Solve an equilibrium problem (using an ICE table) to calculate the pH of each solution. c. a mixture that is 0.15 M in HF and 0.15 M in NaF

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Hey everyone, we're asked to use an ice chart to determine the ph of a solution that contained 0. molar hydrogen cyanide and 0.25 molar sodium cyanide and they provided us the K. Of hydrogen cyanide first. Let's go ahead and write out our reaction. So we have hydrogen cyanide and this is going to react with water. When these two react, we end up with our cyanide ion. And the reason why we don't add the sodium ion is because it is a spectator ion. And we also get our hydro ni um ion. Now creating our ice chart, we were told that initially we had 0.25 molar of our hydrogen cyanide. Our water won't be included in our expression since it is a liquid for our cyanide ion. We were told that we had 0.25 molar initially. And initially we had zero of our hydro knee um ion. For a change, it's going to be a minus X on our reactant side. Since we're losing reactant and a plus X on our product side. Since we're gaining products at equilibrium on our react inside, we will have 0.25 minus X. And in our product side we will have 0.25 plus X under our cyanide ion and an X under our hydro ni um ion. Now we've learned that our K. A. Is equal to our products over our reactant. In this case it will be the concentration of our cyanide ion times the concentration of our hydro ni um ion divided by the concentration of our hydrogen cyanide. Now let's go ahead and plug in these values. So we were told our K. A. Is 6.17 times 10 to the negative 10. This is going to be equal to 0.25 plus X. Times and X divided by 0. minus X. Now we can go ahead and check if we can omit our plus and minus X. To do so all we need to do is take 0. and divide this by our K. A. Of 6.17 times 10 to the negative 10. Now if we get a value greater than 500 then we can omit 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 this as our exes are negligible, simplifying this a bit further, we have 6.17 times 10 to the negative 10 is equal to 0.25 X divided by 0.25, solving for X. We get 6.17 times 10 to the negative moller which is going to be the concentration of our hydro ni um ion solving for R. P. H. We can then take the negative log of the concentration of our hydro knee um ion in this case it is 6.17 times 10 to the negative 10 moller Solving for R. P. H. We end up with a ph of 9.21, which is going to be our final answer. Now, I hope this made sense and let us know if you have any questions.

Solve an equilibrium problem (using an ICE table) to calculate the pH of each solution. c. a mixture that is 0.15 M in HF and 0.15 M in NaF

Verified Solution

Key Concepts

Weak Acids and Conjugate Bases

ICE Table (Initial, Change, Equilibrium)

Henderson-Hasselbalch Equation