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Activity and the Systematic Treatment of Equilibrium

In this section we now include activity coefficients when calculating the pH values of solutions.

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So remember that ph is just the negative log of H plus or H. +30. Plus for example, wanted to ask us if at 50 degrees Celsius, the ionization of pure water Kw is 7.94 times 10 to the negative 14. What is the ph of a neutral solution? Well for a question like this we're dealing with pure water with no free floating ions within it. Here we're dealing with neutral solution and in a neutral solution your H plus concentration is equal to euro H minus concentration. Since we don't know either one of them, we're gonna say they're both equal to X. Remember both of these ions are related to each other by the fact that H plus or H. +30 plus times O. H minus equals kW. And here since both are X. And they're multiplying that's going to be X squared and that equals 7.94 times 10 to the negative 14. Remember your K. W. Like all other equilibrium constants is temperature dependent. Once the temperature is no longer 25 degrees Celsius, R K. W will not be 1.0 times 10 to the negative 14 it becomes a brand new number. So here we're going to take the square root Of both sides to isolate X. So that's gonna give me 2.81 7. 8 times 10 to the -7 Moller. Now that equals H plus as well as O. H minus. But since we're looking for ph we're going to focus only on the H plus. So P H equals negative log of H plus Which equals negative log of 2.8178 times 10 to the -7. So that gives me 6.55 as my concentration after rounding. So that gives us option B. Now that's when we're dealing with pure water, what would happen when we're dealing with uh figuring out the ph still in pure water, but now it has free floating ions in it. We have to take into account what does this do to both our ionic strength as well as our activity coefficient. Think of those things as you read example to you could attempt it on your own but if you get stuck, don't worry, just come back and see how I approach that same example to question.

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Alright so here we're dealing with um the same pure water. So we're gonna say because of the same pure water that means that H plus still equals O. H minus and they both still equal X. But now we're dealing with free floating ions as well. Now an A plus and B. R minus have nothing to do with H plus and O H O H minus. And because they don't have a common ion effect involved, that means they're gonna have some form of ionic strength and some influence on these two ions. So again if they're giving you ionic compounds and those ions don't match up with the ions that you're investigating, That means ionic strength is in play. And if ionic strength is in play that means that our activity coefficients are also in play. So here our equation becomes a bit different. It becomes H plus times the activity coefficient of h plus times O H minus times the activity coefficient of O H minus equals K. W We're still dealing with this 50 degrees Celsius from earlier, so we're still dealing with the same kw Now we know that these two are X. But we need to know what the activity coefficients are for them in order to figure out our correct H plus ion concentration. Now to do that, we're going to figure out what the ionic strength of sodium bromide is. Once we get that ionic strength we're gonna look it up on the activity coefficients table and see what the activity coefficients are for H plus and O H minus plug those into find our final answers. Alright, so we're gonna need room guys. So let me take myself out of the image. So we're gonna say here ionic strength equals this breaks up into N. A plus one and B. R minus one. So this is half the concentration of each ion times charge squared plus the concentration of the other ion times charge squared. So the ionic strength is .005. Look on the activity coefficients table, look for H plus ion and O H minus. What are their values When the ionic strength is this number? If you look carefully you'll see that for H plus it's 0.933 and for O H minus its 00.9 to six. Remember K. W is the same K. W. That we saw previously which is 7.94 times 10 to the negative 14 when the temperature is 50 degrees Celsius. So we're gonna plug that in. Now we're going to multiply these two numbers together. So we multiply both of them together. It gives me .863958. And then we have x variables multiplying. So that's x squared. Now we're going to divide both sides by .863958. So it's gonna give me X squared X squared equals 9.19 times 10 to the -14. Take the square root of both sides here. So that gives me X X equals 3.03 times 10 to the -7 Moller. That's gonna give me the concentration of both H. Plus and O. H. Minus. But we're only concerned with the H plus because we're looking for ph so up above we said P H. Equals negative log of H plus. But remember we're dealing with ionic strength on and because of that we're gonna have to deal with the activity coefficient. So input the activity coefficient of H. Plus. Again now we're gonna do negative log the concentration is 3.3 times 10 to the negative seven times the activity coefficient of H. Plus which is 70.933. Now what's happening here is you're gonna multiply these two numbers together and then you're going to take uh the negative log of that because remember both of these things together represent the activity of H plus ion so equals negative log of. When I multiply those two numbers together, I get 2.828 times 10 to the negative seven Which equals 6.54. Okay so our activity coefficient had a small impact on the ph or initially I got 6.55 from my P. H. I now have gotten 6.54 realized here that if we increased the concentration and or if we increased the charges of the ions the impact on the ph would have been much greater than just a difference of 0.1. This just goes to show you that if we have non common ions floating around, we have to take into account ionic strength and therefore the activity coefficients of the ions involved. Now that we've seen this, take a look at example, three left on the bottom. The same basic premise is being involved here. We're dealing with the K. S. P. Of barium hydroxide, but we have involved lithium nitrate, non common ions. So again, that means that your ionic strength is in play, which means activity coefficients are in play. So remember what we just did and see if you can solve this question.

