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Ch.4 - Reactions in Aqueous Solution

Chapter 4, Problem 92a

The following three solutions are mixed: 100.0 mL of 0.100 M Na2SO4, 50.0 mL of 0.300 M ZnCl2, and 100.0 mL of 0.200 M Ba(CN)2. (a) What ionic compounds will precipitate out of solution?

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Hello everyone today. We have the following problem. If 100 mL of 2.2 50 Mueller of strontium chloride, 1 50 million liters of 500.175 Mueller of copper chlorate and 100 mL of 1000.300 molar of potassium sulfate are mixed in a solution. What precipitates will the solution produce? So the first you want to do is you want to draw each of these dissociating essentially. So let's say we have strong team chloride. We're going to draw dissociating into one Strong team two plus ion and two chloride minus ions. We're going to have our copper chlorate dissociating into copper plus and chloride ions. And lastly our potassium sulfide I misspoke earlier it is actually sulfide would associate into two potassium plus ions as well as one Sulfide to -1. And so it's important to note that all potassium and chlor eight ions or assaults are soluble. And so what would make these insoluble? So let's say we pair each of these ions here. Each of these salts with sulfate and chloride. So we have our strontium two plus combining with sulfate 2 -2. Give us a strong team of sulfate and a general rule so this is going to form a solid. Why? Well sulfates are soluble accept and there are exceptions except when it is paired with calcium strontium barium or any two group to a ions. And so we're going to pair copper and our potassium next Sophie have to copper atoms combining with our sulfate ion. This will form copper two sulfate but this will be in the Aquarius form since sulfates are soluble except when paired with the ones that we mentioned before. And so lastly we have our two potassium ions with sulfate. This one is going to be sybil once again because you mentioned that all potassium ions and salts are soluble so so far we just have our strontium sulfate here as a precipitate. Next we need to combine these ions with chloride. Sophie do strontium, we will get strontium chloride and this is going to be a quiz since we mentioned before that Hey lights are indeed soluble and hey lives are going to be our chlorine are bro means or bromides and our iodide will accept when they are paired with gold, lead, copper or mercury. And so this first combination of strong team and chlorine is going to be soluble because it is not paired with any of those other ions. When we take copper and we get copper chloride, this is actually going to form a precipitate. Why? Because we are taking chloride chloride. Hey lied comparing it with copper making it insoluble. So this is our second precipitate formed copper chloride. And lastly of course we have our potassium forming with chloride giving us potassium chloride a very famous assault. And this is of course going to be a quiz. So our answers for which precipitates will form will be our strong team sulfate and our copper chloride you know that we've answered the question overall I hope this helped and until next time.
Related Practice
Textbook Question

How could you use a precipitation reaction to separate each of the following pairs of cations? Write the formula for each reactant you would add, and write a balanced net ionic equation for each reaction. (a)

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Textbook Question

How could you use a precipitation reaction to separate each of the following pairs of cations? Write the formula for each reactant you would add, and write a balanced net ionic equation for each reaction. (b)

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Textbook Question
How could you use a precipitation reaction to separate each of the following pairs of anions? Write the formula for each reactant you would add, and write a balanced net ionic equation for each reaction. (a)
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Textbook Question

The following three solutions are mixed: 100.0 mL of 0.100 M Na2SO4, 50.0 mL of 0.300 M ZnCl2, and 100.0 mL of 0.200 M Ba(CN)2. (b) What is the molarity of each ion remaining in the solution assuming complete precipitation of all insoluble compounds?

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Textbook Question

A 250.0 g sample of a white solid is known to be a mixture of KNO3, BaCl2, and NaCl. When 100.0 g of this mixture is dis-solved in water and allowed to react with excess H2SO4, 67.3 g of a white precipitate is collected. When the remaining 150.0 g of the mixture is dissolved in water and allowed to react with excess AgNO3, 197.6 g of a second precipitate is collected. (a) What are the formulas of the two precipitates?

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Textbook Question

(b) What is the mass of each substance in the original 250 g mixture?

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