH-] from either the approximation or the quadratic formula.
- Compute pOH and then pH.
Using a more rigorous quadratic treatment changes the answer only slightly because x is tiny relative to 0.10. That is why the square root method is commonly taught. Still, the calculator above lets you compare both methods. In formal chemistry reporting, it is good practice to check whether the approximation is valid by confirming that x is less than about 5 percent of the initial concentration. Here, x is far below that threshold, so the approximation is excellent.
| Quantity | Typical value at 25 degrees C | Meaning in this problem |
|---|---|---|
| Ka of acetic acid | 1.8 × 10-5 | Measures how strongly acetic acid donates H+ |
| Kw of water | 1.0 × 10-14 | Relates [H+] and [OH–] in water |
| Kb of acetate | 5.56 × 10-10 | Measures how weakly acetate accepts H+ |
| Calculated [OH–] at 0.10 | 7.45 × 10-6 approximately | Produced by hydrolysis of acetate in water |
| Calculated pH at 0.10 | 8.87 approximately | Shows the solution is mildly basic |
How concentration changes the pH
A stronger formal concentration of sodium acetate gives more acetate available to hydrolyze, so the hydroxide concentration rises and the pH increases. However, the increase is not linear. Because the calculation uses a square root relationship for weak base hydrolysis, a tenfold concentration increase does not raise the pH by a full unit. This is a good reminder that pH is logarithmic and that weak electrolyte equilibria often scale with roots and logs rather than simple direct proportions.
| NaC2H3O2 concentration | Approximate [OH-] | Approximate pH |
|---|---|---|
| 0.001 | 7.45 × 10-7 | 7.87 |
| 0.010 | 2.36 × 10-6 | 8.37 |
| 0.10 | 7.45 × 10-6 | 8.87 |
| 1.00 | 2.36 × 10-5 | 9.37 |
Common mistakes when solving this problem
- Using Ka directly instead of Kb. Acetate is a base in water, so you need Kb, not Ka.
- Assuming the salt is neutral. Only salts from a strong acid and strong base are typically neutral.
- Forgetting the pOH step. The equilibrium gives [OH-], so calculate pOH first, then pH.
- Confusing molality and molarity. In many basic exercises they are treated similarly for dilute aqueous solutions, but they are not exactly the same unit.
- Ignoring temperature. Ka and Kw change with temperature, so the final pH can shift if you move away from 25 degrees C.
Quadratic method versus approximation
The approximation method is fast and usually accurate when Kb is small and the concentration is not extremely low. The quadratic method is more rigorous because it does not ignore x in the denominator. For this particular sodium acetate problem, both methods produce practically the same pH to ordinary reporting precision. If you are taking general chemistry, your instructor may expect the approximation unless specifically asked to solve exactly. If you are in analytical chemistry or physical chemistry, the quadratic form is a better habit.
Exact setup using the quadratic equation
Starting from:
Rearrange to:
Then solve:
The physically meaningful root is the positive one. Once x is found, it is again equal to [OH-]. Then compute pOH and pH in the standard way.
Where these values come from
Acid dissociation constants and related water ion-product values are established from experimental equilibrium measurements. Chemistry students often use rounded textbook values such as Ka = 1.8 × 10-5 for acetic acid because they are convenient and sufficiently accurate for learning. If you compare different textbooks or reference tables, you may see slight numerical differences like 1.75 × 10-5 or 1.76 × 10-5. That can shift the final pH by a few hundredths, which is normal.
For more background and reliable reference material, consult authoritative sources such as the National Institute of Standards and Technology, chemistry course materials from university chemistry education resources, and educational pages from institutions like university chemistry departments. If you need strictly .gov or .edu references, excellent starting points include webbook.nist.gov, chem.wisc.edu, and chemistry.mit.edu.
Final answer for the classic problem
If you are asked, “calculate the pH of 0.10 m NaC2H3O2 solution,” the standard answer at 25 degrees C using Ka = 1.8 × 10-5 is:
This value reflects mild basicity caused by acetate hydrolysis. It is not strongly basic because acetate is only a weak base. The calculator above can help you verify the answer instantly and explore how the result changes if concentration, Ka, or Kw are adjusted.