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Ionic Equilibrium question

2010 · Shift 0 · Q7
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Ionic Equilibrium question

2010 · Shift 0 · Q7

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
Three reactions involving H2PO4−H_2PO_4^−H2​PO4−​ are given below : (i) H3PO4H_3PO_4H3​PO4​ + H2OH_2OH2​O →\to→ H3O+H_3O^+H3​O+ +H2PO4−H_2PO_4^−H2​PO4−​(ii) H2PO4−H_2PO_4^−H2​PO4−​+ H2OH_2OH2​O →HPO42−\to HPO_4^{2−}→HPO42−​+ H3O+H_3O^+H3​O+ (iii) H2PO4−H_2PO_4^−H2​PO4−​+ OH−OH^-OH− →\to→ H3PO4H_3PO_4H3​PO4​ + O2−O^{2-}O2− In which of the above does H2PO4−H_2PO_4^−H2​PO4−​ act as an acid?
  1. A
    (ii) only
  2. B
    (i) and (ii)
  3. C
    (iii) only
  4. D
    (i) only
View written solutionFree

Correct answer: A

  1. Identify what it means to act as an acid

    By Brønsted–Lowry definition, an acid is a proton donor.

    So, we check in each reaction whether H2PO4−H_2PO_4^-H2​PO4−​ donates a proton.

  2. Check reaction (i)

    H3PO4+H2O→H3O++H2PO4−H_3PO_4 + H_2O \to H_3O^+ + H_2PO_4^-H3​PO4​+H2​O→H3​O++H2​PO4−​

    Here, H3PO4H_3PO_4H3​PO4​ loses a proton and becomes H2PO4−H_2PO_4^-H2​PO4−​.

    So H2PO4−H_2PO_4^-H2​PO4−​ is the conjugate base here, not the acid.

    Hence, in (i), H2PO4−H_2PO_4^-H2​PO4−​ does not act as an acid.

  3. Check reaction (ii)

    H2PO4−+H2O→HPO42−+H3O+H_2PO_4^- + H_2O \to HPO_4^{2-} + H_3O^+H2​PO4−​+H2​O→HPO42−​+H3​O+

    Here, H2PO4−H_2PO_4^-H2​PO4−​ changes to HPO42−HPO_4^{2-}HPO42−​.

    That means it has lost one proton:

    H2PO4−→HPO42−+H+H_2PO_4^- \to HPO_4^{2-} + H^+H2​PO4−​→HPO42−​+H+

    Therefore, H2PO4−H_2PO_4^-H2​PO4−​ is acting as an acid in reaction (ii).

  4. Check reaction (iii)

    H2PO4−+OH−→H3PO4+O2−H_2PO_4^- + OH^- \to H_3PO_4 + O^{2-}H2​PO4−​+OH−→H3​PO4​+O2−

    As written, this reaction is not a normal acid-base reaction in water. If we still inspect the role of H2PO4−H_2PO_4^-H2​PO4−​, it is becoming H3PO4H_3PO_4H3​PO4​ by gaining a proton.

    So H2PO4−H_2PO_4^-H2​PO4−​ would be acting as a base, not an acid.

    Hence, in (iii), it does not act as an acid.

  5. Conclusion

    H2PO4−H_2PO_4^-H2​PO4−​ acts as an acid only in reaction (ii).

    Therefore, the correct option is:

    A: (ii) only\boxed{\text{A: (ii) only}}A: (ii) only​

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