Chem
posted by Lena on .
The question is telling me that NaOH (24.00 grams) is diluted in water (750 mL). Find the pH and pOH. This is what I did so far:
NaOH > Na+ + OH
NaOH:
mass = 24.00 grams
v= 750 ml = 0.750 L
M= 40.00 grams/mol
n= 0.6000 mol
c= 0.800 mol/L
Since NaOH dissociates 100%, the concentration of OH = the concentration of NaOH. Therefore,
pOH = log [0.800]
= 9.69 x 10^2
[pH][pOH] = 1 x 10^14
pH = 1 x 10^14 / 9.69 x 10^12
= 1.03 x 10^3
Did I do this correctly?

[pH][pOH] = 1 x 10^14
pH = 1 x 10^14 / 9.69 x 10^12
No,
pH = 1 x 10^14 / 9.69 x 10^2
= 1.03 x 10^13
about Ph of 13, which is about 4% NaOH
You know the Ph has to be much higher than 7, nearly 14, because NaOH is a BASE ! 
The problem here, Lena, is that
[pH][pOH]=1 x 10^14 is not true.
Two errors.
pH + pOH = pKw = 14 which is the log form. If you wish to use the regular form it is
or (H^+)(OH^)=Kw = 1 x 10^14
You mixed the two. 
Sorry. It's still early for me. DrBob222 came out DrGog222.