Look up the values of Ka in Appendix C for C6H5OH, HNO3,
CH3CO2H...

Question

Look up the values of Ka in Appendix C for C6H5OH, HNO3, 
CH3CO2H, and HOCl, and arrange these acids in order of: 
(b) Decreasing percent dissociation.

Accepted Answer

Welcome back, everyone. We need to write the correct arrangement. Ben's OIC acid with a acid dissociation constant or K equal to 6.5 times 10 to the negative fifth power. Cloris acid HCL 02 with A K equal to 1.1 times to the negative second power and hydrochloric acid with A K A of 1.3 times to the sixth power were also given final C seven H H with A K A of 4.8 times 10 to the negative 11th power. And we need to arrange these compounds in order of decreasing percent dissociation out of all of these compounds which are acids mentioned. We can recognize that from our memorized list of strong acids. Hydrochloric acid HCL is a part of that list. Recall that hydrochloric acid is going to dissolve in water. And because it's a strong acid, it's going to undergo complete dissociation where we have a regular reaction, arrow pointing towards our dissociated products where we have our chloride, an ion and our acid being a proton donor is going to also result in a second product being hydro knee. Um H 30 plus which forms because we want to begin with the assumption that each of our acids Have a concentration of one molar. And we know that are only given acidic compound HCL. In this case is the only one that undergoes a direct complete association. We can understand that our concentration of our strong acid HCL is going to equal our concentration of its conjugate base cl minus. And therefore, we can say that our concentration of cl minus chloride is going to equal our concentration of hydro ni. Um this is because we can see that we have a 1 to 1 ratio between HCL and water on our react inside which produced a 1 to 1 ratio of our products. And thus are concentrations of our products must be equal to one another since they're produced in equal amounts. And assuming that all of our listed assets have a concentration of one molar, we would understand that our concentration of HCL equal to one molar. So we can say in brackets, one molar of HCL is equal to one moller of chloride, its conjugate base. And so therefore, the concentration of chloride equal to the concentration of hydro ni um And just making more room here are equal to one molar. So let's go ahead and continue on to our next acid. So for any one of our next assets, let's just choose the one that has the next highest K A and that would be the K A that is the least negative, which we could see for Cloris assets. So we'll focus on that acid next. So let's write out its partial dissociation because we call that Cloris acid is not one of our memorized strong assets. And so therefore, dissolved in water, it's going to undergo a partial dissociation where we have an equilibrium aero separating our reactant from our products. So we'll produce the conjugate base of Cloris acid which donates a proton to water. And as a result, we form our chloride ion where we have C L 021 minus, we also will form of course, hydro ni um as our second product. Now, in this case, we have a partial dissociation, meaning that the dissociation of our weak acid Cloris acid is going to be dependent upon the given K a value from the prompt 1.1 times 10 to the negative second power. So we can't say that although we have a 1 to 1 ratio of our reactant and products that are concentrations of our products are going to be equal, we're going to have to create an ice chart, which we recall is I C E where we have its initial assumed value for our weak acid course acid, which we stated is going to be one moller for all of our acids. And then recall that in ice charts, we do not include liquids and then for our products will have initial concentrations of zero because we need to figure that out or rather because they haven't formed yet since their products. But what we are going to figure out is their equilibrium concentrations. Now, for the change we're going to have. So the change is going to be minus X for our reactant Cloris acid. And for our products, the change is going to be plus X based on their geometric ratios which have coefficients of one. Now, for our equilibrium, we just bring everything down from initial and change. So we have one minus X for Cloris acid, X for hydro ni um and our chloride an ion. Now we can utilize the given K A for Cloris acid in order determine our equilibrium concentrations. So recall that R K A is equal to the concentration of our products over the concentration of our reactant. And that would equal. In this case, we will write it separately. We'll say that we have are given K A for Cloris acid from the prompt as 1.1 Times 10 to the negative second power. This is set equal to our concentration of our products which according to our ice table is X times X. And then in our denominator, we have a concentration of our reactant 1 -1 according to our ice chart. And so in this case, we can check to see if we can neglect minus X. We ultimately are going to assume that our concentration of course acid HCL 02 at equilibrium is approximately equal to our initial concentration of one molar. And based on this or based on this relationship, we can understand that therefore minus X, the change of course acid is going to be negligible compared to one Moeller. And so we can neglect X from our denominator, meaning that we will now have the following setup where we have our K A on the left, 1.1 times 10 to the negative second power equal to our quotient. Where in our numerator, we now have X squared. And in our denominator, we just have the value one. And this will actually easily simple out to or sorry, simplify out to 1.1 times 10 to the negative second power equal to X squared. So we need to get rid of that square term, meaning we're going to raise both sides to a power of one half. And actually let's use a different color so that it's clear what we're doing. So we're raising both sides to a power of one half so that it cancels out our two on the right hand side. And on the left side of our equation will now come up with a value where we can say that X which represents our equilibrium concentration of hydro knee. Um H 30 plus is equal to a value of 0.10 and will stop at three sig figs moller. And through this calculated concentration of hydro NI. Um we can understand that our Cloris acid and its dissociation in water is dependent on not only its K A but its equilibrium concentration of hydro ni. Um So we'll actually add that in 2.10 moller which so far is less than our equilibrium concentration of hydro knee um or rather not equilibrium concentration of hydro ni. Um