[ParaChase]

I have some data of a reaction at different concentrations. (FeCl3 + KI) What I need to know is if I add water to a mixture of FeCl3 and KI, how does that affect molarity? I realize the entire solution will be come diluted, but exactly how does water affect each concentration? Should I set up the equation as if the water affects each solution individually without considering the volume of the other? For ex. should I be concerned with finding the dilution of KI and H2O then finding the dilution of FeCl3 and the same H2O amount?

[confusedstud]

Water would dilute both reactants. So if initially i have 1 mole of FeCl3 and 2 mole of KI and 1 dm^3 of water and after adding 1 dm^3 of water, the concentration of FeCl3 will change from 1mol/dm^3 to 0.5mol/dm^3. But I am quite unsure what new products would be formed from this reaction. Aren't both iron (iii) iodide and potassium chloride soluble in water?

[ParaChase]

Yeah, they are both soluble. So if I calculate the diluted solution of the FeCl3 to .5M, does that mean the 1L of water is already used up? So the concentration of KI remains the same as it originally was?

[confusedstud]

I think you have some misconceptions here. Molarity is the numb of moles of solute divided by the total volume of water. When I dissolve something in water, the water isn't used up. So the concentration for both solutions would decrease. I hope this helps.

[ParaChase]

Ok, thanks. So just one last question, would both concentrations decrease as if the water was added to it individually? So using your example from earlier, the individual concentrations would be 1M FeCl3 diluted with 1L of water, and 2M KI diluted with 1L of water? Thanks

[ParaChase]

Never mind, I got it. Thanks though