Ionic Strength of Soluble Salts

So let's take a look at the association of silver bromide. So, silver bromide is an ionic solid when we throw it into solution, it's going to basically disassociate into its ions. But in the process an equilibrium is established that just means that very little of these ions are produced during this process. Now connected to the solid ability of any ionic solid is R. K. S. P. Value are solid ability product constant. Remember we talked about this in the past when it comes to the K. S. P. Values for ionic solids, they're typically much less than one because these ionic solids have minimal solid ability within a solvent. Now let's say to this pure solution of water that has this ionic compound dissolving. We will produce x amount of these ions. Now let's say we took that same silver bromide and instead of putting it in pure water we decided to put it into 0.10 moller sodium bromide, realize here that sodium bromide and silver bromide have something in common. They both have bromide ion involved. Remember when it comes to K. S. P. Of ionic solids, we have initial change in equilibrium involved here. The common ion we have is bromide. So initially this would be 0.10 moller this has no common ion right now. So we'd say that zero. Remember we called this the common ion effect but we have an initial amount of water, both ions depending on the solution present. Remember this type of situation helps to decrease the overall cell stability of my ionic compound. So remember think of settlers principle, we're adding more of this. I on here. So according to the settlers principle, when I add product I have to move in reverse to get rid of it. This will cause my reaction to favor the reverse direction towards the solid. Once again. And remember, sorry ability is how much of this ionic solid can we get to dissolve if it's moving backwards it's not dissolving, it's crystallizing. Same thing would happen if I had .25 molar of silver acetate here. The common ion is silver. So in that solution would have an initial amount of .25 molar of silver. Again falling into common iron effect which again would lower my overall psy ability of my ionic compound. So with these common ions we have the overall psy ability decreasing as a result of the common ion effect. But let's say we didn't put it in a solution that had a common eye on, let's say instead we put it in .01 molar sodium per chlorate. Now here sodium is the positive ion. And chlorate is the negative ion. Neither one of those ions matches up with the ions associated with my K. S. P. Value. So here there is no common ion effect involved. What would happen is that this positive ion here would surround this bromide ion. And they basically it would in a way decrease the amount of free floating bromide ions within the solution. The same way here this per chlorate ion would surround this silver ion. As a result, I need to make more of these ions because they're being surrounded by these non common ions here. So to make more of the ions, the ionic solid has to move forward by the settlers principle to re two remake more of those free floating product ions. So here if we put in ions that are not common to our equation, the overall sai ability of my ionic compound will increase as a result of ionic strength. So ionic strength is just the measurement of all the ions in the quickest solution. And ionic strength itself has an equation here we're gonna say ionic strength represents the interactions between ions and water and ions in the solution, ionic strength which is new, equals half the summation of the concentrations of the ions times their charges squared. So if you have two ions, it'd be the concentration of the first ion times its charge squared plus the concentration of the second ion times its charge squared. And if we add additional ions we just keep going until we found the ionic strength. So remember the common ion effect helped to decrease the overall suitability of an ionic compound. But non common ions actually helped to increase sought ability because of ionic strength, they surround those free floating ions. And so your ionic solid adjust by moving in the four direction to create even more of those ions. Now that we know the basic understandings of ionic strength. Take a look at example that's left below. Once you're done, click over to the next video and see how I approach and calculating the ionic strength of that particular compound.

