Problem

Show by suitable net ionic equations that each of…

01:08

Problem 5 Medium Difficulty

Show by suitable net ionic equations that each of the following species can act as a Bronsted-Lowry base:
$$
\begin{array}{l}{\text { (a) } \mathrm{H}_{2} \mathrm{O}} \\ {\text { (b) } \mathrm{OH}^{-}} \\ {\text { (c) } \mathrm{NH}_{3}} \\ {\text { (d) } \mathrm{CN}^{-}} \\ {\text { (e) } \mathrm{S}^{2-}} \\ {\text { (f) } \mathrm{H}_{2} \mathrm{PO}_{4}^{-}}\end{array}
$$

Answer

A. $\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+O H^{-}(a q)$
B. $\mathrm{OH}^{-}(a q)+H_{2} O(l) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)+O H^{-}(a q)$
C. $N H_{3}(a q)+H_{2} O(l) \rightarrow N H_{4}^{+}(a q)+O H^{-}(a q)$
D. $C N^{-}(a q)+H_{2} O(l) \rightarrow H C N(a q)+O H^{-}(a q)$
E. $S^{2-}(a q)+H_{2} O(l) \rightarrow H S^{-}(a q)+O H^{-}(a q)$
F. $H_{2} P O_{4}^{-}(a q)+H_{2} O(l) \rightarrow H_{3} P O_{4}(a q)+O H^{-}(a q)$

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Video Transcript

problem. Five from Chapter 14 is asking us two right net ionic equations for the following molecules that are acting as a bronze, said Lowry Base. Um, so it is important. It's too first define What is it? Brown stood Laurie base. And that means that it is a proton except er So only write these equations, we are going to have to introduce a source of protons for that these Bronson glory bases to accept. Um, And again, these equations have to be net ionic, meaning that they have to have the same charge on either side of the equation. So let's get started with water. So h 20 as no charge. We're going to introduce a source of protons in this case just for simplicity. Secrets can use the hydrogen ion. This water molecule is going to you except the hydrogen ion making it a brown, said Larry Bass. And it's going to have a charge of plus one. So it becomes a hydra. Roni, Um, I am. We have net Ionic is a net ionic equation because we just have a plus one charge on either side. Next we have hydroxide hydroxide ion and never gonna introduce a source of protons in this case, just the hydrogen ion in here. Hi. Jock Side accepts the proton and becomes water. The positive and negative charge on the left hand side of the equation. Cancel out giving us an extra zero. And that equals. And that the charge on the right hand side of the equation of water, which is zero. Next, let's do a mony. Ammonia? Yes, ammonia and h three. Introducing a source of protons and ammonia is going to accept this proton and become ammonium. And each four and ammonium has a positive charge. Sorry, equation is net. Ionic next molecule. We have cyanide, which has a negative charge introducing a source of protons and then sign it is going to accept this proton and become hydrogen cyanide, which has no charge. So this equation is net Ionic. Next we have a sulfide ion. That's s to minus introducing our proton source. And the sulfide is going to accept this proton and become bi sulphide. And our equation is net ionic because this has a net charge of one minus one. Excuse me. And this hide bi sulphide has a charge of minus one as well neck. Lastly, we have di hydrogen phosphate, which may take me a little while right out. But try and be quick, he o whore minus and introducing our source of protons. The hydrogen ion. The di hydrogen phosphate is going to accept this proton and become phosphoric acid. It's an H three people for which has no charge. So this last equation is also net ionic and that is the answer for a problem.

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