Hi! So I'm having a little trouble with figuring out how to start solving G.21 in the textbook. The question reads:
A solution is prepared by dissolving 0.500 g of KCl, 0.500 g of K2S, and 0.500 g of K3PO4 in 500. mL of water. What is the concentration in the final solution of (a) potassium ions; (b) sulfide ions?
My main confusion is how I would set up the equation. I was trying to set it up using the MinitalVinitial = MfinalVfinal equation, but I can't figure out if I should be combining the masses/moles of the different solutes or if I should leave them separately. If anyone knows how to set up/start the problem I would love some help!
G.21 Homework Problem
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Re: G.21 Homework Problem
Start off by converting the grams of each solution to mol of K ions using dimensional analysis. Remember to account for the mole to mole ratio of K ions in each compound. Then I believe you would simply add up the mol of K, divide by L and that will get you the molarity.

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Re: G.21 Homework Problem
For this problem, I start by converting the 0.500 grams of KCl, K2S, and K3PO4 all into moles by using their molar masses. Once I had the moles, I knew the moles of KCl = moles of K, 1 mole of K2S = 2 mole K, and 1 mole K3PO4 = 3 mole K. I used these conversion factors to convert from moles of each of the substances to the amount of moles of K in each substance. Then you add these together and divide by total liters to find the molarity. To find concentration of sulfide ions, I used the moles of K2S I calculated and because moles of K2S = moles of S in this molecule, I just divided this number by the total liters to get molarity. Hope this helps!

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Re: G.21 Homework Problem
Hi!
How I started out was finding out the amount of moles of K in each of the molecules and then adding them together and dividing by the total L at the end. For the sulfide I found the amount of moles of sulfide (there's only one molecule that has sulfide) and divided it by the final L amount! Hope that helps!
How I started out was finding out the amount of moles of K in each of the molecules and then adding them together and dividing by the total L at the end. For the sulfide I found the amount of moles of sulfide (there's only one molecule that has sulfide) and divided it by the final L amount! Hope that helps!

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Re: G.21 Homework Problem
The goal of the problem is to find the molarity of K and S respectively.
Start by finding the molar masses of the given compounds (KCl, K2S, & K3PO4). Convert the molar masses into moles of K and S. Substitute these values into M = n/V where V will be the 500. mL (.500 L) of water. I think the final answer should have 3 sig figs.
For finding the moles of K and S from the compound, look at the coefficients in the compounds:
ie. KCl is a 1:1 ratio of K per mole, K2S is a 2:1 ratio of K per mol, & K3PO4 is 3:1 of K per mol
Hope this helps. :)
Start by finding the molar masses of the given compounds (KCl, K2S, & K3PO4). Convert the molar masses into moles of K and S. Substitute these values into M = n/V where V will be the 500. mL (.500 L) of water. I think the final answer should have 3 sig figs.
For finding the moles of K and S from the compound, look at the coefficients in the compounds:
ie. KCl is a 1:1 ratio of K per mole, K2S is a 2:1 ratio of K per mol, & K3PO4 is 3:1 of K per mol
Hope this helps. :)

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Re: G.21 Homework Problem
Since you will need moles of K + and S 2 to determine their molarities with the molarity equation (M=n/V), start by converting the masses of KCl, K2S, and K3PO4 to moles with their respective molar masses.
The moles of K2S can directly be used to determine the molarity of S 2 because it is the only molecule that contains S 2. Moreover, only 1 S 2 appears in K2S, meaning 1 mol K2S=1 molS 2, so the value for mol K2S can be used without adjustment.
For determining moles of K +, you'll need to add together the moles of all molecules with k + (which is all of them is this case). However, before doing so, the moles of K2S must be multiplied by 2 (because 1 mol K2S has 2 mol K +) and the moles of K3PO4 must be multiplied by 3 (because 1 mole K3PO4 has 3 mol K +). With these adjusted figures added together, you can these use the molarity formula to find the molarity of K +.
Hope this helps!
The moles of K2S can directly be used to determine the molarity of S 2 because it is the only molecule that contains S 2. Moreover, only 1 S 2 appears in K2S, meaning 1 mol K2S=1 molS 2, so the value for mol K2S can be used without adjustment.
For determining moles of K +, you'll need to add together the moles of all molecules with k + (which is all of them is this case). However, before doing so, the moles of K2S must be multiplied by 2 (because 1 mol K2S has 2 mol K +) and the moles of K3PO4 must be multiplied by 3 (because 1 mole K3PO4 has 3 mol K +). With these adjusted figures added together, you can these use the molarity formula to find the molarity of K +.
Hope this helps!

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Re: G.21 Homework Problem
For the textbook readings and exercises: we do these for practice, but we don't turn it in online? Or is there somewhere we are supposed to upload our work to?

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Re: G.21 Homework Problem
For this problem, I did a conversion first. I converted .5 g of KCI, K2S and K3PO4 from grams to moles using the molar mass of each as followed.
.500g/74.55 g/mol= .0067 mol KCI, .500g/110.26g/mole=.0045 mol K2S, .500g/212.27g/mole=.0024 mol K3PO4
Then I multiplied each by the mole ratio of K to the element. (e.g. 1 mole of K2S=2 mole K)
After that I added all the products and divide it by the total liters to find molarity.
M=.0029/.500L = .0458
.0458 is the concentration in the final solution of potassium ions.
.500g/74.55 g/mol= .0067 mol KCI, .500g/110.26g/mole=.0045 mol K2S, .500g/212.27g/mole=.0024 mol K3PO4
Then I multiplied each by the mole ratio of K to the element. (e.g. 1 mole of K2S=2 mole K)
After that I added all the products and divide it by the total liters to find molarity.
M=.0029/.500L = .0458
.0458 is the concentration in the final solution of potassium ions.
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