Solve an equilibrium problem (using an ICE table) to calculate the pH of each solution. a. a solution that is 0.195 M in HC 2 H 3 O 2 and 0.125 M in KC 2 H 3 O 2 b. a solution that is 0.255 M in CH 3 NH 2 and 0.135 M in CH 3 NH 3 Br

Hey everyone, we're asked to use a nice chart to determine the ph of a solution that contained 0.30 molar hydrochloric acid and 0.20 molar sodium azide. And they provided us the K. A. Of our hydro's OIC acid first. Let's go ahead and write out our reaction. So we have our hydrochloric acid and this is going to react with water. When these two react, we get our azide ion. And the reason why our sodium is not included here is because it is a spectator ion. And we're also going to get our hydro ni um ion. Now that we've written out our reaction and check that everything is balanced. We're going to go ahead and create our ice chart. So initially we had 0.30 Moeller of our hydrochloric acid. Our water won't be included in our expression since it is a liquid. For as I die on, we were told that we had 0.20 moller and we had zero of our hydro knee um ion initially for our change in our react inside we will have a minus X. And the change in our product side is going to be a plus X. For both. Since we're losing reactant and gaining products At equilibrium for our react inside, we will have 0.30 -1. 0.20 plus X. For our product side under our azide ion and an X. Under our hydro knee um ion. Now let's go ahead and use our K. A. Which we know to be our products over our reactant. In this case it will be the concentration of her azide ion times the concentration of our hydro knee um ion divided by the concentration of our hydro's OIC acid. Now let's go ahead and plug in our values. So we were told our K. Is 1.9 times 10 to the negative five. This is going to be equal to 0.20 plus X. Times and X divided by 0.30 minus X. Now we can go ahead and check if our plus and -1 can be omitted. In order to do so we're going to take 0.30 and divide this by our K. A. Of 1.9 times 10 to the negative five. Now if we get a value greater than 500 then we can safely disregard our plus and minus X in our expression and in this case we do get a value greater than 500. So we can safely omit these two exes. Now let's go ahead and rewrite our expression. So we have 1.9 times 10 to the negative five is equal to 0.20 X divided by 0. solving for X. We get x equals 2.85 times to the negative five molar. Which is also going to be the concentration of our hydro ni um ion. Now we can go ahead and determine our ph and we can do so by taking the negative log of the concentration of our hydro ni um ion, which is 2.85 times 10 to the negative five. So when we calculate this out, we end up with a ph of 4.55, which is going to be our final answer. Now, I hope this made sense and let us know if you have any questions.

Ch.17 - Aqueous Ionic Equilibrium

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Ch.17 - Aqueous Ionic Equilibrium

Problem 30

Chapter 17, Problem 30

Solve an equilibrium problem (using an ICE table) to calculate the pH of each solution. a. a solution that is 0.195 M in HC₂H₃O₂ and 0.125 M in KC₂H₃O₂ b. a solution that is 0.255 M in CH₃NH₂ and 0.135 M in CH₃NH₃Br

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Identify the type of solution: Both solutions are buffer solutions. Solution (a) is an acetic acid (HC2H3O2) and acetate ion (C2H3O2-) buffer, while solution (b) is a methylamine (CH3NH2) and methylammonium ion (CH3NH3+) buffer.

Use the Henderson-Hasselbalch equation for buffer solutions: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \) for acidic buffers and \( \text{pOH} = \text{pK}_b + \log \left( \frac{[\text{B}]}{[\text{BH}^+]} \right) \) for basic buffers.

For solution (a), find the \( \text{pK}_a \) of acetic acid (HC2H3O2) using a reference table. Then, substitute \([\text{A}^-] = 0.125 \text{ M}\) and \([\text{HA}] = 0.195 \text{ M}\) into the Henderson-Hasselbalch equation to find the pH.

For solution (b), find the \( \text{pK}_b \) of methylamine (CH3NH2) using a reference table. Then, substitute \([\text{B}] = 0.255 \text{ M}\) and \([\text{BH}^+] = 0.135 \text{ M}\) into the Henderson-Hasselbalch equation to find the pOH.

Convert the pOH to pH for solution (b) using the relation \( \text{pH} = 14 - \text{pOH} \).

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Key Concepts

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

Equilibrium and ICE Tables

An ICE table (Initial, Change, Equilibrium) is a tool used to organize the concentrations of reactants and products in a chemical equilibrium problem. It helps in calculating the changes in concentration that occur as the system reaches equilibrium. Understanding how to set up and interpret an ICE table is crucial for solving equilibrium problems, including those involving weak acids and bases.

pH and Acid-Base Chemistry

pH is a measure of the acidity or basicity of a solution, defined as the negative logarithm of the hydrogen ion concentration. In the context of weak acids and their conjugate bases, the pH can be calculated using the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the concentration of the acid and its conjugate base. This concept is essential for determining the pH of solutions containing weak acids and their salts.

Buffer Solutions

A buffer solution is a system that resists changes in pH upon the addition of small amounts of acid or base. It typically consists of a weak acid and its conjugate base or a weak base and its conjugate acid. Understanding how buffers work is important for calculating the pH of solutions that contain both a weak acid and its salt, as seen in the given problem.