00:01
If we want a 0 .1 molar acetate buffer, then that means we want 0 .1 molar as the total concentration of both acetate and acetic acid.
00:11
So what we need to do is take the acetic acid that's a 0 .1 molar and the sodium acetate that's a 0 .1 molar and mix them in such a ratio that we get a liter with the total concentration that's still 0 .1 molar, but the ph is now 4.
00:28
We'll use the henderson -hasselbalch equation where ph is equal to p -k -a plus the log of the moles of base, so the moles of acetate, over the moles of the acid acid.
00:48
So we need to take x liters at 0 .10 molar of acetate, and then we need to take, because the total volume, will be 1 liter, and we'll assume the volumes are additive.
01:09
We'll take 1 minus x of the acetate at a concentration of 0 .1 molar, and now we solve for x.
01:20
If we subtract this from both sides, and then take the anti -log -base 10, we get 0 .17378 is equal to, taking the anti -log base 10 of this, just leaves us with the ratio...