But our standard concentration of hydro knee um for the dissociation, the complete association of HCL, which we said is one molar. So right now, we can already have an order where we know that HCL should come before our Klores acid in our ranking because we're listing from decreasing percent association. So we can already rule out choice C as well as choice A. So let's keep going to figure out the correct answer between B and D. Now let's pick our second or rather third largest K A which would be for Ben's OIC acid 6.5 times 10 to the negative fifth power. So we're going to have to do the same process we did for Cloris acid for Ben's OIC acid. So in this case, we have and we'll separate the work here for Ben's OIC acid C six H five C 02 H. This is Aquarius and undergoes partial dissociation since it's a weak acid in water where we have an equilibrium separating our products. Once Ben's OIC acid donates a product, we have our benzoate an ion C six H five C 021 minus as well as of course, Hydro Knee um as our second product and sorry, the label for Hydro Knee um should be Aquarius. Okay. So moving forward, our next step is to of course, assume our initial concentration of our weak asset is going to be one moller. We don't again consider water in our ice chart and our initial concentration of our products are zero are change is going to be again minus X for our reactant and plus X for our two products carrying everything down, we have one minus X for our reactant and X for our products. And now we're going to set up our K equation. So from our prompt, we see that the K A of Ben's OIC acid or weak acid is equal to 6.5 times 10 to the negative fifth power. And this is set equal to our concentration of our products, which is our, according to our ice chart, we have the conjugate base C six H five C 02 minus R benzoate an ion multiplied by our concentration of hydro diem H 30 plus in our numerator, this is divided by our concentration of reactant, which is our Ben's OIC acid in our denominator. So C six H five C H. So plugging in those concentrations from our ice table. In our numerator, we have X times X again. And then in our denominator, we have one minus X. And just as before, we're going to neglect X because we're going to assume that its value is going to be significantly smaller than the initial concentration of Arbenz OIC acid. And so neglecting X means that we'll have now X squared Over in our numerator, this is over one equal to RKA of 6.5 times 10 to the negative fifth power. And so simplifying this, what we'll have is the same K A 6.5 times 10 to the negative fifth power on the left hand side equal to X squared. And we want to take both sides to a power of one half that's going to cancel out the two on the right hand side or the square term rather. And what we'll have is our X value is now isolated and equal to our concentration of hydro nian, which tells us our percent association of our weak acid benzoate acid in this case. And this is equal to a value of 8.1 times 10 to the negative third power moller. So now this leads us to our last asset to calculate the percent association for. And that would be for our final where we have assumed one molar of fennel. So C seven H 70 H reacting with water or dissolved in water. So it's in equilibrium because fennel is a weak acid. So it undergoes partial dissociation. And in an equilibrium, we form its conjugate base R C seven H seven oh minus an ion plus hydro knee. Um of course, since again, final donates a proton to water. So, going into our ice chart, we follow exactly the same process as before we have one molar of our weak acid phenol. We do not consider our second reactant water and we have zero initially, of our products. We have minus X for our reactant and plus X for our products, we have one minus X for phenol, our reactant and then X for our two products. And now again, we write out the, the K equation where we have our K A for final given in the prompt as 4. times 10 to the negative 11th power set equal to our concentration of our products, which is our conjugate base C seven H seven oh minus times the concentration of hydro ni um divided by, in our denominator, we have our concentration of phenol C seven H 70 H. So this would be translated to in our numerator based on our equilibrium values X times X for our products. And in our denominator one minus X in which again, we neglect X. So simplifying this, we have our K A 4.8 times 10 to the negative 11th power equal to X squared divided by one. Again, we have 4.8 times 10 to the negative 11 power in the next line equal to just X squared in which now we will take both sides to a power of one half to cancel out the square term. And this will result in our X term completely isolated, equal to a value of our concentration of hydro ni um Which equals a value of 6.9 times 10 to the -6 power moller. And so, based on all of our calculations above, for our concentration of hydrogen, we can understand that the lower our concentration of hydro ni um at equilibrium will correspond to a lower percentage of dissociation for a weak acid. We can also interpret that the higher our K A value for our weak acid that will correspond to a greater value of percent dissociation or hydro ni um associated. So a greater concentration of hydro ni um dissociated at equilibrium. And so, based on our calculations, we can confirm that our hydrochloric acid will be the most associated associated weak acid with its concentration of hydro knee um being dissociated to one molar at equilibrium. This was greater than our concentration of Cloris acid that associated which at equilibrium, the percent association was 10.10 moller for hydro ni. Um This was even greater than our percent association of our compound Ben's OIC acid C six H five C 02 H in which we found that at equilibrium, the concentration of hydro ni um was associated to 8.1 times 10 to the negative third power Moeller and just making more room, we have the last compound to add into this outline is our final where we have C seven H H. And we just calculated above that the percent of hydro diem dissociated was 6.9 times 10 to the negative six power moller at equilibrium, which was our smallest value. And so from this order, we've determined our final answer to complete this example in order of decreasing percent association for each of our weak acids, this corresponds to choice being the multiple choice as our final answer. So let us know if you have any questions and I'll see you in the next video.

Look up the values of Ka in Appendix C for C6H5OH, HNO3, CH3CO2H, and HOCl, and arrange these acids in order of: (b) Decreasing percent dissociation.

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