posted by Anonymous .
A solution is made by dissolving 26.9 g of NaOH in approximately 450 mL of water in a volumetric flask. The solution becomes quite warm, but after it is allowed to return to room temperature, water is added to total 500 mL of solution. Calculate the pH of the final solution. Report pH to 2 decimal places.
I'm super confused about this one. I got the moles to be .6748 then divided that by .5 to get 1.3496 M. But then when I got the -log it was a negative answer. Why?
Oh, and this one: Calculate the pOH of 1.62 M HI.
I thought it was a simple matter of getting the -log and then subtracting from 14 to get 13.8 but that wasn't right.
just an update: i got the first one but am uber confused about the second.
mols NaOH = grams/molar mass = 26.9/40 = about 0.67 and that divided by 0.5 = about 1.35M
pH = -log 1.35
pH = -(0.13) = -0.13 approximately.
Remember 1.0M = zero pH.
b. And it is that simple. It's that negative pH again.
HI = 1.62M
(H^+) = 1.62M
pH = -log(1.62) = about -0.21
pH + pOH = 14
-0.21 + pOH = 14
pOH = 14-(-0.21) = 14.21.