00:01
When we add sodium hydroxide to a solution that contains a weak acid, such as acetic acid, which i'll just represent as ha, it produces the conjugate base of the weak acid and water.
00:16
As long as we don't add more of the sodium hydroxide than we have of the weak acid, we'll still have weak acid, but we would have made some weak base.
00:27
Thus, we have a buffer solution.
00:29
To calculate the ph of a buffer solution, we use the henderson -hasselbalch equation.
00:35
Ph equals pka, which will be the negative log of the ka value for acetic acid, which they didn't give us, but in the table i have, it's 1 .8 times 10 to the negative 5.
00:49
We then add to that the log of the moles of the base formed, divided by the moles of the acid still left in solution.
01:00
Sometimes people will use a ratio of molarities rather than a ratio of moles, but in most cases it's easier to use a ratio of moles...