I know this is a very basic problem, I just wasn't sure how to calculate it with 2 different molarities in different proportions, if someone could explain how to quickly solve this it would be greatly appreciated. Thanks.
2006-10-01 16:12:41 · 2 answers · asked by stylesofbeyond40 1 in Science & Mathematics ➔ Chemistry
regardless of the different acids, IF this is a problem for very basic level chemistry, then treat both acids the same because they are both monoprotic.
so you basically have a solution that is 3 parts 0.4 M H+ conc and one part 0.1 M H+ conc...
(3(0.4) + 1(0.1)) / 4 = 0.325 M
thus ph = -log[H+] the rest is straightforward...
Now if you take in consideration the activity coefficient, then this gets a little messy since ph is truly -log(activity of H+) then the hydrated radius plays a part... which is kinda complicated
2006-10-01 19:08:17 · answer #1 · answered by kb27787 2 · 0⤊ 0⤋
photo pink 1
2006-10-01 23:20:51 · answer #2 · answered by Lady E 1 · 0⤊ 0⤋