Two 20.0-mL samples, one 0.200 M KOH and the other
0.200 M CH3NH2...

Question

Two 20.0-mL samples, one 0.200 M KOH and the other
0.200 M CH3NH2, are titrated with 0.100 M HI.
c. Which titration curve has the lower initial pH?

Accepted Answer

Hello everyone today. We are being given the following question and asked to solve for it. So it says a 0.115 Mueller of H. I solution is used to titrate to separate based samples. The first one is a 20 millimeter 200.145 Mueller lithium hydroxide solution and the second one is a 25 millimeter 250. moller of CH three ch two. NH two solution were then asked to find which of the two track workers will have a higher ph at the start. The first thing I want to know is that we're working with lithium hydroxide which is a very strong base. And so therefore the concentration of lithium hydroxide is going to equal the concentration of hydroxide ions to solve for the ph which you must do first solve for the initial ph of each solution. We must use the P. O. H. Equation which says that the P O H. Is equal to the negative log of our hydroxide ions plugging in our hydroxide ions which we have the concentration of. We have 00.145 molar and that gives us a P O. H of 0.839. Of course P H plus P L H is equal to 14. So therefore P H is going to be equal to 14 minus R P O H which is 0.839. Giving us a ph of 13.16. Next we must form an ice table. And so we have to show the dissolution of CH three CH two. NH two and water. So we're gonna say this is going to react with water to give us CH three CH two, NH three plus plus some hydroxide ions. We're gonna set up our I. C. E. Chart. Of course, our initial concentration for this ammonia, ammonia containing compound is going to be .150 moller for water, it's going to be a liquid. So all of the concentrations are going to be um equal to X. And then we have zero concentration for hydroxide and for the ammonia containing compound for the products. Since we're tight trading it, we're going to go ahead and make the change which is representative see being minus X. For the left hand side, but on the product side it's going to be plus X. Since we're adding a specific concentration And he is essentially going to be I plus c. So we're gonna have .150 -1. And they were just going to have X&X for the products. And so he said that the KB of ethyl amine or the ethyl group is equal to 4.5 times 10 to the negative fourth. And so when we solve, we solve for our KB, we're going to say that this KB is equal to X squared or the products, the two X is multiplied, become X squared over 0.50 minus X. Or our reactant since X is negligible, it's so small, we can go ahead, ignore it and we can go ahead and just say that that's going to be X squared over 0.15. So solving for X, we get X squared is going to be equal to 0.123440675. And then X is going to be equal to 0.822 molar. As we solve for P. O. H. Before or ph before we're going to use P. O. H. We're going to say the P. O. H. Is equal to the negative log of our hydroxide concentration. We're gonna plug in negative log of our 0.00822, which will give us a P. O. H of 2.8 as before ph plus P O. H. Is going to give us 14 and so does offer ph we get you that ph is equal to 14 subtracted by our P. O. H, which is two point oh eight. And we're left with a ph of 11.92. And so we can see that the lithium hydroxide will produce the higher ph at the start Being 13.16 and therefore lithium hydroxide will have a higher ph at the start. I hope this helped And until next time

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