Two 20.0-mL samples, one 0.200 M KOH and the other 0.200 M CH 3 NH 2 , are titrated with 0.100 M HI. b. Is the pH at the equivalence point for each titration acidic, basic, or neutral?

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

Ch.17 - Aqueous Ionic Equilibrium

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Tro 4th Edition

Ch.17 - Aqueous Ionic Equilibrium

Problem 63b

Chapter 17, Problem 63b

Two 20.0-mL samples, one 0.200 M KOH and the other 0.200 M CH₃NH₂, are titrated with 0.100 M HI. b. Is the pH at the equivalence point for each titration acidic, basic, or neutral?

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Determine the nature of the titration: KOH is a strong base, and CH_3NH_2 is a weak base. HI is a strong acid.

For the KOH titration: At the equivalence point, KOH will be completely neutralized by HI, forming water and K^+ ions. Since K^+ is a neutral ion, the solution will be neutral at the equivalence point.

For the CH_3NH_2 titration: At the equivalence point, CH_3NH_2 will be completely neutralized by HI, forming CH_3NH_3^+ ions. CH_3NH_3^+ is a weak acid, which will make the solution acidic at the equivalence point.

Conclude that the pH at the equivalence point for the KOH titration is neutral, while for the CH_3NH_2 titration, it is acidic.

Remember that the nature of the ions formed at the equivalence point determines the pH: neutral ions result in a neutral pH, while acidic or basic ions result in an acidic or basic pH, respectively.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Titration and Equivalence Point

Titration is a quantitative analytical method used to determine the concentration of a solute in a solution. The equivalence point occurs when the amount of titrant added is stoichiometrically equivalent to the amount of substance in the sample. At this point, the reaction between the acid and base is complete, and the pH of the solution can be determined based on the nature of the reactants.

Acid-Base Properties of Salts

When a strong acid reacts with a weak base, the resulting solution at the equivalence point tends to be acidic due to the formation of a conjugate acid. Conversely, a strong base reacting with a weak acid results in a basic solution at the equivalence point. Understanding the strength of the acids and bases involved is crucial for predicting the pH at the equivalence point.

pH Scale and Measurement

The pH scale measures the acidity or basicity of a solution, ranging from 0 (very acidic) to 14 (very basic), with 7 being neutral. The pH is determined by the concentration of hydrogen ions (H+) in the solution. At the equivalence point of a titration, the pH can indicate whether the solution is acidic, basic, or neutral, depending on the nature of the reactants involved.