Equilibrium NEET 2025: Phosphoric Acid Question Solved

What is the answer to the Equilibrium NEET 2025 phosphoric acid question?

Option D: statements A, B and C are correct in this Equilibrium NEET 2025 Chemistry question on ionisation of polyprotic acids. Keep the labels separate: option D is correct, while statement D is false.

Phosphoric acid releases three protons through successive equilibria. The stepwise constants and combined equilibrium constant are:

K_{a 1}, K_{a 2}, K_{a 3}; K (combined)

The question asks which claims hold:

Statement A: the logarithm of the combined constant equals the sum of the logarithms of the three stepwise constants.
Statement B: acid strength decreases from phosphoric acid to dihydrogen phosphate to hydrogen phosphate.
Statement C: successive acid ionisation constants decrease.
Statement D: the constants obey an arithmetic-mean relationship. The supplied record does not contain its exact expression, so it is not reconstructed here.

The choices group those statements as follows:

How do you derive the overall constant before taking logarithms?

Adding equilibrium reactions multiplies their constants. Start with the three dissociations: write each concentration expression, multiply them and cancel the intermediate species. The logarithms add only after applying the logarithm to that product.

The first proton comes from phosphoric acid. Its dissociation and concentration expression are:

H_{3} P O_{4} ⇌ H^{+} + H_{2} P O_{4}^{-}

K_{a 1} = \frac{[H^{+}] [H_{2} P O_{4}^{-}]}{[H_{3} P O_{4}]}

Dihydrogen phosphate donates the second proton:

H_{2} P O_{4}^{-} ⇌ H^{+} + H P O_{4}^{2 -}

K_{a 2} = \frac{[H^{+}] [H P O_{4}^{2 -}]}{[H_{2} P O_{4}^{-}]}

Hydrogen phosphate donates the third:

H P O_{4}^{2 -} ⇌ H^{+} + P O_{4}^{3 -}

K_{a 3} = \frac{[H^{+}] [P O_{4}^{3 -}]}{[H P O_{4}^{2 -}]}

When these reactions are added, dihydrogen phosphate appears once on each side and cancels. Hydrogen phosphate also appears once on each side and cancels. Three hydrogen ions remain on the product side:

H_{3} P O_{4} ⇌ 3 H^{+} + P O_{4}^{3 -}

Using conventional NEET concentration notation, the combined constant is:

K = \frac{[H^{+}]^{3} [P O_{4}^{3 -}]}{[H_{3} P O_{4}]}

Now multiply the three stepwise expressions. Both intermediate concentrations appear in a numerator and a denominator:

K_{a 1} K_{a 2} K_{a 3} = \frac{[H^{+}] [H_{2} P O_{4}^{-}]}{[H_{3} P O_{4}]} \frac{[H^{+}] [H P O_{4}^{2 -}]}{[H_{2} P O_{4}^{-}]} \frac{[H^{+}] [P O_{4}^{3 -}]}{[H P O_{4}^{2 -}]}

Regrouping makes both cancellations explicit:

= \frac{[H^{+}]^{3} [P O_{4}^{3 -}]}{[H_{3} P O_{4}]} \underset{1}{\underset{⏟}{\frac{[H_{2} P O_{4}^{-}]}{[H_{2} P O_{4}^{-}]}}} \underset{1}{\underset{⏟}{\frac{[H P O_{4}^{2 -}]}{[H P O_{4}^{2 -}]}}} = K

Therefore:

K = K_{a 1} K_{a 2} K_{a 3}

Take logarithms in the same base throughout:

\log K = \log (K_{a 1} K_{a 2} K_{a 3}) = \log K_{a 1} + \log K_{a 2} + \log K_{a 3}

Statement A is true. Do not add the constants themselves: adding reactions produces a product of constants.

Why are statements B and C true but statement D false?

Successive proton removal becomes harder because the proton donor becomes more negatively charged. For phosphoric acid, the first proton is lost most easily. The next two donors already carry negative charge, so removing another positively charged proton is less favourable.

The donor charges progress as follows:

H_{3} P O_{4} (0), H_{2} P O_{4}^{-} (- 1), H P O_{4}^{2 -} (- 2)

The first constant belongs to phosphoric acid, the second to dihydrogen phosphate and the third to hydrogen phosphate. Since each successive donation is harder:

K_{a 1} > K_{a 2} > K_{a 3}

A larger acid dissociation constant means a stronger proton donor. The matching acid-strength order establishes statement B, while the constant order establishes statement C:

H_{3} P O_{4} > H_{2} P O_{4}^{-} > H P O_{4}^{2 -}

Statement D fails for the reason given in the official solution: successive ionisation constants do not obey the proposed simple arithmetic-mean rule. The decreasing order does not establish an averaging relationship.

