G.21

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arisawaters2D
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G.21

Postby arisawaters2D » Fri Oct 09, 2020 1:17 pm

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?

what are some of the methods you guys used for solving this? thanks!

Chem_Mod
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Re: G.21

Postby Chem_Mod » Fri Oct 09, 2020 2:10 pm

When asked to find concentration, it's generally best to calculate molarity, which is the moles of solute over the total volume of solution. Since you are given the total volume of solution, you should work on calculating moles of solute. You can do this by using dimensional analysis to calculate the total moles of potassium in each sample and then adding them up. Then divide by the total volume of the solution in liters to find the molarity.

rachelhchem1I
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Re: G.21

Postby rachelhchem1I » Fri Oct 09, 2020 2:13 pm

So to do this problem, I first found the number of moles in each .5g sample. Then, you will want to observe the number of potassium and sulfur ions in each compound and multiply the moles by that factor. For example, if there was .008 moles in .5g of K2S, you would multiply .008 by 2 because there are 2 moles of K in that compound. You do this for each sample, add them together, and divide by the 500 ml of water for both potassium ions and sulfur ions. Hope this helps!!

Kayla Law 2D
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Re: G.21

Postby Kayla Law 2D » Fri Oct 09, 2020 2:23 pm

Hi! I started by converting the 500 mL of water into .5 L. Next, I converted the .500 g of KCl, K2S, and K3PO4 into moles by dividing .500 g by each of their molar masses.
For KCl, I found that there were 0.00671 moles.
For K2S I found that there were 0.00453 moles.
For K3PO4 I found that there were 0.00236 moles.
Next, I found the moles of potassium ions in KCl, K2S, and K3PO4.
Since there is only one potassium in KCl, the moles of potassium ions would also be 0.00671 moles.
Since there are 2 potassium in K2S, I multiplied 0.00453 moles by 2 and got 0.00907 moles.
Since there are 3 potassium in K3PO4, I multiplied 0.00236 by 3 and got 0.00707 moles.
Next, to find the concentration, or molarity, of potassium ions in the final solution, you add up each of the moles we solved for above and divide by the volume, .5 L. So 0.00671 + 0.00907 + 0.00707 = 0.02285 moles. 0.02285 moles / 0.5 L = 0.0457 M, or 4.57 x 10^(-2) M. (The answer key says 4.58 x 10^(-2) M but hopefully my answer is close enough to count lol).
For part B, you do essentially the same thing but with sulfide ions instead of potassium ions.
We only find sulfur in K2S, and there is only one sulfur in K2S, so there will be 0.00453 moles of sulfide ions in the final solution. 0.00453 moles / .5 L = 9.06 x 10^(-3) M. (If you use unrounded numbers for calculations you'll get 9.07 x 10^(-3) M which is the answer in the textbook).
Hope this helps! :)

Melanie Lin 3E
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Re: G.21

Postby Melanie Lin 3E » Fri Oct 09, 2020 3:48 pm

I would first try to find the moles of each solute first and convert to the moles of Potassium in each. For example, with K2S you'd double the number of moles of K2S since there are 2 K+ ions in a molecule of K2S. From there you can add up all the moles of potassium and divide it by 0.500 L (500. mL).


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