00:01
First, we recognize that hypochloric acid, hclo, reacts in a one -to -one fashion with potassium hydroxide in order to create potassium hypochlorite and water.
00:18
The k -a value for hypochloric acid was not provided, but we can look it up.
00:24
In the table i have, hypochloris acid has a value of 3 .5 times 10 .5.
00:32
To the negative 8.
00:37
So as we add moles of potassium hydroxide, we consume the hypochlorous acid and make the hypochlorite.
00:46
Every mole of sodium hydroxide we add consumes a mole of hypochloric acid and creates a mole of hypochloride.
00:56
So if we want to create a buffer with the hypochloric acid and the hypochlorite that will result from the chemical reaction, we can use the henderson -hasselball equation to calculate or to relate ph to the concentrations.
01:15
We want to prepare a buffer that has a ph of 7 .210.
01:21
According to the henderson -hasselbalch equation, that'll be equal to p -ka, which will be the negative log of the k .a value for hypochloric acid 3 .5 times 10 to the negative 8, plus the log of, it could be concentration base over concentration acid or moles base over moles acid, and it's going to be easier to use moles in this case.
01:46
So it'll be the moles of the base hypochlorite, which will be equal to the moles of the strong base added potassium hydroxide, divided by the moles of acid.
01:59
We're starting with 150 milliliters, that's 0 .150 liters, liters of hypochloric acid at a concentration of 0 .459 moles per liter.
02:13
Multiplying these two together gives us the moles of hypokloric acid we start with before adding the potassium hydroxide.
02:23
We then subtract off x, which will be the moles of potassium hydroxide we add.
02:28
Every mole that we add consumes a mole of hypochlorite, so we subtract it off, but consumes a mole of hypochloric acid, but makes a mole of hypochlorite.
02:39
So we add it to the numerator.
02:41
Now we can solve for x...