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Part AIf 0.6108 g of sodium oxalate, NazC204, requiresof a 28.74 mL KMnOa solution t0 reach the end point; what is the molarity of the KMnOa [email protected]

Question

Part AIf 0.6108 g of sodium oxalate, NazC204, requiresof a 28.74 mL KMnOa solution t0 reach the end point; what is the molarity of the KMnOa [email protected]

Part A If 0.6108 g of sodium oxalate, NazC204, requiresof a 28.74 mL KMnOa solution t0 reach the end point; what is the molarity of the KMnOa solution? @ Value M



Answers

What mass of sodium oxalate $\left(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)$ is needed to prepare 0.250 $\mathrm{L}$ of a $0.100-M$ solution?

Let's calculate the mouths of Saudi marks. Sleet needed to prepare 0.250 leaders of a 0.1 Mueller solution. And so, starting with, uh, sorry. Uh, wait 250 leaders times, uh, preparing a 0.1 Moeller solution and then times theme older mass of sodium ox elites 134.0 grams per mole. And this would I need to prepare the solution. 3.35 grams of the sodium obsolete.

In this problem, we were given that we tie traded five milliliters of an unknown concentration solution of sodium oxalate with 25.8 no leaders of 0.0 to 140 Moeller potassium from a nanny on giving this reaction here, um, which is our Redox citation reaction. And we're us to determine, um, the number of grams of sodium Oxlade president in one leader of this solution. So the first thing we should do is determine the concentration of our unknown solution. And first, we need to balance our redox reaction so that we know the more issues of each species. So if we read out a 2/2 reactions, we first have our, uh, our permanganate half reaction and we can balance this. First we have four oxygen's on this side. We can balance our oxygen's by adding four waters on the sun. Now that we have eight, hydrogen is on this side. We need to add eight hydrogen ions over here and now we need to bounce our electric charge. We have eight plus minus one. So seven plus on this and two plus over here. So we'll need five more electrons on this side. Five electrons added to the side, and now our charges are equal on both sides. So now if we look at the Aqsa, late half reaction we have are obsolete, going to carbon dioxide. First, we should balance our carbons. And at two over here, now we have our carbons and our oxygen's balance to four. Oxygen's on both sides, and all we need to do is bounce are electron. So we have two minus on this side of the reaction. So we'll need to add two electrons over here. So now we ever, uh, to half reactions will need to balance our electrons. So here we have five elections here we have to, So we'll need to multiply this entire half reaction by to and this entire half reaction by five so that we have rooms this entire half reaction by five so that we have 10 electrons in each half reaction. We do that and combine our reactions. We get our total balanced reaction is as follows. And this is from multiplying each species here by two and each species here by five and then adding both sides of the reactions together. So we get our 16 hundreds here are to remain units and then our electrons on east Side will cancel out and they won't appear in the final reaction. So this is our final balance Redox reaction. And now we can go back to see that we had. We used 25.8 milliliters of our 0.0 to 14 Moeller purveying innate solution. And we can use our circuit metric ratios here to determine the concentration of our sodium oxalate solution so we can start with the concentration of the permanganate solution used. This was our polarity 0.0 to 14 moles of per 1,000,008 for leader and we can multiply by the volume used which was 25.8 mil leaders or 0.258 leaders. This will give us moles of priming unit. Our leaders will cancel and we'll be left with moles on. Then we can multiply by our mobile issue. Um, we have two moles of permanganate for every finals of obsolete. So we have five moles of oxalate for every two moles of Germany And then now that we have moles, these will cancel. This will cancel We have moles of oxalate we can divide by the volume used to get the concentration will have moles per leaders. So if we divide by our five milliliters or 0.5 leaders now we'll have moles of oxalate per leader, which is units of malaria T. And this gives us that the concentration of the sodium oxalate sample is point 276 Mohler sodium. Obsolete? Um, it's 0.27 16 Moeller in both Oxley and sodium oxalate. Since there's a 1 to 1 ratio, there's only one mole of oxalate in every bowl of sodium obsolete. Now, using this, we want to find the mass present in one leader of solution. So if we have Syria 0.276 moles of sodium Locksley Oh, sorry, this should be. That's a nun charged. No. So we have 0.276 Moles from either well, multiply by our Moeller mess, which is 134 grams her one wall, and this tells us that we have your Mosul cancel out here and we'll be left with grams per leader, which is what we want because we want the number of grams in one leader of solution and this gives us that we have 37.0 grams of sodium obsolete in one leader of solution

Mm Yeah 2.335 ground potash um Dich um it is dissolved in enough water to make a 500 millimeter solution. Here we have to find out the Mueller concentration of this solution. Also we have to find out the concentration of potassium iron and that guy commit I'll yeah now we find out the Mueller mass of oh K two cr 207. Now It is equal to two times Atomic mass of Patricia which is 39.09 eight Plus two times atomic mass of chromium is 51.996 plus seven times Atomic mass of oxygen is 15.999. So if we calculate all together will get 294.181 g par more. Now we have to find the Mueller concentration of the solution we know more clarity yes number up more of the solid. Yes and volume of solution in leader so number of more absolute will take The mask given 2.335g potassium document divided by a smaller mask. That is to 94.181. This is the number of more of the Diagram it and divided by the volume in later which will be 0.500 later and hear more And if we calculate this the similarity will be 0.016 more this for a and we have to now calculate the concentration of potash potassium ion and I commit to Now we know in one molecule er potash um die Clomid there are two potassium ions and one did I commit on cr 207 to minus. Therefore concentration of K plus will be two times of concentration of the solution. That is the The commerce solution. So two times 0.016 smaller and it will be 0.032 smaller. And the concentration of the dye commit ion cr 207 two minus. That one will be equal to concentration of K two cr 207 solutions, which is 0.16.

Yeah. All right. So let's say that we have um 2.335g of potassium die Chrome eight In 500 ml of solution. And let's try to figure out what the mola parity is for this. So remember more clarity. His moles of solute over leaders of solution. And so we know in this particular case we have grams we can convert that into molds for our salute And our solutions 500 which we can convert into leaders. So We have our 2335 g In 500 ml. And so let's convert grams to moles by using molar mass Mueller mass here is two 94 0.1 85 g. We know that there's 1000 ml in a leader. And so our grams cancel out our milliliters cancel that are left with moles per liter, which is more clarity. And so we have .0159 moles per liter of our potassium die chrome eight. Now, what does that mean for the ions in solution? Because potassium are soluble compounds. So when we put this into solution we're going to get to potassium ions oh and the dick romain ion stays together. So if I want to know the polarity right? If I have point No 159 moles per liter, that must mean for the potassium I'm going to have twice that amount. And the decree mate coefficients one. So it's still the same amount. Yeah so the concentration of the potassium is point 0318 moles per liter. And the concentration of the cr cr two oh seven ion mm It's still a .0159 moles per liter. Yeah. Yeah.


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