Calculate Ph At The Equivalence Point For Maleic Acid

Use this interactive chemistry calculator to estimate the pH at the first or second equivalence point when maleic acid is titrated with a strong base. The tool also plots a simplified titration curve and shows the underlying stoichiometric checkpoints.

Calculator

Titration Curve Preview

This chart shows a simplified pH versus added base volume profile for titrating maleic acid with a strong base. It marks the first and second equivalence points and updates automatically after each calculation.

For the first equivalence point, the dominant species is the amphiprotic hydrogen maleate ion, so the standard approximation is pH ≈ (pKa1 + pKa2) / 2. For the second equivalence point, the maleate ion behaves as a weak base.

Expert Guide: How to Calculate pH at the Equivalence Point for Maleic Acid

Maleic acid is a classic example of a diprotic acid, which means it can donate two protons in two separate acid-base steps. When you are asked to calculate pH at the equivalence point for maleic acid, the most important thing to identify first is which equivalence point is being discussed. That distinction matters because the chemistry at the first equivalence point is very different from the chemistry at the second equivalence point.

In a strong-base titration, such as maleic acid titrated with sodium hydroxide, the first equivalence point occurs after exactly one mole of hydroxide has reacted per mole of maleic acid. At that point, all of the original maleic acid, H₂A, has been converted into hydrogen maleate, HA^–. The second equivalence point occurs after two moles of hydroxide have reacted per mole of maleic acid, converting everything into maleate, A^2-. Those two solutions do not have the same pH behavior, so using the right formula is essential.

Key idea: At the first equivalence point, maleic acid becomes an amphiprotic species and the pH is commonly approximated with pH = (pKa1 + pKa2) / 2. At the second equivalence point, the maleate ion acts as a weak base, so you must use hydrolysis through Kb = Kw / Ka2.

Step 1: Understand the acid system

Maleic acid can be written as H₂A. Its stepwise dissociation reactions are:

1. H₂A ⇌ H⁺ + HA^– with Ka1
2. HA^– ⇌ H⁺ + A^2- with Ka2

For many classroom and laboratory calculations at 25°C, values near pKa1 = 1.92 and pKa2 = 6.23 are often used. These values show that the first proton is much more acidic than the second. That gap between pKa values also helps justify common titration approximations.

Step 2: Find the stoichiometric equivalence volume

Before you calculate pH, you usually need to know how much base must be added to reach the equivalence point.

• Initial moles of maleic acid: n(H₂A) = C_acid × V_acid
• First equivalence point: n(OH^–) = n(H₂A)
• Second equivalence point: n(OH^–) = 2n(H₂A)

If the base concentration is known, the corresponding titrant volumes are:

• V_eq1 = n(H₂A) / C_base
• V_eq2 = 2n(H₂A) / C_base

Suppose you begin with 25.0 mL of 0.1000 M maleic acid and titrate with 0.1000 M NaOH. The initial acid moles are:

n(H₂A) = 0.1000 mol/L × 0.0250 L = 0.00250 mol

So the first equivalence point requires 0.00250 mol OH^–, which is 25.0 mL of 0.1000 M NaOH. The second equivalence point requires 0.00500 mol OH^–, which is 50.0 mL of base.

Step 3: Calculate pH at the first equivalence point

At the first equivalence point, the dominant species in solution is hydrogen maleate, HA^–. This ion is amphiprotic, meaning it can both donate and accept a proton. For amphiprotic species derived from a diprotic acid, the common approximation is:

pH ≈ (pKa1 + pKa2) / 2

Using pKa1 = 1.92 and pKa2 = 6.23:

pH ≈ (1.92 + 6.23) / 2 = 4.075

So the first equivalence point pH is approximately 4.08. This result is often surprisingly independent of concentration, provided the solution is not extremely dilute and activity effects are ignored. That is why many analytical chemistry problems involving diprotic acids use this elegant midpoint formula.

Why the first equivalence point is not neutral

Students sometimes expect every equivalence point in a titration to have a pH of 7. That is only true for strong acid-strong base systems. Maleic acid is a weak diprotic acid, so its equivalence-point solution contains ions that still participate in proton-transfer reactions with water. At the first equivalence point, hydrogen maleate is neither a strong acid nor a strong base. Instead, it sits in the middle and establishes an amphiprotic equilibrium, producing a pH well below neutral but much higher than the initial acid solution.

Step 4: Calculate pH at the second equivalence point

At the second equivalence point, all of the acid has been converted to A^2-, the fully deprotonated maleate ion. This species acts as a weak base in water:

A^2- + H₂O ⇌ HA^– + OH^–

Its base dissociation constant is found from the second acid dissociation constant:

Kb = Kw / Ka2

If pKa2 = 6.23, then:

Ka2 = 10^-6.23 ≈ 5.89 × 10^-7

Kb = 1.00 × 10^-14 / 5.89 × 10^-7 ≈ 1.70 × 10^-8

Now compute the concentration of A^2- at the second equivalence point. In the earlier 25.0 mL example, the total volume at the second equivalence point is 25.0 mL + 50.0 mL = 75.0 mL = 0.0750 L. Since the original maleic acid moles were 0.00250 mol, the concentration of A^2- is:

C = 0.00250 / 0.0750 = 0.0333 M

For a weak base, a standard approximation is:

[OH^–] ≈ √(KbC)

Substituting values:

[OH^–] ≈ √((1.70 × 10^-8)(0.0333)) ≈ 2.38 × 10^-5

pOH ≈ 4.62

pH ≈ 14.00 – 4.62 = 9.38

So the second equivalence point pH is basic, commonly around 9.4 for this concentration example.

