00:01
First, we need to recognize that when we add the koh, the hydroxide of the koh will react with the acetic acid, which i'll represent as ha, producing more acetate and water.
00:14
As long as we don't add more koh than we have of acetic acid, we will still have ha and its conjugate base, so we'll have a buffer solution.
00:25
And ph can be calculated for a buffer solution using the henderson -hasselbalch equation, where ph equals pka plus the log of the moles of the weak base divided by the moles of the weak acid.
00:42
A lot of people will do molarity base over molarity acid, but it's easier in most cases to do a ratio of moles rather than a ratio of molarities.
00:52
So we want to achieve a ph of 6 .43, knowing that acetic acid has a pka value of 4 .76, so we'll add to that the log of the moles of acetate formed will be equal to the moles of koh added.
01:14
Every mole of koh we add creates a mole of acetate, so that we need to figure out, how many moles of koh we need to add.
01:24
We'll then divide that by the moles of acetic acid still left in solution, which will be the volume of the acetic acid, 830 milliliters or 0 .830 liters at a concentration of 0 .743 moles per liter.
01:44
But every mole of hydroxide we add consumes a mole of ha, so we need to subtract off x to get that ph.
01:53
Now we solve for x, we take 6 .43 and we subtract 4 .76 from both sides and we get 1 .67 equal to the log of x divided by 0 .83 times 0 .743 or 0 .6167 minus x...