Calculate the pH of a 1.60 M NaCH3CO2 Solution
Use this interactive calculator to find the pH of sodium acetate, NaCH3CO2, by modeling acetate ion hydrolysis in water at 25 degrees C. The default setup is preloaded for a 1.60 M solution.
How to calculate the pH of a 1.60 M NaCH3CO2 solution
Sodium acetate, written as NaCH3CO2 or sometimes NaC2H3O2, is the sodium salt of acetic acid. In water, sodium ions act mainly as spectator ions, while the acetate ion, CH3CO2–, behaves as a weak base because it is the conjugate base of the weak acid acetic acid. That means a solution of sodium acetate is not neutral. Instead, it hydrolyzes water to produce a small amount of OH–, which raises the pH above 7. For the specific case of 1.60 M NaCH3CO2 at 25 degrees C, the pH is mildly basic and comes out to approximately 9.47 to 9.48 depending on the exact constant values used.
The chemistry behind this problem is standard weak-base hydrolysis. Once NaCH3CO2 dissolves, it dissociates essentially completely:
NaCH3CO2(aq) → Na+(aq) + CH3CO2–(aq)
Next, the acetate ion reacts with water according to the equilibrium:
CH3CO2–(aq) + H2O(l) ⇌ CH3CO2H(aq) + OH–(aq)
This second equation is the one that controls pH. Since acetate is a weak base, we do not assume complete reaction. Instead, we relate its basicity to the acid dissociation constant of acetic acid. The key connection is:
Kb = Kw / Ka
At 25 degrees C, a common textbook value for acetic acid is Ka = 1.8 × 10-5 and for water Kw = 1.0 × 10-14. Therefore:
Kb = (1.0 × 10-14) / (1.8 × 10-5) = 5.56 × 10-10
Step-by-step setup with an ICE table
If the initial acetate concentration is 1.60 M, then an ICE table for the hydrolysis reaction looks like this:
- Initial: [CH3CO2–] = 1.60, [CH3CO2H] = 0, [OH–] = 0
- Change: [-x, +x, +x]
- Equilibrium: [CH3CO2–] = 1.60 – x, [CH3CO2H] = x, [OH–] = x
Now substitute into the equilibrium expression for Kb:
Kb = [CH3CO2H][OH–] / [CH3CO2–] = x2 / (1.60 – x)
Because Kb is very small, the amount hydrolyzed is tiny compared with 1.60 M, so the common approximation is:
1.60 – x ≈ 1.60
This gives:
x = √(KbC) = √[(5.56 × 10-10)(1.60)] = 2.98 × 10-5 M
Since x is the hydroxide concentration, we calculate:
- [OH–] = 2.98 × 10-5 M
- pOH = -log(2.98 × 10-5) = 4.53
- pH = 14.00 – 4.53 = 9.47
That is the standard answer most instructors expect for the pH of a 1.60 M sodium acetate solution at 25 degrees C when using Ka = 1.8 × 10-5. If you use a slightly different literature value for acetic acid, such as Ka = 1.75 × 10-5, the pH changes slightly but remains essentially the same to two decimal places.
Why NaCH3CO2 is basic instead of neutral
Students often wonder why a salt can make water acidic or basic. The answer depends on the parent acid and parent base used to make the salt:
- A salt from a strong acid + strong base is usually neutral.
- A salt from a weak acid + strong base is usually basic.
- A salt from a strong acid + weak base is usually acidic.
- A salt from a weak acid + weak base requires comparing Ka and Kb.
Sodium acetate comes from acetic acid (weak acid) and sodium hydroxide (strong base). The sodium ion does not significantly affect pH, but acetate does. Because acetate can accept a proton from water, it generates OH– and pushes the solution basic.
Shortcut using pKa and pKb
You can also reach the same answer using pKa and pKb. For acetic acid, pKa is about 4.74 to 4.76 depending on the source. Since pKw is 14.00 at 25 degrees C:
pKb = 14.00 – pKa ≈ 9.24 to 9.26
For a weak base, the rough relation
pOH ≈ 1/2 (pKb – log C)
works well when dissociation is small. For C = 1.60 M and pKb ≈ 9.26:
pOH ≈ 1/2 [9.26 – log(1.60)] ≈ 4.53
so pH ≈ 9.47.
