strong acid+ strong base PH calculation

Can someone please show me how to work through question 1? I should be able to do the rest, I just need some guidance on how to solve it.
Thanks

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image-577308ae-562e-4701-992f-3aaa6e8d98c7-1145865603-compressed.jpg.jpeg

* ive worked out moles for OH and H. minused them to give me the excess OH.
where do I go on from here?

(edited 6 years ago)

Reply 1

Bulletzone

Great!

You've calculated the moles of OH- ions, so you need to work out the concentration of OH- by using:
Moles = concentration x volume (check units!)

Does this look familiar?
Kw = [H+][OH-]

The value of Kw = 1x10^-14 (You don't have to remember this, it's given to you).

Re-arrange above equation to get [H+].

Then finally use pH = -log[H+]

(edited 6 years ago)

Reply 2

Lia22

Original post by Bulletzone

Great!

You've calculated the moles of OH- ions, so you need to work out the concentration by using Moles = concentration x volume (check units!)

Kw = [H+][OH-]
The value of Kw = 1x10^-14 (You don't have to remember this, it's given to you).

Re-arrange above equation to get [H+].

Then finally use pH = -log[H+]

Ah this is helpful, Thank you!

Reply 3

Bulletzone

Original post by Lia22

Ah this is helpful, Thank you!

Absolutely no problem :smile:

Feel free to PM me if you're stuck on anything chemistry related, I'll do my best to help as it is possibly my favourite subject.
It's the only science that makes sense to me. :woo:

Reply 4

Pigster

Instead of using Kw to calculate the pH when an alkali is in excess, try using pOH.

pOH = - log [OH-]

pKw = - log Kw (at 25 oC, i.e. when Kw = 1.00 x 10¹⁴, pKw = 14.00)

pH = pKw - pOH, i.e. = 14.00 - pOH