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Ch.4 - Reactions in Aqueous Solution

Chapter 4, Problem 132

Alcohol levels in blood can be determined by a redox reaction with potassium dichromate according to the balanced equation What is the blood alcohol level in mass percent if 8.76 mL of 0.049 88 M K2Cr2O7 is required for complete reaction with a 10.002 g sample of blood?

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Hey there. Welcome back. Alright, so we have a solution of acetic acid and acetic acid dissolved in the water. Makes vinegar. So here we have 3.42 g of vinegar. Right. And we're told that it was neutralized by 23.6 mL of 0.103 molar sodium hydroxide. And the question here is what is the percent by weight of just the acidic acid in the vinegar? So percent by weight or mass percent. Right. Same thing. So mass percent, let's say of acetic acid in vinegar is going to be what was going to be massive? Just the acidic acid of acetic acid mm hmm. Divided by the total mass of the solution, which is the vinegar vinegar is acidic acid in water. Right? So it's just a total total mass. Oops, total maus Of that solution. And we have that mask it is 3.42. Right? So I'm gonna write here 3.42 grams. So we have that and it's just times 100. Right? So that will give us the mass percent or percent by weight of acetic acid. So the only thing that we don't have here is the massive acidic acid and grams. So that is going to be our goal here is to find mass of acetic acid. Okay, so let's go ahead and take a look at what we have here. So we have acidic acid and sodium hydroxide. So let's go ahead and write out acetic acid. So remember the formula chemical formula for acidic acid is Ch three C. O. A. Trade. So that's acetic acid. Obviously this is in water. So it's Aquarius and then we have sodium hydroxide and H. O. H. So honestly here we don't need to write a reaction. We're going to just to see that there really is a 1 to 1 mole ratio between sodium hydroxide and acidic acid because that is important. Right? So we're gonna go ahead and and draw that. So sodium hydroxide here is a strong base. We're going to have a full one era going forward. So acidic acid is going to obviously it's an asset is going to donate an H plus two to the base. So this is just going to be CH three C. O minus and then H plus, Right and the sodium hydroxide is the base. It's going to break up into sodium plus one and hydroxide ion minus one. So here we're going to have acidic assets without its proton. So it's just acetate. So CH three como minus, still Aquarius. And then here we're going to have this O. H minus, going to combine with the protons and basically make water H 20. And we are going to have that sodium iron in there floating around. Now it might combine with acetate and then it's going to break up again. So basically this is our chemical equation. Again, it's not super important to have this written out as long as you understand that there is a one mole to one mole ratio between the reactant. right? If you take a look at here, this reaction here, everything is balanced, Right? So that is the only important thing that we wanted here. So now we're gonna go ahead and take a sodium hydroxide. And we're going to work with that because that is basically the only thing that we we have that's given to us. So we have its volume and we have its polarity. Right? So from that we're going to be able to find mass of this, right? Or actually I'm sorry, not mask but moles of that. I assume I'm going to find moles of this Because we have all of the information to do that. And then we're going to go from moles of this. Two moles of that because it's 1-1 mole ratio. So basically moles of sodium hydroxide are going to give us the number of moles of the acetic acid and once we have moles of the acetic acid. Well, we just need the molar mass of acidic acid rate and then we can figure out the the gramps the mask. So let's go ahead and do that. So we have we're starting with 23 0. mL. Right, Alright, so we're going to go ahead and take the volume and we're going to multiply by the polarity of sodium hydroxide? But remember polarity is in units of what moles per liter. Right? So we actually cannot use this milliliters, we do need to convert these milliliters into leaders first. So we're going to go ahead and put millimeter on the bottom and leader on top so that this can cancel out millions of metric prefix. We always put one in front of that and millie is going to be 10 to the negative three liters. So once you do that conversion, you're going to get .023, 6 leaders of sodium hydroxide and now we can go ahead and multiply it by the polarity, which is point So 0.103 Molar or moles Per one L. Right, Alright, so notice that here are leaders cancel out and now we have just moles of sodium hydroxide. So now we can go ahead and do a multiple comparison and like we said, it's just a 1-1 mole ratio. So For every one mole of sodium hydroxide, We are also using one