Calculate The Ph Of A 0.800 M Nach3Co2 Solution

This premium calculator solves the pH of sodium acetate solution by treating acetate as a weak base in water. Enter your concentration and acid dissociation constant, then compare the exact quadratic result with the common square root approximation used in general chemistry.

Expert Guide: How to Calculate the pH of a 0.800 m NaCH3CO2 Solution

Sodium acetate, written as NaCH3CO2 or more fully NaC2H3O2, is the sodium salt of acetic acid. When it dissolves in water, it dissociates essentially completely into sodium ions and acetate ions. The sodium ion is a spectator in acid-base chemistry, but the acetate ion is the conjugate base of a weak acid. That means acetate can react with water to produce a small amount of hydroxide, making the solution basic. If you are asked to calculate the pH of a 0.800 m NaCH3CO2 solution, the chemistry is not a strong base problem, even though the solution ends up with a pH above 7. It is a weak base hydrolysis problem.

The core idea is simple: acetic acid is weak, so its conjugate base is strong enough to pull some protons from water. This creates OH- and pushes the pH upward. The correct setup starts with the hydrolysis reaction:

CH3CO2- + H2O ⇌ CH3CO2H + OH-

From there, you calculate the base dissociation constant, Kb, using the relationship between Ka and Kb at 25 degrees C:

Kb = Kw / Ka

For acetic acid, a widely used value is Ka = 1.8 × 10^-5. With Kw = 1.0 × 10^-14, the acetate ion has:

Kb = (1.0 × 10^-14) / (1.8 × 10^-5) = 5.56 × 10^-10

Once you have Kb, use the equilibrium expression for the weak base reaction. For a formal concentration of 0.800, the setup is:

Kb = x^2 / (0.800 – x)

Here, x is the amount of OH- generated at equilibrium. Because Kb is very small, x will be tiny relative to 0.800, so many textbooks allow the approximation 0.800 – x ≈ 0.800. That gives:

x = sqrt(KbC) = sqrt((5.56 × 10^-10)(0.800)) = 2.11 × 10^-5 M

Then calculate pOH:

pOH = -log(2.11 × 10^-5) = 4.68

And finally:

pH = 14.00 – 4.68 = 9.32

So the pH of a 0.800 m NaCH3CO2 solution is approximately 9.32 at 25 degrees C, assuming the molality is treated close to a formal concentration for introductory calculation purposes. In many classroom and homework contexts, that is the expected answer. A more advanced treatment would consider ionic strength, activity corrections, and possibly the distinction between molality and molarity, but the fundamental acid-base logic remains the same.

Why sodium acetate gives a basic solution

A salt can produce an acidic, neutral, or basic solution depending on the strengths of the parent acid and base from which it forms. Sodium acetate comes from:

• NaOH, a strong base
• CH3CO2H, a weak acid

The sodium ion does not appreciably alter pH, but acetate does. Because acetate is the conjugate base of a weak acid, it has enough basicity to react with water and generate hydroxide. That is why sodium acetate solutions are basic rather than neutral.

Step by step method for the 0.800 m problem

1. Write the hydrolysis equation: CH3CO2- + H2O ⇌ CH3CO2H + OH-.
2. Use the known Ka of acetic acid, often 1.8 × 10^-5.
3. Calculate Kb = Kw / Ka = 5.56 × 10^-10.
4. Set the initial acetate concentration to 0.800.
5. Write the equilibrium expression Kb = x^2 / (0.800 – x).
6. Either solve the quadratic exactly or use the square root approximation.
7. Find [OH-], then pOH, then pH.

Exact solution versus approximation

For this problem, the approximation works extremely well because the reaction is weak and the equilibrium shift is tiny. Still, it is useful to compare the two methods. The exact quadratic solution uses:

x = (-Kb + sqrt(Kb^2 + 4KbC)) / 2

With Kb = 5.56 × 10^-10 and C = 0.800, the exact value of x is essentially the same as the approximation to the number of significant figures normally reported in chemistry classes. That is why both methods produce a pH of about 9.32.

