Complete Ionic Equations - Online Tutor, Practice Problems & Exam Prep

Complete Ionic Equations

Complete ionic equations show aqueous compounds as associated ions. Now remember, in a molecular equation we have our aqueous compounds, but we also possess solids, liquids, and gases. When it comes to solid, liquids, and gases, they themselves never break up into ions. It's only the aqueous compounds. And remember, we use the solubility rules to determine if something is aqueous or not. Now when it comes to the complete ionic equation, we're going to say it comes from the molecular equation, and because of that, you're going to have to remember to distribute the coefficient of each compound to determine the correct number of ions. So as we go into going from the molecular equation to our complete ionic equations, we'll see what that entails.

Complete Ionic Equations Example 1

In this example question, it says, convert the following molecular equation into a complete ionic equation. So here we have 3 moles of calcium bromide aqueous reacting with 2 moles of lithium phosphate aqueous, producing 6 moles of lithium bromide as an aqueous compound plus 1 mole of calcium phosphate solid. All right. So remember, we can only break up aqueous compounds. So only these 3 compounds will break up. If you are a solid, a liquid, or gas, you will not break up in the complete ionic equation. Also, remember that the coefficient gets distributed to each one of these compounds.

Now, what exactly does that mean? Well here, that means that this 3 is going to get distributed to the number of calciums and the number of bromides. So it gets distributed to Ca and Br. When this breaks up into its ions, we know that we're going to have calcium ion aqueous plus bromide ion aqueous. Then we distribute the coefficient, so that's going to give us 3 calcium ions. And then this is 3 times 2, 6 bromide ions. Plus, also remember that when we have an ion in aqueous phase within a solution, this 2 is going to get distributed here and here, so we know that's going to give us 6 lithium ions because it is 2 times 3 plus 2 phosphate ions. Remember phosphate is a polyatomic ion. 6 lithium ions plus 6 bromide ions because 2 gets distributed to lithium and bromide as well. This is a solid, so it stays together. It does not break up into ions. So plus 1, not 6, 1 calcium phosphate solid. So this here represents our complete ionic equation. Remember to break up only aqueous compounds and remember to distribute the number of coefficients to each one of those ions that you form.

Complete Ionic Equation:

3Camtext(aq)+ 6Br−(aq)+ 6Li+(aq)+ 2PO43−(aq) → 6Li+(aq)+ 6Br−(aq)+ Ca3(PO4)2(s)

Complete Ionic Equations

Now, a net ionic equation shows the ions participating in the chemical reaction by removing what we call the spectator ions. These spectator ions represent compounds that are both reactants and products. So, they're found on both sides of an equation, but they are not part of a net ionic equation. Now, when it comes to the net ionic equation, we're going to say it comes from the complete ionic equation. So remember, we started out with our molecular equation. From there, we're able to determine our complete ionic equation, and then from our complete ionic equation, we now are looking at our net ionic equation. So, this is the path you have to take to get to our net ionic equation. Now that we know how they're connected, let's move on to the next video and take a look at an example question.

Complete Ionic Equations Example 2

Based on the given reactants, provide both the molecular equation and the complete ionic equation. So here we have Ammonium Sulfate reacting with chloride. So for step 0, we're just going to follow steps 1 to 4 that we've learned in the past to first give the molecular equation. Now remember, when it comes to these steps, we're first going to break up each of these compounds into their ionic forms. So ammonium sulfate breaks up into the ammonium ion and the sulfate ion. Don't worry about the fact that we have 2 ammoniums here. We'll discuss that later on as we're going into the complete ionic equation. And then calcium chloride breaks up into calcium ion and chloride ion. Now remember, we're gonna swap ionic partners because opposite charges attract. This plus one charge is attracted to this minus one charge. So when they connect together, remember when the numbers are the same, they just simply cancel out. So here that will give me NH₄Cl⁺, then we're gonna hav

Provide the net ionic equation that occurs when the following aqueous compounds are mixed together: 

Copper (II) Bromide and Lithium Hydroxide

Here in this practice question, it says to provide the net ionic equation that occurs when the following aqueous compounds are mixed together. Alright. So they're telling us here that copper(II) bromide and lithium hydroxide are aqueous compounds. So here, copper(II) means we have Cu2+. Bromide is in group 7A, so it's -1. We have lithium which is in group 1A, so it's +1. And then we have hydroxide which is a polyatomic ion so it's OH-1. For our reactants, what we initially have is CuBr2 (aqueous) plus 2 LiOH (aqueous). Here, we have to determine what our products will be. So here we have to switch partners. This -1 and this +1 are attracted to one another. Because the numbers in their charges are the same, when they combine, they just cancel out. So here we have lithium bromide. Based on our solubility rules, we know that anything connected to group 1A ion is aqueous. Then plus, we have copper(II) ion and we have a hydroxide ion combining. The numbers in their charges, now though, are different, so they don't just simply cancel out. Remember, when the numbers are different, they crisscross. So here that would give us Cu(OH)2. Remember, hydroxide is normally a precipitate. The only way for it to become aqueous is if it's connected to anything from group 1A, acetate, or nitrate, or if it's connected to calcium, barium, or strontium. So here it's going to be a solid. If you don't quite remember your solubility rules, make sure you go back and take a look at them. This represents our molecular equation. We still have to balance it. We're going to say here we have 2 bromines, so we're going to put a 2 here, but then that gives me 2 lithiums so I have to put a 2 here. It's balanced out because this 2 would distribute to this OH, which is how we have 2 OH's here. This represents my balanced molecular equation. Next, we are going to change our balanced molecular equation into a complete ionic equation. In a complete ionic equation, we break up only the aqueous compounds and make sure you distribute these coefficients to the number of ions that you create. So here, we're going to have: Cu2++2Br-+2Li++2OH-→2Li++2Br-+Cu(OH)2(s) This represents our complete ionic equation. Lastly, we need to give the net ionic equation. Remember, in the net ionic equation, you remove spectator ions. The ions that look exactly the same because they exist as both products and reactants. So here these are spectator ions. They will not be brought down. So what's left would be: Cu2++2OH-→Cu(OH)2(s) This represents our net ionic equation based on the 2 aqueous compounds given to us initially.

Which of the following reagents could be used to separate the two anions from a solution containing magnesium nitrate and cesium hydroxide?

• A. NH₄CN
• B. NaCl
• C. KNO₃
• D. ZnBr₂
• E. CsBrO₃

Which of the following reagents could be used to separate the two cations from a solution containing Lead (IV) acetate and cesium permanganate?

• A. Sr(NO₃)₂
• B. TiC₂H₃O₂
• C. K₂S
• D. NaClO₄
• E. KNO₃