91影视

The objective is to solve the following equation for the Bronsted-Lowry acid-base pairs.

a)$H_{3}P{O}_4$contributes $H^{+}to{H}_{2}O$in the statement. The conjugate base was $H_{2}P{O}_4^{-}$, and the conjugate acid was $H_3{O}^{+}$. The conjugation acid-base pair includes ${\mathrm{H}}_3{\mathrm{PO}}_4$and ${\mathrm{H}}_2{{\mathrm{PO}}_4}^{-}$moreover ${\mathrm{H}}_{2}O$and ${\mathrm{H}}_3{O}^{+}$.

The task is to solve the following equation for the Bronsted-Lowry acid-base pairs.

b) $H_{2}O$donates $H^{+}toC{O}_3^{2-}$throughout the reaction. The conjugate base is obtained, and the conjugate acid is obtained as $HC{O}_3^{-}$. ${\mathrm{H}}_{2}O$and $O{H}^{-}$moreover $C{O}_3^{2-}$and $HC{O}_3^{-}$Consider the acid-base pair in conjugation.

The task is to solve the given equation for the Bronsted-Lowry acid-base pairs.

c) In the process Phosphoric acid (${\mathrm{H}}_3{\mathrm{PO}}_4$)releases ${\mathrm{H}}^{+}$and ${\mathrm{NH}}_3$receives it . The conjugate base of phosphoric acid is${\mathrm{H}}_2{\mathrm{PO}}_4^{-}$, and the conjugate acid of Ammonia is${\mathrm{NH}}_4^{+}$. In this instance, the conjugate acid-base pairs are 鈦�${\mathrm{H}}_3{\mathrm{PO}}_4$and ${\mathrm{H}}_2{\mathrm{PO}}_4^{-}$along with${\mathrm{NH}}_3$and${\mathrm{NH}}_4^{+}$