Statement A: true.
Statement B: true.
Statement C: true.
Statement D: false.

Select option D, statements A, B and C only.

How can a method error lead to option B?

Option B includes statements A and C but excludes statement B. One route to this wrong answer is treating negatively charged or amphiprotic species as outside acid-strength comparisons. That accepts the constant order but wrongly rejects the matching ranking of proton donors.

Dihydrogen phosphate and hydrogen phosphate can both donate a proton. Their ability to accept a proton does not cancel that acidic behaviour. Each has an acid dissociation constant and can be compared with phosphoric acid.

Accepting decreasing constants while rejecting the matching acid-strength order is inconsistent. Choosing B earns minus one mark instead of four, a difference of: $4 - (- 1) = 5 marks$

Write the proton donor beside every acid dissociation constant before ranking acidity. Compare what each species does in the stated dissociation, not just its charge or category.

How do you apply this method to three related questions?

Multiply constants only for the steps included in the required reaction. For acidity scales, the negative logarithm reverses the order. These are original related practice questions, not verified past-paper questions, with each worked answer directly below its question.

What is the overall constant for a diprotic acid?

The overall constant is the product of its two stepwise constants. Start with the two proton-loss steps:

H_{2} A & ⇌ H^{+} + H A^{-} H A^{-} & ⇌ H^{+} + A^{2 -}

The intermediate appears on opposite sides and cancels, giving:

H_{2} A ⇌ 2 H^{+} + A^{2 -}

The concentration expressions confirm the same cancellation:

K_{a 1} K_{a 2} = \frac{[H^{+}] [H A^{-}]}{[H_{2} A]} \frac{[H^{+}] [A^{2 -}]}{[H A^{-}]} = \frac{[H^{+}]^{2} [A^{2 -}]}{[H_{2} A]} = K

Hence:

K = K_{a 1} K_{a 2}

What constant applies when dihydrogen phosphate loses two protons?

The required constant is the product of the second and third stepwise constants. Add only those phosphoric acid dissociations:

H_{2} P O_{4}^{-} & ⇌ H^{+} + H P O_{4}^{2 -} H P O_{4}^{2 -} & ⇌ H^{+} + P O_{4}^{3 -}

Hydrogen phosphate cancels, leaving:

H_{2} P O_{4}^{-} ⇌ 2 H^{+} + P O_{4}^{3 -}

Therefore:

K = K_{a 2} K_{a 3}

The first constant is excluded because the required reaction starts with dihydrogen phosphate, not phosphoric acid. Including the first dissociation changes the starting species and the overall reaction.

How does decreasing acid dissociation strength translate into the pKa order?

The order reverses because the definition contains a negative sign. Smaller acid dissociation constants therefore correspond to larger values on this scale: $p K_{a} = - \log K_{a}$

Given:

K_{a 1} > K_{a 2} > K_{a 3}

Taking logarithms preserves the order, but multiplying by minus one reverses it:

p K_{a 1} < p K_{a 2} < p K_{a 3}

Before selecting an option, check two things on paper: which reactions were added, and which species donated each proton.

Next step: photograph a doubt on NEET JEEnius AI and photograph any question you are stuck on and get a step-by-step solution across Physics, Chemistry and Biology (20 free a month).

For a worked example of the same idea, see Dual Nature NEET 2012: Threshold Frequency Solution.

Frequently asked questions

What is the answer to the NEET 2025 phosphoric acid equilibrium question?

Option D is correct: statements A, B and C only. Statement D is false, so do not confuse the option label with the statement label.

How do you find the overall equilibrium constant for phosphoric acid?

For the overall loss of three protons from H3PO4, multiply the three stepwise constants: K = Ka1 × Ka2 × Ka3. Adding the dissociation reactions cancels the intermediate species and multiplies their constants. Taking logarithms then gives log K = log Ka1 + log Ka2 + log Ka3.

Why is Ka1 greater than Ka2 and Ka3 for phosphoric acid?

Each successive proton donor is more negatively charged, making further proton removal less favourable. Therefore, Ka1 > Ka2 > Ka3. The corresponding acid-strength order is phosphoric acid > dihydrogen phosphate > hydrogen phosphate.

What is the pKa order for phosphoric acid?

The order is pKa1 < pKa2 < pKa3. Since pKa = -log Ka, the decreasing order Ka1 > Ka2 > Ka3 reverses on the pKa scale.