Question

Write the formula for the conjugate base for each of the following acids:

a.${\mathrm{HCO}}_3$

b.${\mathrm{CH}}_3{{\mathrm{NH}}_3}^{+}$

c.${\mathrm{HPO}}_4^{2-}$

d.${\mathrm{HNO}}_2$

Short answer

The conjugate bases for the following acids are

(Part a) ${\mathrm{CO}}_3^{-2}$

(Part b) ${\mathrm{CH}}_3{\mathrm{NH}}_2$

(Part c) ${\mathrm{PO}}_4^{-3}$

(Part d) ${\mathrm{NO}}_2^{-}$

Step-by-step solution

Step 1- Introduction (Part a) 

According to the Bronsted-Lowery theory, an acid-base bond pair consists of molecules with differing $\mathrm{H}+$ions. When an acid loses $1$${\mathrm{H}}^{+}$, it forms a conjugate base, and when an acid gains $1$${\mathrm{H}}^{+}$, it forms a conjugate acid.

Example: $HF$loses one ${\mathrm{H}}^{+}$and the conjugate base $\mathrm{F}¯$is formed.

Step 2- Explanation (Part a)

a.

Conjugate base is fashioned from the acid through lose of $\mathrm{H}+$ion.

${\mathrm{HCO}}_3^{-}$lose one proton and forms ${\mathrm{CO}}_3^{-2}$it's far the conjugate base for${\mathrm{HCO}}_3^{-}$.

${\mathrm{HCO}}_3^{-}\rightleftarrows {\mathrm{H}}^{+}+{\mathrm{CO}}_3^{-2}$

localid="1652599695975" $(acid)(conjugatebase)$

Step 3- Explanation (part b)

b.

Conjugate base is shaped from the acid with the aid of using lose of $\mathrm{H}+$ion.

${\mathrm{CH}}_3{{\mathrm{NH}}_3}^{+}$lose one proton and forms ${\mathrm{CH}}_3{\mathrm{NH}}_2$it's miles the conjugate base for ${\mathrm{CH}}_3{\mathrm{NH}}_3^{+}$.

${\mathrm{CH}}_3{\mathrm{NH}}_3^{+}\rightleftarrows {\mathrm{H}}^{+}+{\mathrm{CH}}_3{\mathrm{NH}}_2$

$\left(acid\right)(conjugatebase)$

Step 4- Explanation (part c)

c.

When the $\mathrm{H}+$ion is lost, a conjugate base is formed in the acid. role="math" localid="1652600316101" $\mathrm{HPO}{¯}_4^2$loses one proton to form role="math" localid="1652600340924" ${\mathrm{PO}}_4^{¯3}$.

This is the conjugate base of ${\mathrm{HPO}}_4^{-2}$.

${\mathrm{HPO}}_4^{-2}\rightleftarrows {\mathrm{H}}^{+}+{\mathrm{PO}}_4^{-3}$

$\left(acid\right)(conjugatebase)$

Step 5- Explanation (Part d)

d.

Conjugate base is formed from the acid by lose of $\mathrm{H}+$ ion.

${\mathrm{HNO}}_2$lose one proton and forms ${\mathrm{NO}}_2^{-}$it is the conjugate base for ${\mathrm{HNO}}_2$.

${\mathrm{HNO}}_2\rightleftarrows {\mathrm{H}}^{+}+{\mathrm{NO}}_2^{-}$

$\left(acid\right)(conjugatebase)$