The following pictures represent solutions at various stages in t...

Question

The following pictures represent solutions at various stages in the titration of a weak base B with aqueous HCl. (Cl- ions and solvent water molecules have been omitted for clarity.)

. (a) To which of the following stages do solutions 1–4 correspond? (i) The initial solution before addition of any HCl (ii) Halfway to the equivalence point (iii) At the equivalence point (iv) Beyond the equivalence point

Accepted Answer

Welcome back everyone. The different stages of the tight rations of a weak base be with acquis nitric acid, H and 03 are represented in the following images are nitrate ions and water were omitted. For clarity, identify the stages represented by the images. Let's begin by writing out our reaction. We have our weak base which when dissolved in water is going to partially dissociate into H. B. Plus being the conjugate acid. Because water will act as an acid. Donating a proton to our base, be our weak base B. And that will leave us with also hydroxide as a product. Recall that in the dissociation of nitric acid which we should recall is one of our memorized strong acids from general chemistry this will fully dissociate into and sorry, let's show that it's dissolving in water. So it's going to fully disassociate where we form our N. 03 minus nitrate an ion. And because our nitric acid is going to donate a proton to water, we will form hydro ni um as a product. So in this case nitrate is going to be our conjugate base of nitric acid. And as we stated, H. B plus is going to be the conjugate acid of our weak base B. Now in the reaction between nitric acid and our weak base, our nitric acid will be our proton donor and we will form HB plus as our conjugate base Now of nitric acid and we will also form nitrate. No three minus and sorry. So H. B plus as we stated earlier is our conjugate acid and N. +03 minus nitrate is going to be our conjugate base of our strong acid nitric acid from the nitric acid, donating a proton to our weak base. Now looking at illustration one and looking at our key given in the prompt, our blue atoms represent our weak base. So all we have present in phase one is exactly eight moles of our weak base B. That means that this is going to be the state of our tin tray shin before any nitric acid is added. So no nitric acid added and therefore this is going to be the initial solution. So looking at our answer choices, we can already rule out pretty much all of our choices except for choice D. Which correctly states that image i is the initial solution before any edition of our strong acid nitric acid. Let's keep considering our second illustration where we see we have two moles of hydro nian present, where we have eight moles of our conjugate acid, H. B plus present. Note that we counted two moles of hydro nian present. And so therefore we have excess hydro ni um notice that we outlined below that hydro knee um is generated when our nitric acid dissociates or dissolves in water. And so that means that we therefore have excess or rather instead of excess will say therefore we have more moles of nitric acid than moles of our weak base. And this only occurs when we are after the equivalence point in our hydration, recall that in our equivalence point at in a titillation, our moles of acid equal our moles of base. So looking at choice D. It says that too is after the equivalence point. And we did outline that. So we would agree with the course of choice D. So far so right now, let's highlight our two faiths initial solution for one. And after equivalence point for illustration to now looking at illustration three, notice that the gray atoms bonded to the blue atoms represent our conjugate acid H. B. Plus and that is all we have present in illustration three, we have eight moles of our conjugate acid H. B. Plus present notice that we outlined in our reaction that our weak base generates our conjugate acid H. B. Plus when it dissolves in water in the partial dissociation. And so we also see our conjugate acid H. B. Plus generated from our nitric acid tight trading our weak base B. So with only the presence of our conjugate acid, H. B. Plus our moles of our strong acid will equal our moles of our weak base, meaning that all of our weak base converts to H. B. Plus, which is why we only see it present. And this means that we are definitely at our equivalence point as we outlined so we can move this over so that it matches with illustration three. For that note. Now let's look at illustration for we do see that in choice D. Illustration three is stated to be the equivalence point. But now looking at image four, we have our conjugate acid H. B plus present. And we have specifically for moles of that present, two formals of our weak base present. Since we have a split even relationship between our weak base and conjugate acid H. B. Plus, we see that half of our moles are weak base and the other half of our moles are are conjugate acid. This means that only half of our weak base converts to are conjugate acid H. B. Plus. And this would therefore represent the halfway equivalence point. And looking at choice D. Illustration four is paired up to be the halfway equivalence point. And so to complete this example, that means that choice D. Represents the proper correct stages of our thai tradition of our weak base B with acquis nitric acid. So D. Is the final answer. I hope that everything made sense. And I'll see everyone in the next video

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