Quick comparison of the two equivalence points

Feature	First equivalence point	Second equivalence point
Dominant species	HA^– (hydrogen maleate)	A^2- (maleate)
Chemical character	Amphiprotic	Weak base
Main formula	pH ≈ (pKa1 + pKa2) / 2	Kb = Kw / Ka2, then weak-base hydrolysis
Using pKa1 = 1.92, pKa2 = 6.23	pH ≈ 4.08	Depends on concentration, often around 9.3 to 9.5
Volume of 0.1000 M NaOH for 25.0 mL of 0.1000 M acid	25.0 mL	50.0 mL

Worked example with real numbers

Let us summarize the full process with a realistic titration problem.

1. Take 25.0 mL of 0.1000 M maleic acid.
2. Titrate with 0.1000 M NaOH.
3. Use pKa1 = 1.92 and pKa2 = 6.23.

Initial acid moles: 0.1000 × 0.0250 = 0.00250 mol

First equivalence volume: 0.00250 / 0.1000 = 0.0250 L = 25.0 mL

Second equivalence volume: 0.00500 / 0.1000 = 0.0500 L = 50.0 mL

First equivalence pH: (1.92 + 6.23)/2 = 4.08

Second equivalence concentration of A^2-: 0.00250 mol / 0.0750 L = 0.0333 M

Kb: 1.00 × 10^-14 / 10^-6.23 = 1.70 × 10^-8

[OH^–]: √(1.70 × 10^-8 × 0.0333) = 2.38 × 10^-5

Second equivalence pH: 14.00 – 4.62 = 9.38

Reference values and comparison data

Parameter	Representative value	Why it matters
pKa1 of maleic acid	1.92	Controls initial acidity and first buffer region
pKa2 of maleic acid	6.23	Controls second buffer region and second-equivalence hydrolysis
First equivalence pH	4.08	From amphiprotic approximation
Second equivalence pH for 0.0333 M A^2-	9.38	From weak-base hydrolysis
Difference between pKa values	4.31 pH units	Indicates distinct titration regions and supports approximation methods

Common mistakes to avoid

• Using pH = 7 at equivalence. That is not valid for weak acid-strong base diprotic systems.
• Forgetting that maleic acid has two equivalence points.
• Applying the amphiprotic formula to the second equivalence point. It only belongs to the first equivalence point.
• Ignoring dilution. At the second equivalence point, concentration after mixing affects the hydrolysis pH.
• Mixing up pKa and Ka without converting properly using Ka = 10^-pKa.

When approximations are most reliable

The first-equivalence formula pH = (pKa1 + pKa2) / 2 is widely taught because it works very well for amphiprotic salts in ordinary analytical chemistry conditions. The second-equivalence weak-base approximation also works well when the hydrolysis is limited and the solution is not too dilute. If you are working in a very low ionic strength environment, at unusual temperatures, or with highly precise research-grade calculations, activity corrections and more exact equilibrium solving may be necessary.

How the titration curve behaves

A maleic acid titration curve usually begins at low pH because the first proton is fairly acidic. As base is added, the first buffer region appears around pKa1. At the half-equivalence point to the first step, the pH equals pKa1. Then the curve rises to the first equivalence point near pH 4.08. Between the first and second equivalence points, the HA^–/A^2- buffer system dominates, and at the halfway point of this region the pH equals pKa2. Finally, at the second equivalence point the solution turns basic because A^2- hydrolyzes water to produce hydroxide.

Authoritative chemistry references

If you want to validate constants, acid-base methods, or broader equilibrium chemistry, these sources are excellent starting points:

• LibreTexts Chemistry for detailed explanations of polyprotic acid titrations and amphiprotic species behavior.
• NIST Chemistry WebBook for trusted chemical reference data and related thermodynamic resources.
• U.S. Environmental Protection Agency for broader analytical chemistry and pH measurement guidance in aqueous systems.

Bottom line

To calculate pH at the equivalence point for maleic acid, always identify whether you mean the first or second endpoint. For the first equivalence point, use the amphiprotic expression pH ≈ (pKa1 + pKa2)/2, which gives about 4.08 for maleic acid with common reference values. For the second equivalence point, treat the fully deprotonated maleate ion as a weak base and calculate hydrolysis using Kb = Kw/Ka2, remembering to include dilution from the added titrant. That framework will let you solve most classroom, laboratory, and exam problems involving the titration of maleic acid with confidence.

This page is intended for educational calculation support. Exact values can shift slightly with temperature, ionic strength, and the reference source used for dissociation constants.