Identify the Bronsted-Lowry acid and the Bronsted-Lowry base on the left side of each of the following equations, and also identify the conjugate acid and conjugate base of each on the right side: (a) \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{NH}_{3}(a q)\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons\) \(\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q)\) (c)\(\mathrm{HCOOH}(a q)+\mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons\) \(\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{HCOO}^{-}(a q)+\mathrm{HPO}_{4}^{2-}(a q)\)

Step by step solution

01

On the left side of the equation, we can see that \(\mathrm{NH}_{4}^{+}\) donates a proton to \(\mathrm{CN}^{-}\), which accepts the proton. Therefore, \(\mathrm{NH}_{4}^{+}\) is the Bronsted-Lowry acid, and \(\mathrm{CN}^{-}\) is the Bronsted-Lowry base. #Step 2: Determine the conjugate acid and conjugate base#

On the right side of the equation, the products after the reaction are \(\mathrm{HCN}\) and \(\mathrm{NH}_{3}\). As \(\mathrm{NH}_{4}^{+}\) loses a proton to form \(\mathrm{NH}_{3}\), \(\mathrm{NH}_{3}\) is the conjugate base of the acid, while \(\mathrm{CN}^{-}\) gains a proton to form \(\mathrm{HCN}\), making \(\mathrm{HCN}\) the conjugate acid of the base. (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}(a q)+\mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons\) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q)\) #Step 1: Determine the Bronsted-Lowry acid and base#

02

In this equation, the proton is being transferred from \(\mathrm{H}_{2}\mathrm{O}\) to \(\left(\mathrm{CH}_{3}\right)_{3}\mathrm{N}\). So, the acid is \(\mathrm{H}_{2}\mathrm{O}\), and the base is \(\left(\mathrm{CH}_{3}\right)_{3}\mathrm{N}\). #Step 2: Determine the conjugate acid and conjugate base#

As \(\left(\mathrm{CH}_{3}\right)_{3}\mathrm{N}\) gains a proton and forms \(\left(\mathrm{CH}_{3}\right)_{3}\mathrm{NH}^{+}\), this species is the conjugate acid of the base. \(\mathrm{H}_{2}\mathrm{O}\) loses a proton to form \(\mathrm{OH}^{-}\), which is therefore the conjugate base of the acid. (c)\(\mathrm{HCOOH}(a q)+\mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons\) \(\mathrm{HCOO}^{-}(a q)+\mathrm{HPO}_{4}^{2-}(a q)\) #Step 1: Determine the Bronsted-Lowry acid and base#

03

In this equation, the proton is being transferred from \(\mathrm{HCOOH}\) to \(\mathrm{PO}_{4}^{3-}\). Hence, the acid is \(\mathrm{HCOOH}\), and the base is \(\mathrm{PO}_{4}^{3-}\). #Step 2: Determine the conjugate acid and conjugate base#

The \(\mathrm{HCOOH}\) loses a proton to form \(\mathrm{HCOO}^{-}\), which is the conjugate base of the acid. Meanwhile, the \(\mathrm{PO}_{4}^{3-}\) gains a proton to form \(\mathrm{HPO}_{4}^{2-}\), which is the conjugate acid of the base.

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