Find concentration of ions given more than one solute

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Kessandra Ng 1K
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Find concentration of ions given more than one solute

Postby Kessandra Ng 1K » Tue Oct 02, 2018 3:43 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?

I know how to do these questions when it's only one solute, but since there's three solutes here how do you start?

Schuyler_Howell_4D
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Re: Find concentration of ions given more than one solute

Postby Schuyler_Howell_4D » Tue Oct 02, 2018 3:48 pm

Do you think we convert each compound to moles and then just find percent composition?

Tam To 1B
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Re: Find concentration of ions given more than one solute

Postby Tam To 1B » Tue Oct 02, 2018 5:52 pm

1. First you would determine the amount of moles for KCl, K2S, and K3PO4 by dividing the mass given by the molar mass of each compound.
In that case, you would get 0.00671 mol KCl, 0.004535 mol K2S, and 0.00236 mol K3PO4.

2. In order to calculate the amount of potassium ions, you would take into account of how many moles of it are in each compound. Since KCl has 1 K+ ion, it is 0.00671 mol. K2S has 2 moles of K, so you multiply 0.004535 by 2 and get 0.00907 mol. Lastly, in K3PO4, 0.00236 multiplies by 3 to get 0.00708 mol.

3. Once you have all the amounts of moles of K+, you add them all together 0.00671 + 0.004535 + 0.00236 = 0.02285 mol K. You then take this amount and divide it by .500 L of solution to get the molarity. 0.02285 mol K+ / .500 L = 0.0458 M

The same procedure goes for the amount of sulfide ions except since only K2S has one sulfide ion, it would be 0.004535 mol S- / .500 L = 0.00907 M


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