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Ch.17 - Additional Aspects of Aqueous Equilibria

Chapter 17, Problem 31a

You have to prepare a pH = 3.50 buffer, and you have the following 0.10 M solutions available: HCOOH, CH3COOH, H3PO4, HCOONa, CH3COONa, and NaH2PO4. Which solutions would you use?

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Hello everyone. So in this video we need a buffer solution with a ph of 7.20. And we're trying to see which combination that they have listed is going to be best to prepare that certain buffer and the buffer we want a page again of 7.20. So let's first recognize that are P K. A. And how we get that is going to be equal to the negative log of our K. A value. Another thing to recognize is our buffering range. This could be very important when we go ahead and start solving for these. So we have or we are given all the K values. Let's go ahead and solve our PK using this equation in green. So I'm just gonna go ahead and put that into my calculator. So for the first one I would put that the P K A. Is equal to the negative log of three point or 6. times 10 to the negative for giving me my P K. A value. Eagling to 3.16. So we will do this for every single K value that were given. So for this second one here my PKA value will be 7.52 and my third one will be that the PK is 4.74. So any buffer or in this buffer we have a conjugate base plus a weak acid. So C B is conjugate base W is weak acid. And if we know this the week acid and conjugate base parents that we could possibly have dealing with these possible choices and knowing what we have, we can possibly have our H F N R N A F. We can maybe have a church C I O and N A C I O. And of course another possibility. And the last one is CH three C O H N R C H three C N A. Alright so Our buffering range is going to be plus or - and to be more exact but just put B R is equal to the ph You're going to peek a plus or -1. So plus or -1 is going to be in that range I'm talking about. And if I go ahead and solve that given our P K A let's put this in maybe red. So for my first one if I subtract one from 3.16 the minimum is 2.16 for my range and my maxim is well 3.16 plus one is 4.16. So do that for every single peak a value. So the second one I would have from 6.5228 point 52. And then my last one I will have 3.74 all the way through or to 5.74. So that's my ranges here. So the P K A. That's nearest Ph of 7.20 that's going to be 7.52 of this PK value. So again valley recognize that the best combination possible for this buffer is going to be a h c i O and n a c i O. And this right here is going to be my final answer for this problem. Thank you all so much for watching.