The arsenic in a 1.22-g sample of a pesticide was converted to
As...

Question

The arsenic in a 1.22-g sample of a pesticide was converted to
AsO43- by suitable chemical treatment. It was then titrated
using Ag+ to form Ag3AsO4 as a precipitate. (c) If it took 25.0 mL of 0.102 M Ag+to reach the equivalence point in this titration, what is the mass percentage of arsenic in the pesticide?

Accepted Answer

Hi everyone for this problem. We're told the amount of potassium chloride in a 0.650 g sample was analyzed. The chlorine and the solution was tight traded with silver to form silver chloride. A total of 26.5 mL of 0. moller silver was used to reach the equivalence point, calculate the percent by mass of potassium chloride in the sample. Okay, So we want to calculate the percent by mass of potassium chloride. So let's recall what is our mass percent equation. So our percent mass in this case is going to equal our mass and grams of our potassium chloride over our total mass And grams of our sample. Okay. And this is all going to be multiplied by 100%. And so we know what our total mass and grams of the sample is. It was given in the problem we're told it is 0.650 g. So we need to now calculate the mass and grams of our potassium chloride. Let's start off by writing our reaction. Okay, So we're told that we have corinne reacting with silver to form silver chloride. Okay, and here we see. We have a 12 mol 1 to 1 molar ratio between our corinne and silver. So the first thing that we need to do is we need to calculate our moles of corinne titrate. Id. Okay, because once we figure out our moles of Corinne titrate ID, we can go from moles to grams of potassium chloride. So let's start here. We're told in the problem that we have 26.5 male leaders of our silver. Okay. And we need to go from male leaders to leaders in order for us to use the mill arat e that were given in the problem. So let's go there first. So in one male leader there's 10 to the negative three leaders. And so are male leaders cancel And we have moles in the problem. We're given malaria T. We're told we have 0.115 moller silver. And recall that malaria T. Means moles over leader. So in this case we have 0.115 moles of silver for every liter of solution. So we can use that as a conversion here. So we have 0.115 moles of silver and every one liter of solution. So now our leaders cancel. And we're left with moles of silver and we can go from moles of silver, two moles of chlorine. Using our multiple ratio we have in the reaction that we wrote out. There's one mole of silver for every one more of chlorine. Okay, so our moles of silver cancel and we're left with moles of chlorine. And when we calculate this out we get 3. times To the negative three moles of chlorine. Okay, so now that we know our molds of corinne. We need to calculate our mass of potassium chloride and we'll use our molds of chlorine as a starting point. So we said we have 3. times 10 to the negative three moles of chlorine. We'll use our multiple ratio to go from moles of chlorine to moles of potassium chloride. And that is going to be a 1-1 molar ratio. So in wana mole of chlorine we have one more of potassium chloride and our moles of chlorine cancel. And we have moles of potassium chloride to go from moles of potassium chloride, two g of potassium chloride. We need our molar mass. So in one mole of potassium chloride we have 74. g of potassium chloride. Okay, so our moles of potassium chloride cancel and we have grams. When we calculate this out, we get 0.227g of potassium chloride. That is our missing piece to our mass percent equation. So let's go ahead and plug it in. We have zero g and this is going to be multiplied by 100%. And we're going to get a final answer of 34.9%. And this is the answer to this problem. This is the mass percent of potassium chloride in the sample. That's the end of this problem. I hope this was helpful

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