Sodium nitrite, NaNO2, is frequently added to processed meats as ...

Question

Sodium nitrite, NaNO2, is frequently added to processed meats as a preservative. The amount of nitrite ion in a sample can be determined by acidifying to form nitrous acid (HNO2), letting the nitrous acid react with an excess of iodide ion, and then titrating the I3 - ion that results with thiosulfate solution in the presence of a starch indicator. The unbalanced equations are
(1) 
(2) 
(b) When a nitrite-containing sample with a mass of 2.935 g was analyzed, 18.77 mL of 0.1500 M Na2S2O3 solution was needed for the reaction. What is the mass percent of NO2- ion in the sample?

Accepted Answer

Welcome back everyone. The Winkler method can be used to determine the amount of dissolved oxygen and water samples by converting dissolved oxygen to manganese dioxide with manganese sulfate and reacting it with sodium hydroxide and potassium iodide. The solution is then acidified for the oxidation of our iodide anion to tri iddy anion by manganese dioxide to occur. The amount of tri iddy anion formed is determined by Tiit with our thos sulfate two minus anion. The unbalanced equations are these three equations given in the prompt. And when 0.15 78 g of a water sample is analyzed, the reaction needs 17.55 mL of 0.1300 molar sodium dios sulfate, calculate the mass percent of oxygen in the sample. So noting the fact that they tell us the equations are unbalanced. We want to begin by balancing these three reactions. So this is reaction one, this is reaction two and this is reaction three beginning with balancing reaction one, we have our manganese two plus ion reacting with oxygen gas to form manganese dioxide solid. So what it appears is that manganese atoms are balanced. Now recognize that we're not going to begin by balancing oxygen because we want to balance the non oxygen and non hydrogen atoms first. However, we want to recognize that because this is a redox reaction in redox reactions, electrons lost must equal the number of electrons gained. And that's just because that's how electrons transfers work. We can't just have electrons disappear in our reaction. So looking at the fact that on the reactant side, this manganese ion has a two plus charge recall that this tells us we lose that number of electrons. So we lose two electrons. We're going to use the electron transfer balance method to place the proper coefficients in our reaction so that our atoms are balanced for manganese. So we're going to begin by noting down the oxidation states of our atoms beginning with our manganese ion, we can see that the oxidation state is equal to its ion charge as plus two moving on to oxygen gas. Our second reactant recall that any element in its standard state, which oxygen here is in its standard state as a diatomic gas will have an oxidation state of zero. And then for our product manganese dioxide solid, we don't know the oxidation state of manganese. However, we know that for one of our oxygen atoms, at least each of them individually have an oxidation state equal to minus two normally. So we're going to need to write out an equation to solve for the oxidation state of manganese. And we're going to recall that we would take X for the oxidation state of manganese plus the oxidation state of each of our oxygens being minus two multiplied by the subscript on our oxygens, which we see we have a subscript of two for two oxygen atoms set equal to the overall charge of our product, which we see as a neutral uncharged product. So this equation is equal to zero and solving for X, we would find that the oxidation state of manganese is equal to plus four. And so now that we understand that our oxidation state of manganese is plus four, we see that we have a net loss of four electrons on manganese. And so to counter that with our reactant side, we also need four electrons lost, meaning we need to place a coefficient of two in front of manganese. And then going back to our product, we recognize that as we stated, each of our oxygen atoms, which we have two have an oxidation state of negative two. And so we would need a net gain of four electrons on the oxygen atom, meaning a minus four charge. So in order to get that, we would place a coefficient of two on our product side in front of manganese dioxide. And that's going to allow the minus four charge to cancel out with the plus four charge on our manganese atom. So right now we can confirm that our electrons lost, which we agree is a total of four electrons lost is equal to our electrons gained being a net gain of four electrons on our oxygen atom. However, now with this coefficient on our product side, we've now introduced a total of four oxygen atoms on the product side. Whereas on the reactant side, we only have one mole or sorry, two moles of oxygen based on its subscript. So we're going to next recall that we bounce oxygen using water. So what we would have is our two moles of manganese cati two plus plus our one mole of oxygen gas plus two moles of water. To now give us four moles of oxygen on the reactant side produces or yields two moles of our manganese dioxide solid. Now that we've introduced water to bounce out our oxygens, we've now also introduced hydrogen, specifically formals of hydrogen on the reactant side. And so we're going to recall that we bounce hydrogen using protons. So H plus and so now we're going to have to add to our product side, four moles of protons. Now, at this moment, according to our prompt, we have a basic solution. And so whenever we have a basic solution and we add protons, this is making our solution more acidic. So we're going to need to add hydroxide to bring our solution back to basic. So we're going to add an equivalent amount of hydroxide to our protons and sorry about that. So we also want to make sure we add this amount of hydroxide to both sides of our equation. So we would add four moles of hydroxide here and four moles of hydroxide here. And so, and actually let's make sure that this is all visible. So yes, we have formals of hydroxide on both sides of our reaction. Next, we want