L37

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Emil Velasco 1H
Posts: 87
Joined: Wed Nov 21, 2018 12:19 am

L37

Postby Emil Velasco 1H » Thu Oct 10, 2019 12:01 am

(a) How would you prepare 1.00 L of 0.50 M HNO3 (aq) from “concentrated” (16 M) HNO3 (aq)?

(b) How many milliliters of 0.20 M NaOH(aq) could be neutralized by 100. mL of the diluted solution?

I'm not entirely sure how to begin solving this problem

sarahsalama2E
Posts: 148
Joined: Fri Aug 30, 2019 12:16 am

Re: L37

Postby sarahsalama2E » Thu Oct 10, 2019 12:10 am

This is a dilution problem, which means that we need to use the equations M1V1=M2V2.
For part a, we need to do 16 M (X L)= 1 L (.50 M)--> this gives us the amount of L that we need to use from the 16 M solution. which is 31 mL. You would then dilute this amount with water, until you reach the 1 L mark

part b, you also use M1V1=M2V2.
(.1 L)(.50 M) (this is the diluted solution) = (X L) (.20 M). The answer is 250 Ml

Brooke Yasuda 2J
Posts: 87
Joined: Sat Jul 20, 2019 12:17 am

Re: L37

Postby Brooke Yasuda 2J » Thu Oct 10, 2019 11:48 am

Yes, so for how to approach or attack these problem, you should start off by looking at what is given and what you are looking for. For example, in part a you are given the molarity of your concentrated solution (.50 M). You are also told that the final product you are looking for is 1.00 L of .50 M HNO3. So, looking at the given info, you are given the initial concentration (.50 M), the final volume (1.00 L), and the final concentration (.50 M). What you are solving for is the volume of the initial solution (.50 M) needed to get you 1 L of .500 solution. So, just plug in this info into the equation as stated above and solve for the missing variable.

Shail Avasthi 3C
Posts: 81
Joined: Fri Aug 30, 2019 12:17 am

Re: L37

Postby Shail Avasthi 3C » Sat Oct 12, 2019 4:57 pm

a) Typically for dilution problems you want to look for a way to use M1V1 = M2V2 if possible. In this case we are given 3 knowns (concentrated molarity, final molarity, and final volume) and one unknown (concentrated volume). When we plug these into M1V1 = M2V2, we get the following:

(16 M concentrated HNO3)(V concentrated HNO3) = (0.5 M diluted HNO3)(1 L diluted HNO3) ; V = 0.0312 L = 31.2 mL NaOH

b) To neutralize the HNO3 in solution, we need equal moles of H+ and OH- ions in solution. The ratio of H+ to OH- between HNO3 and NaOH is 1:1, so no adjustment needs to be done in the required moles of NaOH. Therefore, we can use M1V1 = M2V2 again to calculate the required volume of NaOH:

(0.5 M HNO3)(0.1 L HNO3) = (0.2 M NaOH)(V NaOH) ; V = 0.25 L = 250 mL NaOH


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