mole of acetic acid. All right, there we go. And now let's go ahead and find molds of acidic assets. So it is going to be 2. Time stands in the -3 moles of acetic acid. Alright, so there we are. Now we're gonna go ahead and take those moles and we're going to convert them into grams. Okay, so I'm going to rewrite the moles here to . Times 10 to the -3. And next we're just going to go ahead and take the acidic acid formula. And we're going to calculate the molar mass of that. So we have carbons, we have two of them here, right, for hydrogen is we have four of them. And then oxygen's we have to And if you take a look at the periodic table, carbon is going to weigh about 12.1 hydrogen weighs about 1.01, and then oxygen weighs about 16. Right? So let's go ahead and multiply. This will be 24.02. This will be four 04. And then this will be 32. Right? So these are all Ingram's, let's go ahead and add them together now. And this will give us 60.06 grams per mole. So that's the molar mass of acetic acid. And then we can go ahead and take these moles And say that in one mole Of acetic acid, there are 60.06 g of it. Right? That's the molar mass. So moles here cancel out. And now we're going to have grams left over. So let's go ahead and find that number and it should be .146 grams of acetic acid. Now, of course, we're not done yet, because we do want the percent by weight. What we found here was the mouse of acetic acid. And if we take a look at the equation that we had here previously, we said mass percent was mouse of acetic acid divided by the total mass of the vinegar, which we had so massive acetic acid we have now. So we're just gonna go ahead and take this equation here and we're just going to plug in the rest of the numbers and find the mass percent. Okay, so let's go ahead and go down here. And this is the mass of the acetic acid that we found. So finally mass percent or percent by weight is massively acetic acid, which is .146g, Divided by the total mass of that solution, which was Vinegar, which we said was 3. fixes, three grams of vinegar And then of course, times 100 to get the percentage right. So once you do this calculation, the final final final answer is going to be four 0.27 1%. Alright, so that is the percentage of acidic acid in the vinegar. Alright folks, So that is it. We're all done here. I hope you're able to follow along if you have any questions, please let us know and let's move on to the next problem
Related Practice
Textbook Question
Standardized solutions of KBrO3 are frequently used in redox titrations. The necessary solution can be made by dissolving KBrO3 in water and then titrating it with an As(III) solution. What is the molar concentration of a KBrO3 solution if 28.55 mL of the solution is needed to titrate 1.550 g of As2O3? See Problem 4.128 for the balanced equation. (As2O3 dissolves in aqueous acid solution to yield H3AsO3: As2O3 + 3 H2OS 2 H3AsO3.)
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Textbook Question
The metal content of iron in ores can be determined by a redox procedure in which the sample is first oxidized with Br2 to convert all the iron to Fe3+ and then titrated with Sn2+ to reduce the Fe3+ to Fe2+. The balanced equation is: What is the mass percent Fe in a 0.1875 g sample of ore if 13.28 mL of a 0.1015 M Sn2+ solution is needed to titrate the Fe3+?
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Textbook Question
The concentration of the Sn2+ solution used in Problem 4.130 can be found by letting it react with a known amount of Fe2+. What is the molar concentration of an Sn2+ solution if 23.84 mL is required for complete reaction with 1.4855 g of Fe2O3?
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Textbook Question
Calcium levels in blood can be determined by adding oxa-late ion to precipitate calcium oxalate, CaC2O4, followed by dissolving the precipitate in aqueous acid and titrating the resulting oxalic acid (H2C2O4) with KMnO4: How many milligrams of Ca2+ are present in 10.0 mL of blood if 21.08 mL of 0.000 988 M KMnO4 solution is needed for the titration?
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Textbook Question
Assume that you have 1.00 g of a mixture of benzoic acid (Mol. wt. = 122) and gallic acid (Mol. wt. = 170)), both of which contain one acidic hydrogen that reacts with NaOH. On titrating the mixture with 0.500 M NaOH, 14.7 mL of base is needed to completely react with both acids. What mass in grams of each acid is present in the original mixture?
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Textbook Question

A compound with the formula XOCl2 reacts with water, yielding HCl and another acid H2XO3, which has two acidic hydrogens that react with NaOH. When 0.350 g of XOCl2 was added to 50.0 mL of water and the resultant solution was titrated, 96.1 mL of 0.1225 M NaOH was required to react with all the acid. (a) Write a balanced equation for the reaction of XOCl2 with H2O.

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