to recognize that when we have the addition of protons and hydroxide, they're going to combine to form specifically in this case, four moles of water. And so what we would have is on our reactant side, this two moles of water, we can cancel out with the two of the four moles of water leaving us with two moles of water on the product side. Now, and for our overall equation, we'd be left with our two moles of our manganese two plus ion plus our four moles of hydroxide plus our one mole of oxygen gas is going to yield two moles of our manganese dioxide solid plus two moles of water and apologies, the label for water here should also be liquid. Now, we can confirm that we officially have all of our atoms balanced in this reaction. And now we need to make sure that the charges are balanced. So we have on our reactant side, a net charge of minus four and then plus four, which would give us a net charge of zero on the reactant side. And on our product side, we have no charges on our product. So we also have a net charge of zero. And so we can confirm that we have all atoms balanced and all charges balanced. So let's move on to the second reaction because we're gonna need proper coefficients before we can get to our final answer for each equation. And so moving on to reaction two, we're going to begin by bouncing that one. So reaction two is here below. For reaction two, we have our manganese dioxide solid reacting with our IDDY anion to form our manganese two plus ion and our tri iddy anion. So we're going to make note of the fact that according to the prompt, we're oxidizing our iodide anion with manganese dioxide. And so at this point, our solution is going to undergo a more acidic ph. So we'll make note of the fact that we have an acidic solution. And so our first step is to break this out into half reactions. So we have manganese dioxide solid forming manganese two plus as our, this is our first half reaction. Now, we want to keep in mind that for this reaction, this is able to be or this was able to be split up into a half reaction. So we can directly see that our manganese atoms are balanced. However, oxygen is not. And so we're going to as we did earlier balance oxygen using water. So adding two moles of water, we're now going to have introduced two or four moles of protons to the product side, which we need moles of four moles of protons to the reactant side. So we're gonna add four moles of protons here and we're not going to add hydroxide here because as the prompt states, this solution is acidic, so it's fine to have our protons. Now, we just need to make sure a net charge is balanced on both sides of the equation. On the reactant side, we have a net charge of plus four. On the product side, we have a net charge of plus two. So we're going to get to a net charge of plus four by or sorry of plus two on the product side on the reactant side, sorry by adding two electrons as a third reactant. So we would have net charge of plus two on both sides of our reaction. And now that we have a net charge balance, we are going to go into our second half reaction beginning with our I dyed anion which forms according to the reaction, one mole of our tri iddy anion. So beginning with bouncing out our iodine atoms, we're going to place a coefficient of three in front of our iodide reactant on the sorry, on the iodide anion on the reactant side, giving us three moles of our iodide anion on both sides. And now making sure charges are balanced. We're going to recognize we have a net charge of minus three on the reactant side and a net charge of minus one on the product side. So we would need to add two electrons to the second half reaction to have a net charge of minus three on both sides of the reaction. Now that we have net charge and atoms balanced in both reactions. And we can also see that both of these half reactions have matching number of electrons, we can add these two reactions together. And so in adding these together, we'll be able to cancel out the two electrons in each reaction. And there's no other compounds that we can cancel out. So we're just going to carry everything left down that leaves us with our manganese dioxide solid plus four moles of protons plus three moles of our iodide anion yields one mole of our tri anion plus one mole of our manganese two plus ion and plus two moles of liquid water. So this is our second balanced reaction and now we're going to move on to reaction three given in the prompt, that would be the try iodide. And this is our titration step. So try I did, which is titrated by our S two 032 minus dios sulfate anion. And this is going to produce one mole of our I died anion plus are tetra cyanide anion. So now we are going to do as before and split this into two half reactions. For our first half reaction, we'll begin with our one mole of tri iodide, which forms our one mole of iodide as a product. And so now we're just going to see if this is balanced. And we can see that we need a coefficient of three in front of our product side, which will give us three moles of each of the iodide anions. And as far as net charge, we have a net charge of minus three on the product side. So we're going to at a total of two electrons to our reactant side to also give us and sorry, this is two electrons here to our reactant side to also give us a net charge of minus three on the reactant side as well as the product side. So now that we have our charge balance for this first half reaction, as well as the atoms, we move on to our second half reaction where we would have our thos sulfate two minus anion which forms the tetra ate two minus anion as a product. So beginning by bouncing out our sulfur atoms, because we want to begin again with non oxygen and non hydrogen atoms, we'll place a coefficient of two in front of our thos sulfate reactant. And this is going to give us four moles of sulfur on both sides of our reaction. And now also give us six moles of oxygen on both sides of the reaction. And as far as net charge, we have a net charge of negative two times two. So minus four on the reactant side and a net charge of