Respuesta :
Answer: [tex]HCOOH[/tex] and [tex]HCOO^-[/tex]
[tex]HNO_3[/tex] and [tex]NO_3-^-[/tex]
[tex]H_2SO_4[/tex] and [tex]SO_4^{2-}[/tex]
[tex]H_2CO_3[/tex] and [tex]CO_3^{2-}[/tex]
Explanation:
According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.
For the given chemical equations:
1. [tex]HCOOH(aq)\rightarrow HCOO^-(aq.)+H^+(aq.)[/tex]
Here, [tex]HCOOH[/tex] is loosing a proton, thus it is considered as an acid and after losing a proton, it forms [tex]HCOO^-[/tex] which is a conjugate base.
2. [tex]HNO_3(aq)\rightarrow NO_3^-(aq.)+H^+(aq.)[/tex]
Here, [tex]HNO_3[/tex] is loosing a proton, thus it is considered as an acid and after losing a proton, it forms [tex]NO_3^-[/tex] which is a conjugate base.
3. [tex]H_2SO_4(aq)\rightarrow SO_4^{2-}(aq.)+2H^+(aq.)[/tex]
Here, [tex]H_2SO_4[/tex] is loosing a proton, thus it is considered as an acid and after losing a proton, it forms [tex]SO_4^{2-}[/tex] which is a conjugate base.
4. [tex]NaCl[/tex] and [tex]NaOh[/tex] are not conjugate acid/ base pairs.
5. [tex]H_2CO_3(aq)\rightarrow CO_3^{2-}(aq.)+2H^+(aq.)[/tex]
Here, [tex]H_2CO_3[/tex] is loosing a proton, thus it is considered as an acid and after losing a proton, it forms [tex]CO_3^{2-}[/tex] which is a conjugate base.
