Start typing, then use the up and down arrows to select an option from the list. # Analytical Chemistry

Learn the toughest concepts covered in your Analytical Chemistry class with step-by-step video tutorials and practice problems.

Activity and the Systematic Treatment of Equilibrium

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

## pH Revisited Calculations

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Glass Electrodes 2m
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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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Glass Electrodes 6m
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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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Glass Electrodes 6m
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