## Homework Problem G.21 [ENDORSED]

GFolk_1D
Posts: 101
Joined: Fri Aug 09, 2019 12:15 am

### Homework Problem G.21

A solution is prepared by dissolving 0.500g of KCl, 0.500g of K2S, and 0.500g of K3PO4 in 500mL of water. What is the concentration in the final solution of (a) potassium ions; (b) sulfide ions?

Would it be appropriate to group the compounds and their masses together at the beginning of the problem and then solve? As in add up the masses to equal 1.50g and divide by the molar mass of all of the compounds combined? Or is there another way to go about solving this problem?

Thanks!!

sarahsalama2E
Posts: 164
Joined: Fri Aug 30, 2019 12:16 am

### Re: Homework Problem G.21  [ENDORSED]

concentration is moles/liters so this means that we need to find the number of moles in KCl, K2S, and K3PO4--respectively. To find the number of moles, we first need to determine the molar masses of each compound.

molar mass of KCl: 74.55 g/mol
molar mass of K2S: 110.26 g/mol
molar mass of K3PO4: 212.27 g/mol

now to find the total moles of K, because part a is asking for the concentration of potassium ions--we do the following calculation
moles of K in KCl: 0.500 g/ 74.55 g/mol
moles of K in K2S: 0.500 g/ 110.26 g/mol x 2molK/1mole K2S
moles of K in K3PO4: 0.500g / 212.27 g/mol x 3mol K/1 mole K3PO4

adding all of those resultant values up gives us 2.29x10^-2 moles K---this would be the TOTAL number of moles of K in the solution.
then to find the molarity, which is the moles of solute divided by the volume we divide 2.29x10^-2 moles K by .5 L (given from the question). dividing this gives us a molarity of .0458 M. that is the concentration of potassium ions in the solution.

For sulfide ions you would repeat this same process except you would find the moles of sulfur in each compound. In this case it would be much faster as the only sulfur present in the solution is in K2S.

Amy Pham 1D
Posts: 103
Joined: Fri Aug 09, 2019 12:15 am

### Re: Homework Problem G.21

Especially after having done part a, part b can be done very quickly as you have already found the mols of K2S, the only compound in which sulfide ions are present. The mols of K2S are the mols of sulfide ion, and you would simply divide this number by 0.500L.