Problem 12.33 Sixth Edition

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Kate_Santoso_4F
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Joined: Fri Sep 28, 2018 12:29 am

Problem 12.33 Sixth Edition

Postby Kate_Santoso_4F » Sat Jan 19, 2019 2:37 pm

A student added solid Na20 to a 200.0 mL volumetric flask, which was then filled with water, resulting in 200.0 mL of NaOH solution. 5.00 mL of the solution was then transferred to another volumetric flask and diluted to 500.0 mL. The pH of the diluted solution is 13.25. What is the concentration of the hydroxide ion in (a) the diluted solution? (b) the original solution?
I have solved for the concentration of hydroxide ions in the diluted solution. How can I solve for the concentration of hydroxide ions in the original solution?

Vincent Li 4L
Posts: 48
Joined: Fri Sep 28, 2018 12:19 am

Re: Problem 12.33 Sixth Edition

Postby Vincent Li 4L » Sat Jan 19, 2019 11:49 pm

In order to solve for the hydroxide concentration of the original solution, we must work backwards from the dilution that took place. Using the equation M1 * V1 = M2 * V2, we can plug in the values of the original solution for M1 and V1, and the values for the diluted solution go into M2 and V2. The volumes are given as 5.00mL of the original solution being used to dilute to 500.0mL. The molarity of the diluted solution was found, so the last unknown is M1, or the molarity of the original solution.

Nina Do 4L
Posts: 61
Joined: Fri Sep 28, 2018 12:27 am

Re: Problem 12.33 Sixth Edition

Postby Nina Do 4L » Tue Jan 22, 2019 7:54 pm

For me, I divided the final volume (.5 L) by the initial (.005 L) and multiplied it by the original concentration of hydroxide (.18) and I got 18 M which is correct according to the solutions manual.


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