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.
To find the pH and pOH of a solution, you need to consider the dissociation of NaOH in water. NaOH dissociates completely into Na+ and OH- ions. You have correctly determined the initial concentration of NaOH (0.800 mol/L) by dividing the number of moles (0.6000 mol) by the volume of the solution (0.750 L).
To find the pOH, you can take the negative logarithm of the OH- ion concentration, which is the same as the NaOH concentration since they dissociate 1:1. Therefore:
pOH = -log [0.800]
= 0.097
Now, to find the pH, you can use the relationship between pH and pOH:
pH + pOH = 14
So:
pH = 14 - pOH
= 14 - 0.097
= 13.903
Your calculation of the pOH as 9.69 x 10^-2 is correct. However, when finding the pH, you made a calculation error in your final step. The correct calculation should be:
pH = 14 - pOH
= 14 - 9.69 x 10^-2
= 13.903
Therefore, the correct pH value is approximately 13.903.