Calculate The Ph Of 100.00Ml 0.182 M Naac

Use this premium sodium acetate calculator to determine the pH of a 100.00 mL, 0.182 M NaAc solution with step-by-step chemistry. The tool applies weak base hydrolysis for acetate, estimates hydroxide concentration, and visualizes the result on a pH scale chart.

Core chemistry used:
Sodium acetate is a salt of a weak acid and strong base. Acetate hydrolyzes in water:

CH₃COO^– + H₂O ⇌ CH₃COOH + OH^–

K_b = K_w / K_a, then for a weak base solution:
[OH^–] ≈ √(K_b × C)
pOH = -log[OH^–], and pH = 14.00 – pOH

Expert guide: how to calculate the pH of 100.00 mL 0.182 M NaAc

If you need to calculate the pH of 100.00 mL 0.182 M NaAc, the key idea is that sodium acetate is not a neutral salt. It is the sodium salt of acetic acid, a weak acid. Because the acetate ion is the conjugate base of acetic acid, it reacts with water to produce hydroxide ions, making the solution basic. That means the pH will be greater than 7 at standard laboratory temperature.

Chemically, NaAc is commonly written for sodium acetate, although you may also see CH₃COONa or NaC₂H₃O₂. Once it dissolves, sodium ions act as spectator ions for this calculation, while acetate ions participate in hydrolysis. The calculation therefore focuses on the acetate concentration and the acid dissociation constant of acetic acid.

Why sodium acetate gives a basic pH

Sodium acetate dissociates essentially completely in water:

CH₃COONa → Na⁺ + CH₃COO^–

The sodium ion comes from the strong base sodium hydroxide and does not significantly affect pH. The acetate ion, however, is the conjugate base of acetic acid, so it can remove a proton from water:

CH₃COO^– + H₂O ⇌ CH₃COOH + OH^–

Since hydroxide ions are produced, the solution becomes basic. This is why the pH of sodium acetate is above 7, even though there is no strong base added directly.

Step-by-step calculation for 100.00 mL of 0.182 M NaAc

1. Identify the concentration of acetate. Because sodium acetate dissociates completely, the initial acetate concentration is the same as the sodium acetate concentration:
   [CH₃COO^–] = 0.182 M
2. Use the acid dissociation constant for acetic acid. A common value at 25 degrees C is:
   K_a = 1.8 × 10^-5
3. Convert K_a to K_b for acetate.
   K_b = K_w / K_a = 1.0 × 10^-14 / 1.8 × 10^-5
   K_b ≈ 5.56 × 10^-10
4. Apply the weak base approximation. For a weak base with initial concentration C:
   [OH^–] ≈ √(K_bC)
   [OH^–] ≈ √((5.56 × 10^-10)(0.182))
   [OH^–] ≈ √(1.01192 × 10^-10)
   [OH^–] ≈ 1.006 × 10^-5 M
5. Calculate pOH.
   pOH = -log(1.006 × 10^-5) ≈ 4.997
6. Calculate pH.
   pH = 14.000 – 4.997 ≈ 9.003

Therefore, the pH of 100.00 mL 0.182 M NaAc is approximately 9.00 at 25 degrees C when K_a for acetic acid is taken as 1.8 × 10^-5.

Does the 100.00 mL volume affect the pH?

In this specific problem, the listed volume is important for finding the total number of moles, but it does not change the pH as long as the concentration remains 0.182 M. That is because pH depends on concentration, not on the absolute amount, for a simple homogeneous solution like this one.

You can still calculate the moles to better understand the system:

moles NaAc = M × V = 0.182 mol/L × 0.10000 L = 0.01820 mol

This tells you that the sample contains 0.01820 mol of sodium acetate, and therefore initially 0.01820 mol of acetate ion. But because the concentration is already given, the pH computation remains based on 0.182 M, not the volume alone.

ICE table interpretation

Many chemistry students prefer to confirm the weak base setup using an ICE table. For acetate hydrolysis:

• Initial: [CH₃COO^–] = 0.182, [CH₃COOH] = 0, [OH^–] ≈ 0
• Change: -x, +x, +x
• Equilibrium: 0.182 – x, x, x

Then:

K_b = x² / (0.182 – x)

Since K_b is very small, x is much smaller than 0.182, so 0.182 – x ≈ 0.182. This leads to:

x ≈ √(K_b × 0.182)

Here x represents [OH^–] at equilibrium. The approximation is valid because the percent ionization is tiny, on the order of a few thousandths of a percent.