So here we have to calculate the ionic strength of the following ionic compound. So here this compound is copper two sulfate. It breaks up into copper two ion plus the sulfide ion. Now how do I know that copper has a plus two charge? Well you need to remember your poly atomic ion S. +03. Here is sulfide ion, right? So that's the formula sulfide ion. If the overall compound is neutral and it's a 1 to 1 relationship, one comfort of one cell fight then if this is two minus, comprehends to be two plus the charges cancel out and have a neutral ionic compound at the end. So right now we have a 1 to 1 relationship of the ions. If the concentration for the entire ionic compound is 0.10 moller then each ion produced will have that same concentration. So here this is 0.10 moller and this is 0.10 moller. Now we're gonna say that our ionic strength equals half. So the concentration of the first ion which is .010 Times its charge squared. So plus two squared plus the concentration of the second ion times its charge squared. So when we do all that, we'll get an answer of .04 for the ionic strength of 0.10 moller, copper two sulfate. So just follow the steps that we did if there were multiple ions. So let's say that there were two coppers for some reason that we'd have to multiply the concentration by two because we have to take into account all the copper ions with the overall concentrations that they have. But here, since it's a 1 to 1, the initial concentration I have for the compound matches the concentrations for the ions provided. Later on, we'll continue more questions that deal with determining the ionic strength of an overall ionic compound. So again, make sure you remember your poly atomic ions and how to correctly classify uh the relationships between the ions and their charges involved.

So just as we stated in the previous videos when figuring out the ionic strength for any ionic compound, it's best to break it up into its ions to see how many ions we have of each one. So we can figure out their concentrations. So here we have aluminum carbonate, aluminum carbonate breaks up into two aluminum ions Plus three carbonate ions. The concentration of the entire ionic compound is this but the number of ions of each type determines their overall concentration. Here we have two aluminums, so it be two times the original concentration. So the overall concentration of aluminum ions would be .060 moller. Here we have three carbonates, so that's three times the original concentration. So the overall concentration of carbonate is .090 moller. Now we use the formula to determine ionic strength. So ionic strength equals half. Remember it's the summation of we're gonna say the concentration of the ion times its charge squared plus concentration of the next ion times its charge squared. So that'll be half the concentration of my aluminum ion is 0.60 moller times its charge squared. So plus three squared. And then plus the concentration of bicarbonate ion, which is .090 moller Times its charge squared which is -2 squared. So here, if you work out what's in the brackets correctly, you'll get half of .90. So that would mean that my on ionic strength here would be .45. Now that you've seen this example, Try the next one and this one we're asked to figure out the total ionic strength of this solution. So here we have two different ionic compounds mixing together. What effect would that have on the concentration of each ion present to determine the final ionic strength? Try it out for yourself and then come back and see how I approach the same example to question.

So here it asks, what is the ionic strength of a solution? That is 0.1 molar sodium phosphate and 0.5 molar sodium hydrogen phosphate. Alright, so we're gonna break these compounds up into their ions here, sodium phosphate breaks up into three sodium ions plus one phosphate ion, sodium hydrogen phosphate breaks up into two sodium ions plus one hydrogen phosphate ion. Remember the number of each particular ion has an effect on the overall concentration that it has. So hear the overall morality of the entire sodium phosphate molecule or compound is 0.1 molar. But there are three sodium is here, so that's three times 30.1 molar Here. This would just be .1 molar since there's only one here, this is $0.5 for this. So this is two times 20.5 moller since there's two sodium. And here this would just be 20.5 molar. For ionic strength. Remember ionic strength equals half and it's the summation. So here we have the concentration times its charge squared. But here we have a lot of ions. So C two, Z two squared plus C three, Z three squared plus Z four Z four, C four, Z four squared. Now I know that the both of these are sodium. So technically if you want you can combine them together but I'm keeping them separate because they have different concentrations. So this equals half. So the concentration of the first sodium ion, it's three times 30.1, so that's gonna be 0.3 moller times its charge squared. So 0.1 squared plus the concentration of the next ion which is phosphate ion, there's only one of it. So let's just 10.1 molar times negative three squared. Plus we're gonna have the concentration of the next sodium ions, there's two of them. So that's 20.1 molar plus one squared. Plus we have 0.5 moller times plus two squared. Okay, so that's gonna give me here, if you work this out correctly, it's half times everything within there. So we're gonna have half times .150. So that's gonna give me a final ionic strength of .75. Now that you've seen these two examples attempted this attempt. This practice question that's left here on the bottom. I remember the first thing you should do is to split this compound into its ions from there, determine how many you have of each particular ion. And then from that you can determine their concentrations. To help you determine the final ionic strength for the entire ionic compound. So guys attempted on your own. Once you've done that, come back and see how I approach the same exact practice question