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Write Brief Answer · Q3

Q.Identify the conjugate acid–base pair for the following reactions in aqueous solution:

i) HS−(aq)+HF⇌F−(aq)+H2S(aq)HS^-(aq) + HF \rightleftharpoons F^-(aq) + H_2S(aq)
ii) HPO42−+SO32−⇌PO43−+HSO3−HPO_4^{2-} + SO_3^{2-} \rightleftharpoons PO_4^{3-} + HSO_3^-
iii) NH4++CO32−⇌NH3+HCO3−NH_4^+ + CO_3^{2-} \rightleftharpoons NH_3 + HCO_3^-
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Step 1 (reaction i). HS−(aq)+HF⇌F−(aq)+H2S(aq)HS^-(aq)+HF \rightleftharpoons F^-(aq)+H_2S(aq): HFHF loses a proton to become F−F^- (acid/conjugate-base pair HF/F−HF/F^-); HS−HS^- gains that proton to become H2SH_2S (base/conjugate-acid pair H2S/HS−H_2S/HS^-).

Step 2 (reaction ii). HPO42−+SO32−⇌PO43−+HSO3−HPO_4^{2-}+SO_3^{2-} \rightleftharpoons PO_4^{3-}+HSO_3^-: HPO42−HPO_4^{2-} loses a proton to become PO43−PO_4^{3-} (pair HPO42−/PO43−HPO_4^{2-}/PO_4^{3-}); SO32−SO_3^{2-} gains that proton to become HSO3−HSO_3^- (pair HSO3−/SO32−HSO_3^-/SO_3^{2-}).

Step 3 (reaction iii). NH4++CO32−⇌NH3+HCO3−NH_4^++CO_3^{2-} \rightleftharpoons NH_3+HCO_3^-: NH4+NH_4^+ loses a proton to become NH3NH_3 (pair NH4+/NH3NH_4^+/NH_3); CO32−CO_3^{2-} gains that proton to become HCO3−HCO_3^- (pair HCO3−/CO32−HCO_3^-/CO_3^{2-}).

✓Final answer

i) HF/F⁻ and H2S/HS⁻. ii) HPO4²⁻/PO4³⁻ and HSO3⁻/SO3²⁻. iii) NH4⁺/NH3 and HCO3⁻/CO3²⁻.

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