Hey! This is my answer to this question which is correct due to the answer key. But after a while I thought that there was something wrong with it cause the total volume will be 0.9+0.1=1 lit… but the problem is that in this case the ph will be 13 which is not logical; Cause the initial ph of the solution is 12 and after its mixture with the water it should have a lower ph … so 13 can’t be the answer.
Hey,I don’t quiet get it how you solved it, but the way I would solve this question would be like this.
With this formula ,we will end up with two different answers and according to that it need to be logically true,adding water decreases the ph number, as a result ,13 can not be an option as you said so. then i believe 11 should be the answer.
The formula
delta pOH =log V2/V1
PH1 =12 14-12=2
PH2=X 14-X
@Yasy@dorsa_vaezi
To solve it properly you would need a calculator and then find the moles of NaOH to get pH 12 in 100ml of solution, then change the volume by adding the water to the volume, but we don’t need to do that here as we can think about the solution logically.
Assuming that this experiment is done under standard conditions (which we can always assume is unless stated otherwise) The pH of water is 7, and the pH of our solution is 12. This means that if we mix them together, we will get a pH bigger than 7 but smaller than 12, so the answer must be C.
I think you had a mistake in your calculation, the correct calculation would be to find the moles in 100cm^3 of NaOH of pH 12, then dilute it in the total volume.
But remember you don’t need to calculate anything to solve this answer.
@AriHoresh
Thanks for your wonderful solution💫
My calculation was this but I just did it in the
way that my high-school teachers have told us…
100cm3=0.1 lit
10^-12 × 0.1 = 10^-13 moles of the NaOH
900cm3=0.9 lit
The total volum of the solution: 0.9+0.1=1 lit
Finding the concentration of the new solution: 10^-13/1 = 10^-13
Finding the PH: Log10^-13=13 (can’t be the answer)
So what’s wrong with it?
You can’t calculate them together. It’s a base, not an acid. Because we are calculating the pH, you would need to find the moles of NaOH in the solution, then assume the [H+] moles based on how much base you have (base/acid ratio), and only then dilute it in the total volume. You missed these steps and calculated the [OH-] by mistake.
