00:01
To calculate the ph of the buffer solution after the addition of sodium hydroxide, we recognize the sodium hydroxide, represented as hydroxide, is going to react with the base in the system, that'll be the formate, hcoo-, and make a little bit more formic acid and water, the stoichiometry being one to one.
00:27
So as long as we don't add more sodium hydroxide than we have of the formate, we still have a buffer solution, and ph can be calculated using the henderson -hasselbalch equation, where ph equals pka plus the log of the moles of the formate divided by the moles of the formic acid.
00:53
Now sometimes people like to use a ratio of molarities, but in most cases it's easier to use a ratio of moles, so that's what i'll do here.
01:01
So the new ph is going to be equal to pka.
01:08
Pka will be the negative log of the ka value, but when we have a buffer that has equal concentrations of weak acid and conjugate base, ph equals pka, they tell us that when we had equal concentrations, 0 .5 of formate and formic acid, the ph was 3 .77, so that then would also be the pka value.
01:36
We'll then add to that the log of the moles of formate...