G.21

[Gabby Sigal 3H]

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?

Can someone explain how to go about doing this problem?

[Alexandra Lu 2J]

You would first need to convert the mass in grams of each molecule to moles, just to know how many moles there are of each ion. For example, you would divide 0.500 g of KCl by 74.548 g/mol, its molar mass, and therefore know that the number of moles of Cl in the solution is just 0.00670708805 mol, as the ratio of moles of Cl to moles of KCl is 1:1. (For part a of this problem, you would be adding up the number of moles of K that each molecule contributes.) Finally, you would take the number of moles you found and divide it by the solution's volume in liters, because the units for molarity is mol/L, to find a given ion's concentration. Chlorine's concentration, for example, would be 0.00670708805/.5 = 0.0134141761 M.

[Rose Arcallana 2B]

I had trouble with this question too! So the key part to this problem is after taking the moles for each compound, you multiply that mole by the # the ions present in the compound, add it all together then divide by the volume. :)

For K ions:
KCl = 0.006716 moles * 1 (1 K) = 0.006716 moles
K2S = 0.004545 moles * 2 (2 K present) = 0.00909 moles
K3PO4 = 0.002358 * 3 (3 K present) = 0.007075 moles
Total moles = 0.02288 moles
Concentration = 0.02288 moles/0.5L = 0.04576 M