Concentration/molarity question (needing clarification): "If 160 mL of water is added to 40 mL of 1.2 M NaCl solution, what will be the final concentration?" What are they referring to by "concentration"? ******************************** Also: "Will Ca(NO3)2, an ionic solute be soluble in water or not? If yes, explain your answer."
concentration in terms of molarity is a measure of how much solute you have in a given volume of solvent. Remember that if you dilute a solution by adding solvent, the number of moles of solute has not changed. Therefore we can write \[n_{initial} = n_{final}\] As n = cV (molarity times the volume in L) we can write \[c_iV_i = c_fV_f\] Rearranging this will give you the solution. For volume to volume stock changes, you do not need much calculation actually and can do this intuitively. The volume of new solution will be 160mL more than the original 40mL which is 200mL total. This will be 5 times more dilute compared to the 40mL (200/40). Thus the final concentration is 5 times less than the original. For nitrates, they are always soluble (remember the precipitate rules?) This is largely a result of the strong hydrogen bonding that can be created with the free oxygens from the nitrate ion.
@mww I know that nitrates are always soluble, but I think the intent is for a different method of answering that problem... if not, then I guess I have the answer. :\
I thought I had to do some kind of molarity/molality calculations to solve it, but I'm not sure how -- if that even is the right way...
@kittiwitti1 now that I think about it, I do remember a means to indicate solubility. Use the Ksp (K solubility product) to calculate the concentration of solute in water. http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch18/ksp.php This might be difficult to isolate the value but wikipedia lists the solubility at 1212g/L of water. Thus you can dissolve large amounts of calcium nitrate in water. This is not a chemical explanation however.
Alright thank you. @mww
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