Concentration via pH (Q11.35)

[Alexis Baranoff 3B]

Can someone help explain this question to me, focusing on b and c?

A student added solid Na2O to a 200.0mL volumetric flask, which was then filled with water, resulting in 200.0mL of NaOH solution. 5.00mL of the solution was then transferred to another volumetric flask and diluted to 500.0mL. The pH of the diluted solution is 13.25. What is the concentration of hydroxide ion in (a) the diluted solution? (b) the original solution? (c) What mass of Na2O was added to the first flask?

[Helen Leka 1H]

Sorry it's a little messy but here's how to do it. Let me know if you need clarification.

[Christopher Phung 3G]

For part b:
Recall that pOH = -log[^-OH] where log stands for "common log."
Isolating [^-OH] on the right side of the equation, we can find that it equals 10^-pOH.
Since we know pOH from part a (14 - pH = pOH = 0.75), to solve for [^-OH] we calculate 10^-0.75 which is approx. 0.18. For future calculations, I will not be using this rounded value and will instead use the exact value [^-OH] until I arrive at the final answer, at which I will round to the proper amount of significant figures.

For part c:
In order to solve for the mass of Na₂O that was put into the original solution, we need to first find the number of moles of ^-OH in the original solution, and then calculate the amount of Na₂O that is required to result in that many number of moles of ^-OH.
Recall that molarity (M) = moles of solute / liters of solution.
From part b, we know the molarity of [^-OH] in the final diluted solution.
That is, [^-OH]_dilute = moles of ^-OH_dilute / 0.5 L
Solving for moles of ^-OH_dilute, we can find that it equals [^-OH] * 0.5 L
These moles of ^-OH in the dilute solution came from the 0.005 L of the original solution that was transferred over and then diluted.
Since the composition of this 0.005 L of solution that was transferred over is the same as its source, we can say the molarity of this 0.005 L of solution is the same as the molarity of the original solution.
That is, M_original = moles of ^-OH_dilute / 0.005 L = ^-OH_original / 0.2 L
From this, we can conclude that the molarity of the original solution is [^-OH] * 0.5 / 0.005 L.
The original solution has a volume of 0.2 L, so in order to get the moles of ^-OH in the original solution, we multiply the molarity of the original solution by 0.2 L. This comes out to be approx. 3.55 moles.
The balanced equation for the reaction of Na₂O with water is Na₂O + H₂O = 2NaOH.
Since every mole of NaOH dissociates into OH, using mole ratios we can conclude that for every 1 mole of Na₂O reacting with water, 2 moles of ^-OH is formed.
So to get the number of moles of Na₂O that was thrown into the original solution, we divide the number of moles of ^-OH (which we found to be approx. 3.55, but use exact values until the end!) by 2.
To get the mass of Na₂O, multiply this value with the molar mass of Na₂O, which is 61.979 grams / mole.
The calculation results in 110.215... grams of Na₂O
We round to two significant figures, which yields our final answer, 110 grams of Na₂O

[Nan_Guan_1L]

just to add on to the discussion above,
when I was solving this question myself, I think it's helpful to note that concentration x volume = moles. Also, it's helpful to keep in mind of what stays the same and what changes throughout the questions. For example, the concentration of the 5ml extracted is of the same component as the original solution. yet the moles of OH- ions in the 5ml is not the same as in the original. you would have to use the concentration times the volume to find how many moles of OH- ions are in the original solution for question c.

[Erin Woolmore 1C]

Sorry it's a little messy but here's how to do it. Let me know if you need clarification.

Can anyone explain how the answer to part b is 18M?