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Ch.4 - Reactions in Aqueous Solution

Chapter 4, Problem 129

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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Hi everyone. I have a problem giving us the following that ionic equation and it tells us that the amount of iodine three minus ions can be determined by the eye line three minus solution with known concentration of S two minus Aquarius at the sulfate ion. And asking us to calculate the similarity of an iodine three minus solution. Given that it requires 35.8 mL of 0.54 moller N A two S 203. Acquiesce to type four millimeter sample of iodine three minus a quiz. So first let's see what happens when in a two S 203 disassociate. So we're going to get to sodium plus plus one S two minus. And our is going to equal moles over leaders. So we're going to just to get The moles of S203 and then we can use that together as to the malls Of iodine 3 -. So we have zero 0.45 four malls of S her one leader. And we're going to multiply that by 35.8 million L. And that has to be changed to leaders. So we're going to multiply that by 10 to the negative third. Leaders over one m. And that gives us 0.01625 moles of S 203 two minus. Now we want to convert this two moles of red, I'm three minus. So we have our 0. moles Of 2032 -. And then we're gonna multiply that at multiple ratio. So for every one mole of iodine three months we have two moles of s two 032. And that's found on our balanced equation. So that's going to equal 0.0081, 2, 66 malls of iodine a minus. Now, we need to calculate the actual more clarity. So we have our 0. balls of iodine three minus over our leaders. And it gave us 34 ml. So we're just going to divide that by 1000. So to change it to leaders. So 0.034 l And that equals 0.239 Smalling iodine 3 -. And that is our final answer. Thank you for watching. Bye.
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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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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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