minus two on the product side. So we're going to expand our product side and add two electrons as a product to give us net charge of minus four on both sides of the reaction. Yet again, we see that we have matching electrons in both of our half reactions. So we can straightforwardly add up these two half reactions together to come up with a third overall balanced reaction. This is going to yield the two electrons canceling out. And what we're left with is our one mole of tri iodide anion plus two moles of our thos sulfate two minus anion which produces our tetra diana anion. And this is a six here and our three moles of tri of our iodide and ion here. All right. Now that we're done bouncing our reactions, we're going to scroll down for more room because the next part of our prompt is going to be recalling that from the prompt. We are using our sodium cyan eight. So A N A two S 203 and this is our salt that our thos sulfate anion tyrann is sourced from. And it's at a concentration given in the prompt of point 1300 mol per liter. And recall that moles per liter represents molar. So we're going to use this concentration of our salt in a step to determine moles of our tirin, which is our thos sulfate anion. And so beginning with the volume of the solution given in the prompt as 17.55 mL, which was the volume of our water sample or correction the volume of the tyrant needed. We're going to multiply to go first because we can't directly go into leaders. We're going to go from milliliters two liters in the numerator and then we'll be be able to multiply by our concentration of our salt. So recall that our prefix Milly tells us we have 10 to the negative third of our base unit liters, canceling out milliliters will cancel out liters. Next. And now our next step is to go from mo of our salt and a two S 2032 moles of our tirin, which is our dios sulfate anion S 2032 minus. And we can see that in our formula of our sodium thos sulfate, we have a 1 to 1 molar ratio. So we would fill that in canceling out moles of our salts. We now are able to calculate in our calculators carefully, the moles of our tyrant ouro sulfate anion. And this is going to come out to be 0.0022815 moles of our dios sulfate anion, which is our t as we stated. So now, ultimately, our end point is to get that mass percent of oxygen in the sample. And so we're going to or I'm going to paste in our previously balanced equations. That we went through. So we have our equations here pasted in these are our balanced equations. And we want to reference the ratios first between our thos sulfate anion. And that is in the third reaction, we have two moles of thos sulfate. And we're going to look at it at a ratio with another reactant that we find in our second equation here, which is our tri iodide anion. So we have a 2 to 1 ratio here from the third reaction between thos sulfate and tri iodide. And so to begin with finding our massive oxygen, we would begin with the moles, we just calculated of our tyrant, 0.0022815 moles of thos sulfate and multiplying this again by the molar ratio between moles of thos sulfate and moles of our reactant that connects us to equation two, which is our tri aide anion. We see again, as we stated in equation three, we have a one or sorry, a two in the denominator two moles of our dios sulfate anion for one mole of our tri anion that it reacts with. In equation three, these were our bounced coefficients that we used. And so now this allows us to cancel out moles of our thos sulfate. And now we're going to go from moles of our tri iodide and working backwards, recall that our prompt tells us that our dissolved oxygen converts to manganese dioxide. As we see in equation one. And so that means that we're comparing our one mole of our tri anion to manganese dioxide here. Where in equation two, we see that we have a 1 to 1 molar ratio. So one mole of our tri iodide anion for one mole of our manganese dioxide anion. And so canceling out moles of tri iodide anion, we can now directly go from moles of manganese dioxide, two moles of what we want, which is oxygen where then our last step and I'll actually draw this in the line below our last step of our sto geometry. Here is going to be multiplying by moles of oxygen in the denominator and grams of oxygen as our final unit. So that there will be putting in the molar mass. But right now, we're going to plug in that ratio between our manganese dioxide and oxygen, which we can clearly see we have in our first reaction, a 1 to 2 molar ratio. So one mole of oxygen for two moles of manganese dioxide. And so canceling out manganese dioxide anion with our manganese dioxide solid. We're left with moles of oxygen, which we will next refer to our periodic tables for the molar mass of oxygen where we see that for one mole of oxygen, we have a molar mass of 32 g. And so canceling out. Now mold of oxygen, we're left with our final unit, which is our mass of oxygen. And this is going to yield the results equal to 0.018252 g of oxygen. So now that we have our mass, we can finally recall our formula for mass percent. So mass, our mass per cent we recall is calculated from taking our mass of our component. In this case, our mass of oxygen divided by the mass of our sample. So in this case, the mass of our water sample, which given from the prompt in the denominator is cited as 0.1578 g of water. And in our numerator, we just will plug in our massive oxygen that we found as 0.018252 g of oxygen. And so now canceling out or sorry, multiplying by 100%. In the same step, we would cancel out our units of grams. We are left with a percentage and we have a result of 11.57%. And so for our final answer, we found the mass percent of oxygen. So a mass percent of oxygen in our water sample that was reacted using the Winkler method in order to achieve this amount of dissolved oxygen. So this will correspond to choice c in the multiple choice. I hope everything I reviewed was clear. If you have any questions, please leave them down below. And I'll see everyone in the next practice video.

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