3. Determine the acid-base pairs in the example below
CHOOH (aq) + H2O (l) ↔ HCOO- (aq) + H3O (aq)

H2S (aq) + NH3 (aq) ↔ NH4+(aq) + HS- (aq)

  CHOOH (aq)   +    H₂O (l)    ⇄      HCOO⁻ (aq)         +    H₃O⁺ (aq)

   Weak acid             Base          Conj. strong base         Conj. acid

  H₂S (aq)    +    NH₃ (aq)     ⇄           NH₄⁺(aq)        +          HS⁻ (aq)

Weak acid      Weak Base        Conj. strong acid        Conj. strong base

Explanation:

To determine the acid-base pairs in the example below we follow the Brownsted Lowry theory.

An acid releases protons

A base accepts protons

CHOOH (aq) + H₂O (l) ↔ HCOO⁻ (aq) + H₃O⁺ (aq)

Formic reacts to water in order to produce formate and hydroniums

Formic release a proton → Acid

Formate can accepted the proton → Conjugate base

Water will be the base, because it will acepted the proton, so the hydronium will be the conjugate acid, because it will release the proton to become water again.

H₂S (aq) + NH₃ (aq) ⇄ NH₄⁺(aq) + HS⁻ (aq)

The sulfidric behaves as an acid, it release the proton to ammonia.

The acid sulfide, becomes the base. It can accept the proton for being, H₂S again.

Then the ammonia is the base (it can accept H⁺) and the ammonium is the conjugate acid (it release the H⁺)

your answer is c) plasma

~*"auba14"*~

c. electrical charges of ionic bonds.