6B.11 Question

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Stella Nguyen 1J
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Joined: Wed Sep 30, 2020 9:41 pm

6B.11 Question

Postby Stella Nguyen 1J » Wed Jan 13, 2021 5:13 pm

Hi everyone!

I was a little confused about how to start this problem, and I was wondering if you guys could explain it to me?

6B.11: A student added solid Na2O to a volumetric flask of volume 200.0 mL, which was then filled with water, resulting in 200.0 mL of NaOH solution. Then 5.00 mL of the solution was transferred to another volumetric flask and diluted to 500.0 mL. The pH of the diluted solution is 13.25. (a) What is the molar concentration of hydroxide ions in (i) the diluted solution, (ii) the original solution? (b) What mass of Na2O was added to the first flask?

Thank you so much!

Andrew Wang 1C
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Re: 6B.11 Question

Postby Andrew Wang 1C » Wed Jan 13, 2021 5:27 pm

Using the pH, you can find the current (diluted) concentration of OH-. Then you can use M1V1=M2V2 to find the concentration of the original solution. Using the concentration you found, you can figure out the moles of NaOH, then use molar ratios and stoichiometry to find the mass of Na2O added.

Hope this helps!

Benjamin_Hugh_3F
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Re: 6B.11 Question

Postby Benjamin_Hugh_3F » Wed Jan 13, 2021 5:37 pm

We are given the PH(13.25), which means -log[H+]=13.25. Then [H+] would equal 10^-13.25( or 5.6 x 10^-14).

Using that value you can find [OH-] in the diluted solution since Kw=[H+][OH-]=10^-14

After calculating [OH-], [OH-] x (500ml/5ml) would give you the concentration of [OH-] in the original solution.

To calculate the mass of Na2O added in the first flask, multiply the [OH-] of the original solution by the volume of Na2O added to find the moles then from there you can calculate the mass using mole ratio and molar mass.


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