What is observed when equal volumes of 0.1 M aqueous HCl and 0...

Question

What is observed when equal volumes of 0.1 M aqueous HCl and 0.01 M aqueous Na₂SO₃ are mixed?

Accepted Answer

Welcome back everyone. We need to consider the tight rations of celestic acid, a dia protic acid with an acquis potassium hydroxide solution. The diagrams below depict different stages of the thai tradition. For simplicity, potassium ions and water molecules are not shown. We have a key which shows us the representation for each of our molecules in each diagram, and we need to match each diagram to the correct stages of the thai tradition. We have four scenarios to consider. Let's begin with scenario one, we're told to consider the first half equivalence point, recall that at our equivalence point our moles of acid will be equal to our moles of base. So focusing on our cell enic acid H two S. E. 04, we want to recall that this is actually a weak acid. It's not in our list of memorized strong acids. However, celestic acid is unique because it has a very high K. A. Or acid dissociation constant. And so because it has a very high K. A. It can be considered to act as a strong acid, which we would recall undergoes full dissociation so fully associates. And so instead of writing an equilibrium expression, we can say that when we have selected acid react with water, we have a complete right arrow to form the products H. S. E. 04 minus because our celestic acid acts as our bronze ted Lowry acid donating a proton to water, which is our bronze ted Lowry base so it accepts the proton and we would form H plus as our second product. So with this in mind we can see we fully dissociated into the two ions hydrogen cell innate and hydro ni. Um Now we need to take note that our titra according to the prompt is potassium hydroxide and potassium hydroxide we would recall is a strong base and so recall that potassium hydroxide or really any strong base is always going to fully consume our strong acid. Considering our first equivalence point, we would understand that if our potassium hydroxide titrate fully consumes the strongest acid between our hydrogen Selena an ion and hydro knee um hydrogen cell innate is going to be the conjugate acid of celestic acid. And so therefore it's also a it's a weaker acid than Selena acid and hydro nian we recognize as a strong acid. And so at the first equivalence point we would see that potassium hydroxide fully consumes hydro ni um the strongest acid and so therefore only hydrogen cell innate an ion is left over. So again following our line of thought for part one, which is the first half equivalence point, that means that potassium hydroxide only consumes half of our hydro ni um strong acid and thus we can say therefore the concentration of hydro ni um is equal to half the concentration of our hydrogen cell in it. And so going to our diagrams, we would want to choose the diagram that has a 1 to ratio between our molecules of hydrogen cell in it and hide Roni. Um and we can see that in diagram D. We have eight molecules of our hydrogen cell in eight and four molecules of hydro knee. Um And here we can confirm we have a 1 to 2 ratio. So we would say that for part one, diagram D represents the first half equivalence point. So now let's move on to scenario two and determine the diagram that corresponds. So in diagram two or sorry, part two, we're considering the second half equivalence point. So actually we need to consider our second equivalence point first before we consider the second half equivalence point. So for the second equivalence point we would begin with our hydrogen Solenni an ion H S C 04 minus one. We would be reacting with potassium hydroxide where potassium is our spectator ion. So it's not going to really participate. And we would just have O h minus reacting. And hydrogen cell in eight is going to be our bronze ted Lowry acid, donating a proton to proton to hydroxide. So we would be left with S. E. 042 minus which is our cell innate an ion and then we would form hydro ni um And so at the second equivalence point and sorry they should say AQ. Here we have a 1-1 ratio between our hydrogen cell innate an ion and a hydroxide. And so we can say therefore only our cellini an ion S. E. 042 minus remains because hydrogen cell in eight is fully consumed. So technically we've already answered scenario three, which is just considering the second equivalence point where only sell innate remains. And so going to our diagrams, we would see that this corresponds and let's see the molecule for selling eight is just a purple molecule alone. That would be scenario C. And so here for scenario three, we would correspond this to diagram see for the second equivalence point and so continuing on to scenario two for the second half equivalence point which we will consider now that we've gone through our second equivalence point. We begin with selling it or sorry, we begin with hydrogen cell innate. So H S E 04 minus where we said earlier, we have a 1 to 1 ratio with hydrogen cell innate and sell in eight. And at the half equivalence point, only half Of hydrogen cell in eight is consumed by hydroxide. And therefore due to that 1-1 ratio, that means our sorry, our 1-1 ratio, that means our concentration of hydrogen cell in it would be equal to our concentration of selenium an ion. And just so that this is visible, we'll scoot this over. So the concentrations would be equal at the second half equivalence point. And so going to our diagrams, we would see that between our molecules of cell innate and hydrogen cell innate, the ones or the diagram that has an equal amount of moles. We have four moles of hydrogen cell innate and formals of sellin eight and so diagram a would represent the second half equivalence point and now we can go ahead and consider scenario four because we've completed one through three. So four considers after the second equivalence point. So We have scenario four After the 2nd equivalence point, going back to what occurs or what is left at our second equivalence point, which was this here, we determined that only our cellini an ion remains so if only selling an ion remains after the at the second equivalence point, then that means that once our potassium hydroxide consumes hydro ni um our concentration of protons and sorry, this is a bracket. Our concentration of protons are equal to zero and so therefore there's no more acid available to be neutralized by potassium hydroxide. And so we would have excess hydroxide ions. So looking at our diagrams, we have hydroxide represented by this molecule here and we can see that we have a high concentration of our cell in eight to minus an ion, we have specifically eight molecules And we have three moles of excess hydroxide molecules left over. So after the second equivalence point, Scenario four describes Diagram D. And so for our final answers, we can write that out next to each of our scenarios the first half equivalence point we agreed represents diagram D. So let's just write it all next to our diagrams, we have the first half equivalence point. Then for The 2nd equivalence point we went with diagram and we'll actually write it down next to the letter. We went with diagrams see where we only had salon eight ion ions present. So so far we have these two other first two answers. And then for diagram B we agreed that we would have excess hydroxide and xsl innate and ions left over. So diagram B represents after the second half equivalence point, we have no more acid to neutralize and then we have four diagram A. We agreed that that corresponded to the second half equivalence point Where we have a 1- ratio between our cell innate and hydrogen cell innate ions. So everything highlighted in yellow represents our final answers. To complete this example. I hope everything I went through is clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video.

What is observed when equal volumes of 0.1 M aqueous HCl and 0.01 M aqueous Na2SO3 are mixed?

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What is observed when equal volumes of 0.1 M aqueous HCl and 0.01 M aqueous Na₂SO₃ are mixed?