Question

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?

Answer 1

[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 !

Answer 2

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.

Answer 3

Sorry. It's still early for me. DrBob222 came out DrGog222.