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Ch.17 - Additional Aspects of Aqueous Equilibria

Chapter 17, Problem 2a

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. (a) Using Figures 16.8 and 16.9, which solution has a higher pH?

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Welcome back everyone. In this example, we have an indicator, bro methyl blue added to each of the solutions shown below where solution A contains 0.0 20 molar hydrotic acid. And solution B contains a mixture of 0.0 20 molar hydro cyanic acid and 0.0 20 molar sodium cyanide. Using the figures below as reference, identify the solution with a lower Ph. So according to our prompt, we have this first solution solution. A sorry that's solution A which is 0.0 20 molar of our hydrotic acid. And then we want to recall that hydro saic acid is actually one of our weak acids from our memorized list of strong acids. We won't find that on that list. So it's a weak acid but looking at solution two or solution B rather the prompt tells us that this is a mixture of 0.0 20 molar hydro saic acid plus 0.0 20 molar sodium cyanide. So N AC N, so we need to think of the PHS of both of these solutions. So referring to our graph below, we want to find the indicator bro, methylal blue listed from the prompts because that's the indicator that's used in both of these solutions. And because we have a more yellow colored solution, that means that our ph for solution A is either between five to about six as the PH. So we're gonna say the PH is about 5 to 6 based on our bro methylal blue indicator. And for solution B, we have a more blue colored solution according to our be methyl blue indicator. And so the PH four solution B because it's blue, it's most likely around seven to around eight ph. So we'll say the ph here is probably around 7 to 8. Now, we also want to recognize that according to our composition of solution B, we have our weak acid hydro cyanide and then we have our salt, sodium cyanide. And we want to recall that a salt is always going to be dissociating into its ionic parts. So we would form the N A plus C ion and we would form our cyanide anion which we recall has a minus one charge and recall that our cyanide anion is ultimately the conjugate base of our weak acid hydrogen or hydro cyanic acid HCN. And so if we recall our definition of what a buffer is, recall that a buffer solution consists of a weak acid plus its conjugate base. And in this case, solution B does consist of a weak acid and the salt of its conjugate base, which is our cyanide anion. So essentially solution B is a buffer solution. And so the PH is definitely going to be a little bit higher than our ph of just a solution with the weak acid hydro cyanic acid, which is solution A. And so we can confirm that the solution with the lower PH is going to be solution. A since it's not a buffer solution, it only consists of the weak acid hydro cyanic acid. So our final answer is going to be solution. A as the lower ph which we stated is going to be around 5 to 6. So I hope everything I reviewed was clear. What's highlighted in yellow is our final answer. If you have any questions, leave them down below and I will see everyone in the next practice video.