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So here it says find the ph of a saturated solution of barium hydroxide when dissolved in 0.5 moller of lithium nitrate. So here we're dealing with moller lithium nitrate here the K. S. P. Of barium hydroxide is 5.0 times 10 to the negative three. Well we know that we're dealing with barium hydroxide. So this is my ionic solid. So we're talking about how that ionic solid breaks up into its ions. We know that we don't deal with solids here. This would be X. And hear this because there's a two here, this would be two. X. K. S. P. Equals just products because we're ignoring the solid, that's the reactant. So we have B. A. Two plus now realize here that because we're dealing with non common ions, that means ionic strength is in play and therefore the activity coefficient is in play. So it's gonna be activity coefficient of barium to ion times hydroxide ion because the two here it's squared times the activity coefficient of hydroxide also squared. Alright, so we're going to need to determine what our concentration of hydroxide ion is to figure out what our P. O. H. Is. And once we do that we can find our ph but first I need to take into account these non common ions from it, I'll be able to determine my ionic strength. So this is made up of lithium ion and nitrate ion, it's a 1-1 relationship. So the concentrations will not change. So .05 molar for the lithium ion times its charge squared plus the concentration of the nitrate ion times its charge squared. That gives me .05 for the ionic strength. And now that we know what the ionic strength is. We look up on our activity coefficients chart for the activity coefficients for barium ion and hydroxide ion. So when you look those up so we have barry mayan here which is X. When you look up at its activity, cold fishing you're gonna see that it is equal 2.465 Times Hydroxide which is two x. Don't forget that it's squared times the activity coefficient. When you look it up you get .81 squared. Alright. K. S p is 5.0 times 10 to the -3 equals. Alright so we have a lot of numbers that are multiplying with one another. So we're gonna have to take all of those into account. So we have We're gonna have .465 times .81 Squared two Squared is four. So that's when you multiply all those together. That gives me 1.22035. And don't forget we have our X variables. We have X. Here and then X. Here is getting squared. So that's X squared X. X times X squared, gives me X cubed, Divide 1.22035. So X cubed Equals .004097. Take the cube root of both sides here. So when we do that that's gonna give me x equals .160016 Molar. But remember we're looking for the concentration of hydroxide ion O H minus and O h minus does not equal X equals to X. So h minus concentration equals two X. So it's two times this number we just found. So that's .320032 molar. Next we're going to figure out what the activity of hydroxide is. So the activity here would equal the concentration of the hydroxide iron. We just found times its activity coefficient. So the concentration we just found is this And we bring back that activity coefficient we found earlier of .81. So that's gonna give me .2592-5. Now that I have the activity of hydroxide, I can take the negative log of that to find P. O. H. So that's negative log of that number. So that's gonna give me a p.. h. of .586 ph equals 14 minus P. O. H. So my ph at the end is approximately 13.41. So again remember when we're dealing with an ionic compound that's dissolving in solution and then all of a sudden they introduce non common ion effect ions. That means that those are going to help increase the ability of my on ionic compound by ionic strength since ionic strength is in play, that means activity coefficients are in play. So we're gonna have to incorporate them within our calculations to find the final concentrations of the specified ions. So keep in mind some of the techniques we've used here, which is just a continuation of concepts we've learned in terms of calculating PH.

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