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moray-eel-bite · June 25, 2015

I made a few assumptions for this.. Ie equal proportions of vinegar and water with a pH of 7. I also more or less treated acetic acid like a strong acid, which isn't true, but works well enough for this example. The reason your calc. Won't work is that ph 2 is 100000 times more concentrated in hydronium ions than ph 7. So of course the end result is much closer to the vinegar than the water (you're only diluting by two-fold). The equation for ph is ph=-log(hydronium ion). Hence the log scale.

June 25, 2015	#42
moray-eel-bite Tomatovillian™ Join Date: Mar 2014 Location: Ontario, Canada Posts: 47	I made a few assumptions for this.. Ie equal proportions of vinegar and water with a pH of 7. I also more or less treated acetic acid like a strong acid, which isn't true, but works well enough for this example. The reason your calc. Won't work is that ph 2 is 100000 times more concentrated in hydronium ions than ph 7. So of course the end result is much closer to the vinegar than the water (you're only diluting by two-fold). The equation for ph is ph=-log(hydronium ion). Hence the log scale.