Solution Concentration: Molarity

[courtrigrad]

Hello all

I just need confirmation as to whether I am performing this problem correctly:

How would you prepare 1.00 L of a 0.50 M solution of HCl from a 12M stock reagent?

My solution:

1.00 L * (0.5 mol HCl/ L solution) = 0.5 mol HCl

V * (12 mol HCl / L solution) = 0.5 mol HCl

V = 40 mL

is this the correct volume?

any help is appreciated

thanks!

[mrjeffy321]

seems right to me. except, remember to add the rest of the water to it.

you have the right idea, find the amount of moles you need, in this case .5, then find out how much of the original you need to get .5, then just dilute that amount up to the end amount you want, in this case 1 L.

[courtrigrad]

thanks a lot

just have one more question

If i want to prepare 1.00 L of a 0.50 M solution of NiCl2 from the salt NiCl2 * 6H20, how would i go about in solving this?

[so-crates]

The same way, except remember to subtract (or add, depending on how you do it) the weight of the H20 as it will be released when it dissolves in teh solution.

[osskall]

Hello all

My solution:

1.00 L * (0.5 mol HCl/ L solution) = 0.5 mol HCl

V * (12 mol HCl / L solution) = 0.5 mol HCl

V = 40 mL

is this the correct volume?


Yeah, well I would, in this case, skip the 0.5 mol HCl step and just state that I want to dilute the stock solution with a factor 24 (12/0.5), which indeed means taking ~40mL (1/24 of 1000mL) of the stock, if 1L is what you wanted, and add water (and mix!) until the volume is 1L.
Also think about accuracies. How close to 12 M is the concentrated hydrochloric acid and how close to 0.5 M (as well as "how accurately defined", which is a totally different question) do you want the contration of your dilute solution to be? Seems to me you can use the most convenient (and probably least accurate) glassware in this case :cool:

[pack_rat2]

...V = 40 mL...is this the correct volume?
41.7 ml is a better answer, since you can easily read a 50 ml graduated cylinder that closely.

[pack_rat2]

thanks a lot

just have one more question

If i want to prepare 1.00 L of a 0.50 M solution of NiCl2 from the salt NiCl2 * 6H20, how would i go about in solving this?
Here's a tip...first figure out how many moles of solute you need; you need 0.5 mole of NiCl2. That means you need 0.5 mole of the hexahydrate, too, since each mole of it contains a mole of NiCl2.