Identify the conjugate base of {H_2}PO_3^- an | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The conjugate base of an acid is formed by removing one proton $({H^+})$ from the species. For ${H_2}PO_3^-$,removing one ${H^+}$ gives ${HPO_3^{2

Vedclass · Accepted Answer

Conjugate base: ${HPO_3^{2-}}$; Conjugate acid: ${H_2CO_3}$

Vedclass · Answer

(A) The conjugate base of an acid is formed by removing one proton $({H^+})$ from the species. For ${H_2}PO_3^-$,removing one ${H^+}$ gives ${HPO_3^{2-}}$. The conjugate acid of a base is formed by adding one proton $({H^+})$ to the species. For ${HCO_3^-}$,adding one ${H^+}$ gives ${H_2CO_3}$. Therefore,the conjugate base is ${HPO_3^{2-}}$ and the conjugate acid is ${H_2CO_3}$.

Identify the conjugate base of ${H_2}PO_3^-$ and the conjugate acid of ${HCO_3^-}$.

Similar Questions

Identify the acids in the following reaction: $HCl_{(aq)} + H_2O_{(aq)} \rightleftharpoons H_3O^{+}_{(aq)} + Cl^{-}_{(aq)}$. Which of these acids is stronger?

The ionization constants of $HF$,$HCOOH$,and $HCN$ at $298 \ K$ are $6.8 \times 10^{-4}$,$1.8 \times 10^{-4}$,and $4.8 \times 10^{-9}$ respectively. Calculate the ionization constants of their corresponding conjugate bases.

The heat of neutralisation of a strong acid by a strong base is a constant value because

Which of the following is the strongest Lewis base?

The ammonium ion $(NH_4^+)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module