Quantity	Value	Meaning
Sodium acetate concentration	0.182 M	Initial acetate concentration in solution
Volume	100.00 mL = 0.10000 L	Used to calculate total moles present
Moles of NaAc	0.01820 mol	Total dissolved sodium acetate
Ka of acetic acid	1.8 × 10^-5	Standard literature value near room temperature
Kb of acetate	5.56 × 10^-10	Hydrolysis constant of acetate in water
[OH^–]	1.006 × 10^-5 M	Hydroxide formed by base hydrolysis
pOH	4.997	Negative log of hydroxide concentration
pH	9.003	Final solution pH at 25 degrees C

How accurate is the square root shortcut?

The shortcut [OH^–] ≈ √(K_bC) is widely used because it is both fast and reliable when the weak base dissociates only slightly. In this sodium acetate case, it works very well because K_b is much smaller than the initial concentration. If you solved the full quadratic equation, the result would be extremely close to the approximation.

This is a useful lesson in analytical chemistry and general chemistry: not every equilibrium problem needs a full exact numerical solution. Once you recognize a weak conjugate base in moderate concentration, the hydrolysis approximation becomes the standard approach.

Comparison with other sodium acetate concentrations

The pH of sodium acetate rises gradually as concentration increases, but not dramatically, because pH depends on the logarithm of hydroxide concentration. The table below shows approximate pH values using the same K_a assumption and 25 degrees C conditions.

NaAc Concentration (M)	Approximate [OH^–] (M)	Approximate pOH	Approximate pH
0.010	2.36 × 10^-6	5.627	8.373
0.050	5.27 × 10^-6	5.278	8.722
0.100	7.45 × 10^-6	5.128	8.872
0.182	1.01 × 10^-5	4.997	9.003
0.500	1.67 × 10^-5	4.778	9.222
1.000	2.36 × 10^-5	4.627	9.373

Common mistakes students make

• Assuming sodium acetate is neutral because it is a salt.
• Using K_a directly instead of converting to K_b.
• Forgetting that acetate is the conjugate base of a weak acid.
• Confusing moles with molarity when volume is given.
• Using the Henderson-Hasselbalch equation without a buffer pair being present initially.
• Entering 100.00 mL directly as liters when calculating moles.

When would you use Henderson-Hasselbalch instead?

Henderson-Hasselbalch is primarily used for buffer systems where both a weak acid and its conjugate base are present in appreciable amounts. A pure sodium acetate solution is not, by itself, a full acetic acid acetate buffer unless acetic acid is also present. For a pure salt of a weak acid, the hydrolysis approach is conceptually cleaner and usually expected in coursework.

Laboratory relevance of sodium acetate pH

Sodium acetate is widely used in biochemical buffers, crystallization procedures, DNA precipitation workflows, and acid-base equilibrium demonstrations. In many practical systems, sodium acetate is combined with acetic acid to produce acetate buffers spanning the pH range near the pK_a of acetic acid, around 4.76. On its own, however, sodium acetate gives a mildly basic solution. A 0.182 M solution with pH near 9.00 is consistent with the conjugate base behavior expected from equilibrium theory.

Authoritative references for equilibrium constants and acid-base concepts

For verified educational and reference material, consult:

• Chemistry LibreTexts for weak acid and conjugate base equilibrium explanations.
• U.S. Environmental Protection Agency for pH fundamentals and aqueous chemistry context.
• NIST Chemistry WebBook for compound data and scientific reference support.
• Princeton University Chemistry for academic chemistry resources and instructional context.

Final answer summary

To calculate the pH of 100.00 mL 0.182 M NaAc, first recognize that sodium acetate is the salt of a weak acid and strong base. The acetate ion undergoes hydrolysis, so the solution is basic. Using K_a = 1.8 × 10^-5 for acetic acid, you find K_b = 5.56 × 10^-10. Then applying the weak base approximation gives [OH^–] ≈ 1.006 × 10^-5 M, pOH ≈ 4.997, and pH ≈ 9.003.

In short, the pH is approximately 9.00. The 100.00 mL figure helps determine that the sample contains 0.01820 mol of sodium acetate, but the pH itself is governed by the concentration and equilibrium constant rather than volume alone.

Quick takeaway: For a 0.182 M sodium acetate solution at 25 degrees C, the expected pH is about 9.00, confirming that acetate behaves as a weak base in water.