posted by Gerald on .
A 0.1 M solution of acetic acid is titrated with 0.05M solution of NaOH. What is the pH when 60% of the acid has been neutralized? The equilibrium constant (Ka) for acetic acid is 1.8x10^-5
2. Chemistry question:
On average, how far is a molecule of air in the room in which you're sitting from the nearest molecule of air to it, assuming it is an ideal gas. Appropriate assumptions about the temperature and pressure.
I think it is easier if you assume some arbitrary number for the volume of acetic acid. Something like 100 mL.
moles acetic acid = M x L.
moles NaOH when 60 mL acetic acid have been neutralized = M x L.
Then use the Henderson-Hasselbalch equation to solve for pH.
2. I would assume a volume for the room, use PV = nRT to determine the number of moles at some P and T, convert to number of molecules and go from there. Another approach is to take a volume of 1 L, you know the density of air is about 1.29 g/L, convert to molecules/L, then take the cube root to convert to molecules along one side and convert to distance between molecules.