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# A2 Chem: 0.3M of NaOH. calculate the pH when Kw = 1x10^14 at 298K watch

1. 0.3M of NaOH.

Calculate the pH when Kw = 1x10^14 at 298K.

I dont get how to do this.

I know that Kw = [H+] x [OH-]

But since I only know the conc. of the NaOH, I dont understand how to use that identity above to work out the pH of NaOH.

Or am I being stupid and confusing myself over information that has no bearing on the answer?

Thanks
2. use pOH = -log [OH-] then pH + pOH = 14
3. (Original post by EierVonSatan)
use pOH = -log [OH-] then pH + pOH = 14
eh?

surely its pH = -log(OH), so pH = 0.522

just like it is for acids?
4. (Original post by 2strong)
eh?

surely its pH = -log(OH), so pH = 0.522

just like it is for acids?
nope:

potential of hydrogen pH = -log [H+]

potential of hydroxide pOH = -log [OH-]
5. (Original post by EierVonSatan)
nope:

potential of hydrogen pH = -log [H+]

potential of hydroxide pOH = -log [OH-]
what!?

wait hang on.

If there is 1.4 M of H2SO4, what is its pH? -0.442 right?
6. (Original post by 2strong)
what!?

wait hang on.

If there is 1.4 M of H2SO4, what is its pH? -0.442 right?
yeah (well -0.447)
7. (Original post by EierVonSatan)
yeah (well -0.447)
Ok, so how is the method I used to get that answer different to the method I use to get the answer to the question in the opening post?
8. (Original post by 2strong)
Ok, so how is the method I used to get that answer different to the method I use to get the answer to the question in the opening post?
You want to find the pH, and you're given the concentration of NaOH (a strong base so can assume [NaOH] = [OH-]). You cannot find the pH directly from this and so you are given that Kw = [OH-][H+] = 10-14.

So, firstly find the pOH then use pH + pOH = 14 to find the pH.

If pH = -log [OH] then > pH = 0.55 is a bit low for a strong base
9. (Original post by 2strong)
0.3M of NaOH.

Calculate the pH when Kw = 1x10^14 at 298K.

I dont get how to do this.

I know that Kw = [H+] x [OH-]

But since I only know the conc. of the NaOH, I dont understand how to use that identity above to work out the pH of NaOH.

Or am I being stupid and confusing myself over information that has no bearing on the answer?

Thanks
pOH = 13.48
pH = 0.52

Add them together to get a pH of 14.

No?

PS: It's 1 x 10^-14
10. (Original post by EierVonSatan)
You want to find the pH, and you're given the concentration of NaOH (a strong base so can assume [NaOH] = [OH-]). You cannot find the pH directly from this and so you are given that Kw = [OH-][H+] = 10-14.

So, firstly find the pOH then use pH + pOH = 14 to find the pH.

If pH = -log [OH] then > pH = 0.55 is a bit low for a strong base
The underlines pH and the boldend pH are different arnt they?
11. (Original post by 2strong)
The underlines pH and the boldend pH are different arnt they?
The bottom line is your suggestion, and is wrong

This is how you do it;

pOH = -log (0.3) = 0.523

pH = 14 - 0.523 = 13.5

so pH = 13.5

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