Fundamentals G21 [ENDORSED]

[Rida Ismail 2E]

Hey guys. I'm struggling with G21.

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

Can anyone please help me solve this?

[Ashley Kim 3F]

To find the final concentration of potassium ions and sulfide ions, you have to use stoichiometry to find the total moles of K+ and S2- ions.

Add the number of moles of K+ ions in .500g KCl, K2S, and K3PO4, respectively and divide the total moles K+ by .5L H2O to get the final concentration of potassium ions.

Similarly for S2- ions, find the number of moles of S2- ions in .500g K2S and divide by .5L H2O to get the final concentration of potassium ions.

[Jacob Puchalski 1G]

Ashley explained it well, here's part a written out.

.5 grams KCl x (1 mole KCl)/(74.55 g KCl) = .0067 moles of KCl.
For every molecule of KCl, there is one K atom. So there are .0067 moles of K.

.5 grams K2S x (1 mole K2S)/(110.27 g K2S) = .0045 moles of K2S
For every molecule of K2S, there are two K atoms. So there are .0090 moles of K.

.5 grams K3PO4 x (1 mole K3PO4)/(212.3 g K3PO4) = .0024 moles of K3PO4
For every molecule of K3PO4, there are three K atoms. So there are .0072 moles of K.

.0067 + .0090 + .0072 = .0229 moles of K.

.0229 moles K/ .5 L = concentration of .0458 M, or 4.58 x 10^-2 M.

Part b is pretty much the same thing but with sulfide ions.