The pH of an aqueous solution of sodium hydroxide is 12.9. What is the molarity of the solution?

Question

I need an answer to this question, but could you PLEASE provide an explanation? I really need to understand this!

Thanks!

To the first answerer, are you sure that's it? It's asking for the molarity, not the concentration. I know they're both measured in molarity but usually when it asks me for the H+ concentration it specifies it. I have an answer key but it's useless to me if I don't know how to work the problem. I know the answer is .08M, but why?

2nd answerer:

HOLY MOLEY THAT EXPLAINS IT ALL!! THANK YOU SO MUCH~!!!

so simple... yet so hard.....

Answer 1

pH=-log[H+]

[H+] =1.258*10^-13mol/dm3

assuming the temp. is 25 degrees C,
therefore Kw is 1*10^-14

Kw=[H+]*[OH-]

[OH-]=1*10^-14/1.258*10^-13
=0.079 mol/dm3
NaOH= Na+ + OH-
1 : 1

therefor molarity of NaOH=0.079 mol/dm3
molality of NaOH=0.079 mol/kg

Answer 2

Well, I'm answering this off the top of my head, but I think it's right. The following equation is what came immediately to mind:

pH = -log[H+]

That means that the pH of the solution, 12.9, is equal to the negative log of the concentration of H+. However, NaOH (Sodium Hydroxide) forms Na+ and OH- in aquous solution, so instead find the pOH:

14 - pH = pOH
14 - 12.9 = 1.1

pOH = -log[OH-]
1.1 = -log[OH-]

Now, just solve for OH- in that equation. Since [OH-] means it is given in terms of molarity, that will be your answer. It is also the molarity of the solution.

1.1 = -log[OH-]
-1.1 = log[OH-]
10^-1.1 = [OH-]
[OH-] = 0.0794

The molarity of NaOH will be the same as the molarity of either Na+ or OH- individually because it dissociates into each equally.

Cheers

Answer 3

If the pH of a solution is 12.9, it is considered very basic. Because pH = -log [H+], to find the molar concentration, you do 10^-ph (this case ph=12.9) to find the concentration of [H+] ions in it.

Sorry, as for the answer, it appears to be asking for the [OH-] ions. In this case, you can find the pOH, which is 14 - 12.9, which is 1.1 - you will find that 10^-pOH (pOH is 1.1) equals 0.08, the same answer as the lady's below.

Answer 4

If pH = 7.6, then pOH = 14-7.6 = 6.4 Because the pH/pOH is very close to 7.00, it is insufficient to calculate: [OH-] = 10^-pOH = 10^-6.4 = 3.98*10^-7 as being the molarity of the NaOH Water dissociates to produce [OH-] = 10^-7, so this [OH-] must be subtracted from the total [OH-] in order to isolate the [OH-] originating from the NaOH [OH-] from NaOH = (3.98*10^-7) - (10^-7) = 2.98*10^-7 The concentration of the NaOH in solution = 2.98*10^-7M

Answer 5

Sha's answer said molality. The answer is numerically correct, and so is the process, but all of the concentrations are molar (moles per liter of solution). This is how pH and pOH are defined. Molality is moles per kg of solvent.

Answer 6

The pH of solution is 12.9 so the pOH of this solution is 14 - 12.9 = 1.1
Moreover, pOH of a solution is equal to -lg[OH-] with [OH-] is molarity of ion OH-. Therefore, 1.1=-lg[OH-] => [OH-] = 10^(-1.1) = 0.0794 M