Page 1 of 1

HW problem G21

Posted: Wed Oct 02, 2019 5:04 pm
by Simon Dionson 4I
How should I approach this problem? Would it make sense to find the molarity of each substance and then collectively add up the concentration (ex. 5M K + 6M K + 3M K + ...) of the ions based on the molecular formulas, or is there another way to do it?

Re: HW problem G21

Posted: Wed Oct 02, 2019 5:13 pm
by Mansi_1D
I think you can get to the final answer by adding up the molarities you calculate, but I'm not sure. The way I did is that I found the moles of K in each of given the compounds by diving the grams of each samples by its molar mass and multiply the molar ratio of K ions in one mole of each sample. I added the moles of K together and used the moles/volume formula to find the molarity of K.

Re: HW problem G21

Posted: Wed Oct 02, 2019 5:31 pm
by Hannah Lee 2F
I agree with Mansi! The concentration of ions in a solution usually depend on the mole ratios between the dissolved substance in the solution and the cations and/or anions it forms — in this case, the question is asking for K+ and S2-. After dividing the mass of each compound by the molar mass to find the amount of moles, you would use mole ratios to find the amount of moles of each ion. For instance, for K2S, it would be 0.500 g K2S divided by 110.26 g/mol = 0.00453 mol K2S. Because there is a 2:1 ratio of K+ ions to K2S, you would multiply that number of moles by 2 via mole ratios to find # of moles of K+ in K2S: (0.00453 mol K2S) (2 mol K+ / 1 mol K2S). After doing that for the rest of the compounds, you would add them all up to find the total number of moles of that ion present; lastly, you would use M = n / vol to find the actual concentration of ions in the solution.

Re: HW problem G21

Posted: Wed Oct 02, 2019 5:35 pm
by Michelle Chan 1J
Molarity is moles/volume of solution. So to use (b) as an example, first I would find the moles of S ions in 0.500g of K2S with stoichiometry --> 0.500g K2S x 1 mol K2S/110.26g x 1 mol S/ 1 mol K2S. Then convert 500. mL of volume to 0.5 L. Finally, divide the number of moles by volume to get M. Hope that made sense!