Solve an equilibrium problem (using an ICE table) to calculate the pH of each solution. a. a solution that is 0.20 M in HCHO₂ and 0.15 M in NaCHO₂

Video transcript

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.