## 6B 11

ashwathinair
Posts: 113
Joined: Sat Aug 17, 2019 12:17 am

### 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?

When going over the solution for this question, I saw that the amount of liquid the solution was diluted to was divided by the amount of solution used in the dilution (500 ml/5 ml) whereas I believe in other questions it was written the opposite way where the diluted amount was dividing the amount of solution used in dilution (5 ml/500 ml). Can someone explain why the calculation was written this way?

Also, why is the [OH-] molarity equivalent to the molarity of [NaOH]?

nicolely2F
Posts: 149
Joined: Sat Sep 14, 2019 12:17 am

### Re: 6B 11

Since I don't know which exercises you're referring to (when you mention that some exercises do 5/500 instead of 500/5), I'm not sure if there's a particular reason; but it's just a matter of finding a ratio of solvent between the original and the diluted, and as long as you know the magnitude of difference (1:100), either result (10^-2 or 10^2) is ok if you know how to apply it to the calculations that follow. I hope this made sense, let me know if you want clarifications.

The [OH-] molarity is the same as the [NaOH] molarity because in the dissociation reaction (NaOH -> Na+ + OH-) NaOH and OH- have the same ratio, i.e. coefficients.

DHavo_1E
Posts: 118
Joined: Sat Aug 17, 2019 12:17 am

### Re: 6B 11

nicolely2F wrote:Since I don't know which exercises you're referring to (when you mention that some exercises do 5/500 instead of 500/5), I'm not sure if there's a particular reason; but it's just a matter of finding a ratio of solvent between the original and the diluted, and as long as you know the magnitude of difference (1:100), either result (10^-2 or 10^2) is ok if you know how to apply it to the calculations that follow. I hope this made sense, let me know if you want clarifications.

The [OH-] molarity is the same as the [NaOH] molarity because in the dissociation reaction (NaOH -> Na+ + OH-) NaOH and OH- have the same ratio, i.e. coefficients.

Hello,
I am still confused about why they solved for the ratio the way they did. Could you explain this further? Thank you!

nicolely2F
Posts: 149
Joined: Sat Sep 14, 2019 12:17 am

### Re: 6B 11

Yeah of course. I'll give you an example. If I say "A is 100 times bigger than B", it's the same thing as "B is 100 times smaller than A". So what you need to extract from the exercise to solve this part is that the original solution is 100 times more concentrated than the diluted solution ( = the diluted solution is 100 times less concentrated than the original solution). So when you do the math, as long as you don't use 0.01 where you're supposed 100 or vice-versa, you'll get the correct result.

In the context of the exercise, saying that 5mL is 100 times smaller than 500mL is the same thing as saying 500mL is 100 times larger than 5mL. What matters is that there is a 1:100 ratio to the solutions. You need this number to calculate the new concentration: since there is a 1:100 ratio between the old and new volumes, there will be a 1:100 ratio between the old and new concentrations. This tells us that the new concentration is 100*0.18 = 18M. Make sense?