Quantity	Symbol	Value used	Meaning in this calculation
Acetic acid dissociation constant	Ka	1.80 × 10^-5	Measures how strongly acetic acid donates H+ in water
Ion product of water at 25 degrees C	Kw	1.00 × 10^-14	Links Ka and Kb for conjugate acid-base pairs
Acetate base dissociation constant	Kb	5.56 × 10^-10	Determines how much OH- acetate generates by hydrolysis
Formal sodium acetate concentration	C	0.800	Starting concentration of acetate before equilibrium shift
Hydroxide concentration at equilibrium	[OH-]	2.11 × 10^-5	Calculated from weak base equilibrium
pOH	pOH	4.68	Negative log of hydroxide concentration
Final pH	pH	9.32	Basic solution due to acetate hydrolysis

How much acetate actually reacts?

One reason the square root approximation works is that only a tiny fraction of the acetate ion hydrolyzes. The percent hydrolysis is:

(x / 0.800) × 100 = (2.11 × 10^-5 / 0.800) × 100 = 0.00264%

That is much less than 5%, which is the common rule of thumb used to justify approximation methods in equilibrium calculations. In other words, almost all of the acetate remains unreacted, and only a trace amount converts to acetic acid and hydroxide.

Effect of concentration on pH

When the concentration of sodium acetate increases, the pH rises, but not dramatically. That is because hydroxide concentration depends on the square root of KbC for a weak base approximation, not directly on concentration in a one-to-one way. Doubling concentration does not double pH. It changes [OH-] by a square root relationship, and the pH responds logarithmically.

NaCH3CO2 concentration	Kb used	Approximate [OH-]	Approximate pOH	Approximate pH
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
0.800	5.56 × 10^-10	2.11 × 10^-5	4.68	9.32
1.000	5.56 × 10^-10	2.36 × 10^-5	4.63	9.37

Molality versus molarity in this problem

The notation 0.800 m technically means molality, which is moles of solute per kilogram of solvent, while 0.800 M means molarity, moles of solute per liter of solution. In rigorous physical chemistry, that distinction matters because equilibrium constants and concentration terms can be influenced by density and activity effects. However, in many general chemistry settings, problems like this are solved by treating the given 0.800 m value as the formal concentration term in the ICE table. That produces the standard classroom answer of about pH 9.32.

If you needed a highly precise real-world pH, you would want more information:

• Solution density to convert between molality and molarity
• Temperature if Kw differs significantly from 1.00 × 10^-14
• Activity coefficients because ions at moderate ionic strength deviate from ideality

Common mistakes to avoid

• Using Ka directly to compute pH. Since acetate is a base, you should use Kb, not Ka, for the hydrolysis equilibrium.
• Forgetting the pOH step. The equilibrium gives you OH-, so calculate pOH first, then convert to pH.
• Treating sodium acetate as a strong base. It is not. The base behavior comes from hydrolysis of acetate, which is weak.
• Ignoring the unit note. The symbol m is not the same as M, though many educational solutions approximate them similarly for this type of question.
• Dropping significant figures too early. Carry extra digits through Kb and [OH-] before rounding the final pH.

Quick chemistry intuition check

Even before doing the math, you can predict the answer range. Sodium acetate is the salt of a strong base and a weak acid, so the solution must be basic. Because acetate is only a weak base, the pH should not be extremely high. A value around 8.5 to 10 is chemically reasonable, and the calculation confirms a pH close to 9.32.

Final answer summary

To calculate the pH of a 0.800 m sodium acetate solution at 25 degrees C, use the weak base hydrolysis of acetate. Start from acetic acid’s Ka = 1.8 × 10^-5, convert to Kb = 5.56 × 10^-10, solve for [OH-] ≈ 2.11 × 10^-5, then compute pOH ≈ 4.68 and pH ≈ 9.32. That is the accepted general chemistry result for this problem.

Authoritative references

• USGS: pH and Water
• NIST Chemistry WebBook: Acetic Acid Data
• MIT OpenCourseWare: General Chemistry resources

Reference values commonly used in general chemistry at 25 degrees C: Ka for acetic acid = 1.8 × 10^-5 and Kw = 1.0 × 10^-14.