Textbook practice problem 6B.11

[Jazlyn Romero 1I]

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?

Hello! for part a, can someone explain how can we find the molar concentration of the hydroxide ions for the original solution? How do we know which mL of volume to use? Thanks!

[Joseph Hsing 2C]

The pH of the diluted solution is given so set that equal to -log[H30+] and then solve for [H30+]. I got a value of 5.62 * 10^-14.

Now you can use Kw = [H30+][OH-] = 10^-14, substitute and solve for [OH-] and that is the molar concentration of hydroxide ions of the diluted solution of 0.18M.

Next, use dilution equation and isolate so I used M2=M1V1/V2. ( 0.18M*500mL) / 5 mL = 18 M

You use 500mL for V1 because it is the volume associated with the 0.18M diluted solution and then 5mL for V2 because it was the volume of solution that was diluted. You only use the 200mL if u wanted to find the number in moles of -OH in the original solution.

Hope this helped :)