Comparison table: pH of sodium acetate at different concentrations
The concentration matters because more acetate generally means a larger hydroxide concentration, although pH changes on a logarithmic scale rather than a linear one. The following values use Ka = 1.8 × 10-5 and Kw = 1.0 × 10-14 at 25 degrees C with the weak-base approximation.
| NaCH3CO2 concentration (M) | Kb of acetate | Estimated [OH-] (M) | Estimated pOH | Estimated pH |
|---|---|---|---|---|
| 0.010 | 5.56 × 10-10 | 2.36 × 10-6 | 5.63 | 8.37 |
| 0.100 | 5.56 × 10-10 | 7.45 × 10-6 | 5.13 | 8.87 |
| 0.500 | 5.56 × 10-10 | 1.67 × 10-5 | 4.78 | 9.22 |
| 1.00 | 5.56 × 10-10 | 2.36 × 10-5 | 4.63 | 9.37 |
| 1.60 | 5.56 × 10-10 | 2.98 × 10-5 | 4.53 | 9.47 |
| 2.00 | 5.56 × 10-10 | 3.33 × 10-5 | 4.48 | 9.52 |
This table shows a useful pattern: increasing sodium acetate concentration increases pH, but not dramatically. Even when concentration rises from 0.10 M to 1.60 M, pH increases by only about 0.60 units because pH depends on the logarithm of the hydroxide concentration.
Data table: acid-base constants commonly used in this problem
Different textbooks and data sources may round constants slightly differently. These small variations explain why one source may report pH = 9.47 while another may show 9.48.
| Quantity | Common value | Interpretation | Effect on final pH |
|---|---|---|---|
| Ka for acetic acid | 1.75 × 10-5 to 1.80 × 10-5 | Measures how strongly acetic acid donates H+ | Smaller Ka means stronger conjugate base and slightly higher pH |
| pKa for acetic acid | 4.74 to 4.76 | Logarithmic form of Ka | Common for shortcut calculations |
| Kw at 25 degrees C | 1.0 × 10-14 | Water autoionization constant | Used to convert Ka into Kb |
| Kb for acetate | 5.56 × 10-10 to 5.71 × 10-10 | Measures how strongly acetate generates OH- | Directly sets [OH-] and pH |
Exact solution versus approximation
For this problem, the approximation works very well because x is extremely small relative to 1.60 M. You can verify that by checking the percent ionization:
Percent ionization = (x / 1.60) × 100 = (2.98 × 10-5 / 1.60) × 100 ≈ 0.0019%
That is far below the common 5 percent rule, so the approximation is fully justified. Still, the exact quadratic approach is useful for advanced coursework or for very dilute solutions. The exact form comes from:
x2 + Kb x – KbC = 0
so
x = [-Kb + √(Kb2 + 4KbC)] / 2
When you insert Kb = 5.56 × 10-10 and C = 1.60, the exact x is practically identical to the approximation at the precision usually reported in general chemistry.
Common mistakes when solving this pH problem
- Treating sodium acetate as neutral. Because it is a salt, some learners assume pH = 7. This is incorrect because acetate is the conjugate base of a weak acid.
- Using Ka directly instead of converting to Kb. The reacting species is acetate, not acetic acid, so the correct equilibrium constant is Kb.
- Forgetting that sodium is a spectator ion. Na+ does not significantly hydrolyze water in this context.
- Mixing up pH and pOH. Once you calculate [OH–], you first find pOH, then convert to pH.
- Using concentration as [OH-] directly. The full 1.60 M is the starting acetate concentration, not the hydroxide concentration.
Why the final answer is around 9.47
The result makes intuitive sense. Acetate is a weak base, so even in a fairly concentrated 1.60 M solution it creates only a modest amount of OH–. That means the solution is definitely basic, but not strongly basic. A pH of about 9.47 is far below the pH of a strong base such as 1.60 M NaOH, which would be close to pH 14. In other words, sodium acetate raises pH through equilibrium chemistry, not through complete hydroxide release.
Practical interpretation
Sodium acetate solutions are often encountered in buffer chemistry, biological sample preparation, and analytical chemistry. On its own, sodium acetate gives a basic solution. When mixed with acetic acid, however, it forms the classic acetic acid-acetate buffer system. Understanding the standalone pH of sodium acetate helps explain why the buffer resists pH change and how the conjugate base contributes to the equilibrium balance.
Authoritative references for further study
Final answer summary
To calculate the pH of a 1.60 M NaCH3CO2 solution, treat acetate as a weak base. Convert the acetic acid dissociation constant to a base dissociation constant using Kb = Kw / Ka, solve for the hydroxide concentration from the hydrolysis equilibrium, then convert from pOH to pH. Using Ka = 1.8 × 10-5 and Kw = 1.0 × 10-14, you obtain [OH–] ≈ 2.98 × 10-5 M, pOH ≈ 4.53, and pH ≈ 9.47.