Calculate the concentration of all species in a 0.225 M C 6 H 5 NH 3 Cl solution.

Hey everyone we're asked to enumerate the species present in a 0.284 moller methyl ammonium chloride solution with their concentration and polarity. And they provided us the KB. So we know that we have methyl ammonium chloride and this will dissociate into the protein eight version of our methyl amine, plus our chlorine ions. Now let's go ahead and take our protein eight version of methyl amine and react this with water. When we react this with water, we end up with methyl amine plus our hydro ni um ions. Now let's go ahead and create our ice chart. So we were told that we started off with 0.284 moller of our pro name version of methyl amine And we started off with zero of our products. We can go ahead and disregard our water since it is a liquid and our change is going to be a minus X. On our react inside and a plus X on our product side. Since we're losing reactant and gaining products at equilibrium, we have 0.284 minus X in our react inside and an X. And an X in our product side. Now let's go ahead and use our KB to sell for R K. A. Now we know for our K. It will be our K. W divided by R K. B. So plugging in our values, we know that our K. W is one times 10 to the negative 14. And we can go ahead and divide that by the KB they provided which was 4.47 times 10 to the negative four. This will get us to a K. A. Of 2.2371 times 10 to the negative 11. And again we have to use our K. Since we are in our acidic form and we're producing hydro knee um ions. So taking our K. A. And solving for X, we take our 2.2371 times 10 to the negative 11. And this is going to be equal to our products over our reactant. So we have x times x over 0.284 -1. We can go ahead and check if we can disregard our X. And our denominator by taking our 0.284 and dividing it by 2.2371 times 10 to the negative 11. Which was our K. A. If we get a value greater than 500 then we can go ahead and disregard it. And in this case we do get a value greater than 500. So we can safely disregard our X. Now let's go ahead and simplify this a bit further. So we have X squared is equal to 2.2371 times 10 to negative 11. And we multiply both sides by 0.284. In order to get rid of the 0.284 in our denominator. Taking the square root of both sides. We end up with an ex of 2.5206 times 10 to the -6 Molar. Now that we have our X. Let's go ahead and calculate our concentration of our pro Nate version of our methyl amine. And we can do so by taking our 0.284 and subtracting that 2.5 to 06 times 10 to the negative six Mohler. This will get us to a value of 0. Mueller. Next let's go ahead and calculate our concentration of our chlorine ions. Now the concentration of our chlorine ions can be found by taking our 0.2 a four moller of our methyl ammonium chloride. And looking at the multiple ratios so we know that one mole of methyl ammonium chloride contains one mole of our chlorine ion, calculating this out. We end up with a value of 0.284 moller of our chlorine ions. And we can also calculate the concentration of our hydroxide ions by taking our KW and dividing this by our concentration of our hydro ni um ions. So plugging in our values, we get one time sensing negative 14 and we can go ahead and divide this by our X which was 2.5206 times 10 to the negative six Mohler. This will get us to a concentration of 3.97 times 10 to the -9 moller to summarize our concentration of our methyl amine. Came up to 2.5, 2 times 10 to the -6 Mohler. And this was also the concentration of our hydro ni um ions. The concentration of our pro native version of our methyl amine came up to 0.284 moller, and the concentration of our chlorine ions also came up to 0.284 moller. Lastly, the concentration of our hydroxide ions came up to 3.97 times 10 to the -9 moller. And this is going to be our final answer for this question. So I hope this made sense and let us know if you have any questions.

Ch.16 - Acids and Bases

Tro - Chemistry: A Molecular Approach 4th Edition

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Ch.16 - Acids and Bases

Problem 108

Chapter 16, Problem 108

Calculate the concentration of all species in a 0.225 M C₆H₅NH₃Cl solution.

Verified step by step guidance

Identify the components of the solution: C6H5NH3Cl is an ionic compound that dissociates into C6H5NH3+ (anilinium ion) and Cl- (chloride ion) in water.

Recognize that C6H5NH3+ is a weak acid, which can donate a proton to water, forming C6H5NH2 (aniline) and H3O+ (hydronium ion).

Write the equilibrium expression for the dissociation of C6H5NH3+ in water: C6H5NH3+ + H2O ⇌ C6H5NH2 + H3O+.

Use the acid dissociation constant (Ka) for C6H5NH3+ to set up an ICE (Initial, Change, Equilibrium) table to find the concentrations of C6H5NH2 and H3O+ at equilibrium.

Calculate the concentration of Cl- ions, which is equal to the initial concentration of C6H5NH3Cl, since it fully dissociates in water.

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Key Concepts

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

Acid-Base Chemistry

C6H5NH3Cl, or phenylammonium chloride, is a salt formed from the weak base aniline (C6H5NH2) and hydrochloric acid (HCl). In solution, it dissociates into C6H5NH3+ (the conjugate acid) and Cl-. Understanding the behavior of weak acids and bases is crucial for determining the concentrations of species in solution.

Dissociation of Salts

When salts dissolve in water, they dissociate into their constituent ions. For C6H5NH3Cl, it dissociates completely into C6H5NH3+ and Cl-. The concentration of each ion in the solution can be directly derived from the initial concentration of the salt, which is 0.225 M in this case.

Equilibrium and Concentration Calculations

In solutions involving weak acids or bases, equilibrium plays a significant role. The concentration of the conjugate acid (C6H5NH3+) will influence the pH of the solution. To calculate the concentrations of all species, one must consider both the initial concentration and any changes due to dissociation and equilibrium shifts.