Identify the spectator ions, and write a net ionic equation for each reaction in item 6 .

Name: Problem 7, Chapter 8: Chemistry
Uploaded: 2020-06-10T11:11:14-08:00
Duration: 14 min 32 s
Channel: Travis Maslanik
Description: Identify the spectator ions, and write a net ionic equation for each reaction in item 6 .

Identify the spectator ions, and write a net ionic equation...

Key Concepts

Net Ionic Equations

A net ionic equation shows only the chemical species that participate in a reaction. It is derived by first writing a complete ionic equation and then removing the ions that do not change during the reaction (the spectator ions). This equation focuses on the actual chemical change occurring in aqueous solution and is essential in understanding reaction mechanisms in inorganic and analytical chemistry.

Spectator Ions

Spectator ions are the ions present in an aqueous solution that do not participate directly in the chemical reaction. They remain unchanged on both sides of a complete ionic equation and are omitted from the net ionic equation. Recognizing spectator ions is important because it provides clarity by highlighting the species involved in the formation of a product, such as a precipitate, gas, or weak electrolyte.

Chemical Reactions in Aqueous Solutions

Many chemical reactions in aqueous solutions involve interactions between ionic compounds, particularly in precipitation, acid-base, and redox reactions. Understanding how to predict the formation of insoluble products using solubility rules and then expressing only the chemically active ions via a net ionic equation is key to mastering this subject. It helps in simplifying the overall chemical process and emphasizes the transformation of reactants into products.

Transcript

00:01 All right, so this question asks us to identify spectator ions and write net ionic equations for different reactions.

00:08 So when writing net ionic equations, there are some steps that i like to take to make sure that you get the correct net ionic equation.

00:15 The first step is to identify what type of reaction is taking place.

00:19 Typically, when you do neonic equations, you want to make sure that you have either a displacement reaction or a double displacement reaction.

00:25 The second step is to balance the equation.

00:27 In other words, you want to make sure you have the same amount of each atom on both the reactants and product side.

00:35 And then the third step is to write the total ionic equation.

00:38 In order to do this, you want to make sure that you identify the aqueous ionic compounds, and then you can subsequently find out or separate it into its individual ions, aqueous ions.

00:53 The fourth step is to identify the spectator ions.

00:57 Those are easy to identify because there are ions that remain unchanged in the solution as aqueous ions.

01:04 So in other words, they'll show up in both the reactants and product side.

01:08 The fifth step and final step is to actually write your net ionic equation.

01:12 And you do this by canceling out the spectator ions since they do not participate in the reaction.

01:17 And then whatever remains is the net ionic equation.

01:21 So for reaction a, we are given liquid bromine plus aqueous sodium iodide.

01:33 And this goes to aqueous sodium bromide plus solid iodine.

01:44 And the first step here is to identify the type of reaction taking place.

01:47 It looks like we have a displacement reaction, so we should be able to write an ionic equation for this.

01:52 The second step is to balance the equation.

01:55 We do this by putting a two in front of the sodium iodide and a two in front of the sodium bromide.

02:02 And consequently, we should be able to see that we have two of each atom.

02:06 So we have two bromines, two sodiums, and two iodines on both sides of the equation.

02:11 So the equation is balanced.

02:13 The third step is to write our total ionic equation.

02:19 So in this case, we have liquid bromine that remains unchanged.

02:23 And then our sodium iodide actually dissociates into sodium and iodide atoms.

02:27 So we have two sodium aqueous ions.

02:33 And we also have two aqueous iodide atoms.

02:40 And this goes to two aqueous sodium atoms on the product side.

02:45 Plus two aqueous bromide atoms.

02:48 And then we also have our solid iodine, which remains unchanged.

02:57 Our fourth step here is to identify our spectator ions.

02:59 So these are ions that appear on both sides of the equation.

03:02 It looks like we have our sodium ions that appear on both sides of the equation so we can cross those out.

03:10 When writing our net ionic equation, whatever remains is our net ionic equation.

03:14 So in this case, we have our liquid bromine plus two moles of iodine or iodide in the aqueous form.

03:24 And that goes to 2 moles of bromide in the aqueous form plus our solid iodine.

03:34 And for reaction b, we have calcium hydroxide in the aqueous form plus hydrochloric acid in the aqueous form.

03:49 And this goes to the products of calcium chloride aqueous plus liquid water.

03:57 So the first step here is we see that this is a double displacement reaction, therefore we should be able to write netanek equation for this.

04:03 Second step is to balance the equation.

04:05 We do this by putting a two in front of the hydrochloric acid and a two in front of the.

04:08 The water.

04:11 Third step here is to write our total ionic equation.

04:15 So in this case, we have our calcium hydrochloric acid and our calcium chloride.

04:20 They're all aqueous compounds.

04:21 We should be able to write them into ions.

04:23 So we first have calcium 2 plus in the aqueous form, plus our two moles of hydroxide, which is aqueous, plus our two hydrogen ions in the aqueous form, plus our two moles of chloride in the aqueous form.

04:44 And then this goes to the products of 2 plus aqueous.

04:54 This goes to our products of calcium 2 plus aqueous, plus 2 moles of chloride in the aqueous form, plus 2 moles of water in liquid form.

05:07 This remains unchanged.

05:13 Okay, so our fourth step here is to identify our spectator ions.

05:15 It looks like our nitrate is on both sides of the equation.

05:18 So we would identify that, and when we cross it out when writing our net ionic equation.

05:27 Oh, sorry, i was looking at a reaction c, which is the next reaction...