Problem G15 (7TH EDITION) Part B

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Heidi Ibarra Castillo 1D
Posts: 60
Joined: Fri Sep 28, 2018 12:19 am

Problem G15 (7TH EDITION) Part B

Postby Heidi Ibarra Castillo 1D » Thu Oct 04, 2018 12:45 pm

How would I go upon answering the question?
Question: An experiment requires the use of 60.0 mL of 0.50 M NaOH(aq). The stock room assistant can only find a reagant bottle of 2.5 M NaOH(aq). How can the 0.50 M NaOH(aq) be prepared?

klarratt2
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Joined: Fri Sep 28, 2018 12:16 am

Re: Problem G15 (7TH EDITION) Part B

Postby klarratt2 » Thu Oct 04, 2018 1:29 pm

Having a 0.50 M NaOH solution means that there are 0.50 moles of NaOH in 1 liter of solution. So, you will need to dilute the 2.5 M NaOH solution until it is 0.50 M.

You know that in the required solution of 0.50 M, there are 0.50 moles in 1 liter. You would divide this 0.50 moles by the molarity of the reagant solution (2.5 M or 2.5 mol/L). This would give you the volume of 2.5 M solution that contains 0.50 moles of NaOH. To make this calculated volume into a 0.50 M solution, you just add water until the total volume equals 1 liter (0.50 moles of NaOH in 1 liter of solution makes it a 0.50 M solution).

With 1 liter of the new solution you will have enough to use 60.0 mL of 0.50 M NaOH to use in the experiment!

Heidi Ibarra Castillo 1D
Posts: 60
Joined: Fri Sep 28, 2018 12:19 am

Re: Problem G15 (7TH EDITION) Part B

Postby Heidi Ibarra Castillo 1D » Thu Oct 04, 2018 9:43 pm

So, since the concentration basically needs 1/5 of the starting NaOH solution , does it mean I would need to add four volumes of water to one volume of the 2.5 mol*L^-1 solution? I divided the 60 by 5 and got 12.0 mL. So I got that 12.0 mL of 2.5 mol*L^-1 of the solution is added to the 48.0 mL of water. Is that correct?

Chem_Mod
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Re: Problem G15 (7TH EDITION) Part B

Postby Chem_Mod » Thu Oct 04, 2018 10:33 pm

Yes, your reasoning for part b is correct.


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