[Chemistry] [End of tether] Calculating PH

[Jeros]

this is driving me nuts now

I have some questions on calculating PH for a assingment im doing, i must have been ill the day we where doing this in class (im on an access course, 2 years of a level basics wodged into 6 months) but im really stuck on these two

calculate the PH of
0.3 moldm-3 of NaOH
0.8 moldm-3 of KOH

can anyone help or nudge me in the right direction

 

[Zenith.UK]

Okay, start with the easy stuff.
pH is a negative logarithmic scale of hydrogen ion concentration in a solution. If you're putting OH- ions in solution, you're decreasing the number of free H+ ions in solutions as they react together to form water.
Neutral pH has 1E-7 H+ ions and 1E-7 OH- ions in solution, 1E-14 in total.

Calculate pH for 0.3M NaOH:
NaOH is strong base and completely ionises in solution. Therefore, [OH-] = 0.3M
[H+][OH-] = 1E-14
[H+][0.3] = 1E-14
[H+] = (1E-14) / 0.3 = 3.333E-14
pH = -log[H+] = log[3.333E-14] * -1 = 13.48

Calculate pH for 0.8M KOH:
Like NaOH, KOH is strong base and completely ionises in solution. Therefore, [OH-] = 0.8M
[H+][OH-] = 1E-14
[H+][0.8] = 1E-14
[H+] = (1E-14) / 0.8 = 1.25E-14
pH = -log[H+] = log[1.25E-14] * -1 = 13.9

That makes sense as you've got a stronger alkali for the second calculation and so you have a higher pH number. FYI, 1.0M NaOH solution is pH 14.
Hopefully someone else can double check me, but I reckon that's right.

 

[Cadelin]

this is driving me nuts now

I have some questions on calculating PH for a assingment im doing, i must have been ill the day we where doing this in class (im on an access course, 2 years of a level basics wodged into 6 months) but im really stuck on these two

calculate the PH of
0.3 moldm-3 of NaOH
0.8 moldm-3 of KOH

can anyone help or nudge me in the right direction

The first thing you need to convert this into Moles per litre, which requires you to divide it by 1000. If you assume that NaOH completely ionizes, you have 3.0 x 10^-4 M of OH- ions. (similarly for KOH)

To convert this into H+ ions you need to use:
[H+] . [OH-] = kw
where kw = 1.0 x 10^-14

You can then convert this into a pH using:
pH = - log [H+]


Edit: Just saw Zenith answer: We agree with the calculation except for the units. The concentration should be per litre and the units you have given are per m-3, so you should divide by 1000 (which on a log scale is 3). You should check the units given.

I get about pH 10.5 for NaOH and pH 10.9 for KOH.

 

[Zenith.UK]

Curses!
I KNEW there was something missing but I couldn't see what it was.
Thanks for putting me straight Cadelin!