G.21

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Andrea- 3J
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Joined: Tue Nov 14, 2017 3:00 am

G.21

Postby Andrea- 3J » Thu Oct 11, 2018 7:10 pm

A solution is prepared by dissolving 0.500g of kcl, 0.500g of k2s, and 0.500g of k3po4 in 500.ml of water. What is the concentration in the final solution of (a) potassium ions: (b)sulfide ions?

I know that you find concentration by dividing moles by liters.

For instance: convert 0.500g of kcl to moles and then divide by 0.5L.

I don't know what to do next.

Can someone show me step by step how to find a and b?

Jeffrey Xiao 4A
Posts: 35
Joined: Fri Sep 28, 2018 12:28 am
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Re: G.21

Postby Jeffrey Xiao 4A » Thu Oct 11, 2018 7:30 pm

For part a) convert .500g of KCl to moles, convert .500g of K2S to moles then multiply the result by 2 since there are two potassium ions in that compound, and convert .500g of k3Po4 to moles then multiply the result by 3 because there are 3 potassium ions in that solution

Finally, after all the conversions, add them all up to get the moles of solute for potassium ions and divide that answer by .5L to get the molarity since M=n/V

For part b) do the same except with sulfide ions so 0.500g of k2s divided by molar mass of k2s will be moles of sulfide ions and since there is only one sulfide ion, you do not have to multiply it by anything else. Divide moles of sulfide ions by .5L to get molarity.

spark99
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Joined: Fri Sep 28, 2018 12:18 am
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Re: G.21

Postby spark99 » Thu Oct 11, 2018 7:30 pm

First, you convert 0.500 g of KCL, K2s, K3Po4 into moles. Then, you multiply the moles by the mole ratio between the moles of potassium and the compound.

0.500g KCl= .0067 mol *1
0.500g K2S= .0045 mol*2
0.500g K3PO4= .0023 mol*3

Then, you just add the results for the total amount of moles of potassium in the solution and plug it into the concentration equation, M=n/v. The total amount should be .02284.

M=.02284mol K/.25L

That should equal .048, or 4.8*10^-2M.

For b, you should use the same process.


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