00:01
When we add the sodium hydroxide, it's the hydroxide of sodium hydroxide that ends up reacting with the hocl, the hypochlorous acid, to produce hypochlorite, ocl-, and water.
00:19
As long as we don't add more strong base than we have of the acid, we will still have a buffer solution, and we can use the henderson -hasselbalch equation to calculate ph, where ph equals pka plus the log of the moles of the base, hypochlorite, still in solution, divided by the moles of the hocl in solution.
00:45
Now a lot of people will use a ratio of molarities here rather than a ratio of moles, but a ratio of moles is equivalent, and it often makes the calculation easier, so that's what i will be using.
00:58
So ph will be equal to pka, which will be the negative log of the ka value, 3 .5 times 10 to the negative 8, plus the log of the moles of hypochlorite.
01:16
That'll be equal to the moles we start with.
01:19
We started with 100 milliliters, or 0 .100 liters...