G. 21

Juanalv326
Posts: 49
Joined: Fri Sep 29, 2017 7:04 am

G. 21

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 fi nal solution of (a) potassium ions; (b) sulfi de ions?

Swetha Sundaram 1E
Posts: 59
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Re: G. 21

For this problem, in part a) I first converted KCl, K2S, and K3PO4 from grams to moles and then I added each of the moles together which was 2.29 x 10^-2 mole and then I divided that by 0.5 L because Molarity multipled by Volume = moles and got 4.58 x 10^-2M. In part b) I used the grams of K2S to find the mol of the sulfur ions and then divided it by the volume 0.500 L to get 9.07 x 10^-3 M. Hopefully this made sense lol

Lucian1F
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Re: G. 21

You first need to calculate the amount of moles for each compound by dividing the given masses by the molar masses resulting in .00671 mol KCl, .00453 mol K2S, and .00236 mol K3PO4. Therefore, there are .00671 mol K in KCl, .00906 mol K in K2S, and .00472 mol K in K3PO4 for .0205 mol K total and .00453 mol S in K2S. With these values, you can now divide the moles of K and S by the volume to find the concentrations, resulting in .0410M for K and .00906M for S. There is rounding error in my answers but this is the procedure you would do for this problem. I hope this helps!

Haocheng Zhang 2A
Posts: 52
Joined: Fri Sep 29, 2017 7:04 am

Re: G. 21

When you dissovle ionic compounds into water, they exist in form of ions. For example, when you dissovle 1 mole of KCl into water completely, you get 1 mole of potassium ions and 1 mole of chlorine ions. The number of mole of potassium ions is that of potassium. For this question you only have to calculate the number of mole of potassium and sulfur in all solutes, and you get the mole number of potassium ions and sulfur ions. Finally, divide the number of mole by volume of solution and you get the concentration. Hope it helps.