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Ch.17 - Additional Aspects of Aqueous Equilibria

Chapter 17, Problem 2b

The beaker on the right contains 0.1 M acetic acid solution with methyl orange as an indicator. The beaker on the left contains a mixture of 0.1 M acetic acid and 0.1 M sodium acetate with methyl orange. (b) Which solution is better able to maintain its pH when small amounts of NaOH are added? Explain. [Sections 17.1 and 17.2]

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Welcome back everyone in this example we have the flask on the left with a mixture of 0. molar hipAA course acid and 0.2 molar potassium hypochlorite, we're told that this is also mixed with a thermal blue whereas the flask on the right has a solution of 0.2 molar hipAA course acid with just choose the solution that is able to resist drastic changes in ph even a small amount of potassium hydroxide are added and explain why we're given the acid dissociation constant of hipAA course acid as 2.9 times 10 to the negative eighth power. We should recall that hipAA course acid represented as H oh cl is considered a weak acid. We also want to show that potassium hydroxide is one of our memorized strong basis. And when we have a solution that resists drastic changes to BH that solution is considered a buffer solution. So let's show how our potassium hypochlorite is formed. We have HipAA course acid H oh cl reacting with potassium hydroxide according to the prompt and this is going to be an equilibrium reaction will will form the salt of our conjugate base which is going to be our potassium hypochlorite will also form water As our 2nd product. So we know that potassium hypochlorite is the salt of our conjugate base of hipAA course acid because HipAA course assets conjugate base is going to be with the loss of a proton where we would form o cl minus. And so with this potassium adam added on giving us potassium hypochlorite. This we recognize as our salt of our conjugate base. Now, according to our prompt, our second beaker here only consists of hipAA course acid, our weak acid with a thermal blue. And if we look below at this ph chart we can see that from a thermal blue ranges at a ph from about five to about a ph of seven. We want to recall that a buffer solution has a composition where we have a weak acid and the salt of a strong alkali. Recall that strong alkali is just another way of saying strong base and in this case we do have our strong base which is added as potassium hydroxide, meaning that because our first solution has our weak acid and its conjugate base pair being our potassium hypochlorite Solution, one is going to result in a buffer solution where we will have a resistance of changes in ph even when small amounts of this potassium hydroxide is added to the solution. So we would say that solution one is a buffer. And so our final answer is going to be that image one containing our first speaker is a buffer solution and therefore will resist drastic changes to ph even when small amounts of potassium hydroxide is added. And we know that our explanation is due to the fact that it's a buffer solution as to why it's going to resist these drastic changes in ph because our first speaker has the presence of our weak acid, hipAA Cloris acid and the salt potassium hypochlorite which is a salt of its strong base. We're sorry. Of the strong base potassium hydroxide or of the conjugate base. Oh, cl minus so solution one is our final answer. I hope everything I explained was clear. If you have any questions, let them down below and I will see everyone in the next practice video.