dilution question

[gegogi]

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When 27 mL of a 305 mL stock solution containing 2 M BaF2 is diluted to 70 mL, what is the concentration of F- in the new solution?

the answer was 0.231

However, I am not sure how it comes out....anybody who wanna volunteer? 🙂

 

[joonkimdds]

I am not quite sure how it's 0.231, but here is what I tried.
M1V1 = M2V2
2 x 0.027 = M x 0.07
M = 0.7714
but that's the concentration of the entire BaF2 but the question only asked for F-.
So hm...i don't know exactly how to find F- from BaF2 but I just divided it by 3 and got 0.257 but don't think this is the right method. I assumed that for each [Ba] there are 2 of [F-] so i divided it by 3. Anyone??

 

[admap]

Ok this is the way to do it:

Since you have 305 ml of 2M stock solution, you first have to find out the no. of moles present in it = (2mol/1L) * (1L/1000ml) * (305ml) = 0.6mol

Now apply the M1V1=M2V2
0.6 * 0.027 = M2 * 0.07
M2 = 0.231

 

[Andre3k]

.6mol of what? BaF2 or F-.
And M is equal to the molarity not the number of moles.

 

[joonkimdds]

hm....

 

[admap]

So, I think I did a mistake here and maybe the answer provided to the OP should be multiplied by 2 for 2 moles of F- in BaF2. The answer should actually be 0.462

Yeah, I know M is the molarity and not the no. of moles. So, 0.6 is the no. of moles of BaF2. As far as the moles go, if you divide them by 1L, it becomes M. In an equation, if you divide both sides by 1L, it doesn't matter. It will just cancel out.

Anyway, maybe I am right, maybe I am not. I'll check and let you know. Till then nobody else wants to take a shot at it?

 

[soopasteve]

So, I think I did a mistake here and maybe the answer provided to the OP should be multiplied by 2 for 2 moles of F- in BaF2. The answer should actually be 0.462

Yeah, I know M is the molarity and not the no. of moles. So, 0.6 is the no. of moles of BaF2. As far as the moles go, if you divide them by 1L, it becomes M. In an equation, if you divide both sides by 1L, it doesn't matter. It will just cancel out.

Anyway, maybe I am right, maybe I am not. I'll check and let you know. Till then nobody else wants to take a shot at it?

I did the exact same thing but I also assume that its supposed to be multiplied by two assuming that BaF2 solvates into ions. I dont know though, isnt F a ****ty ionizer due to its hardness?

 

[joonkimdds]

Does anybody get this now? :?

 

[GCT]

When 27 mL of a 305 mL stock solution containing 2 M BaF2 is diluted to 70 mL, what is the concentration of F- in the new solution?

the answer was 0.231

However, I am not sure how it comes out....anybody who wanna volunteer? 🙂

Here are the steps

- calculate the molarity of F- in the stock solution , remember there are 2 F- for every BaF2 molecule

- calculate the moles of F- in the 27 mL solution

- calculate the molarity by using above mole value and dividing by the new volume