Please, help me out with this???
How is it done?
2006-08-22 13:56:29 · 5 answers · asked by RED MIST! 5 in Science & Mathematics ➔ Chemistry
The dissociation constant of water is Kw = 1.011 ×10^−14. What this generally means is that [H+] x [OH-] = 1.011 x10^-14. Since you have a [H+] of 3 x10^-4, simple math yields an [OH-] of 3.37 x10^-11. No log tables or complicated procedures necessary.
2006-08-22 14:10:30 · answer #1 · answered by Doctor Why 7 · 1⤊ 0⤋
Take the –log[H+] to get pH. Subtract the pH from 14, which gives you the pOH and then take 10^-pOH
Or another easier way would be to divide the Kw of water (1 x 10^-14) by the [H+] because [H+][OH-]=Kw
2006-08-22 14:07:23 · answer #2 · answered by Anonymous · 0⤊ 0⤋
3x10^-11
The last guy had it on the nose, but you would be able to round off to 3x10^-11 in most high school classes.
2006-08-22 14:50:15 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Keep in mind the [OH-] and [H+] are in balance and pH = -log([H+]). Namely, pH + pOH = 14.
So, calc pH, find pOH, and take the antilog.
2006-08-22 14:04:07 · answer #4 · answered by ChemDoc 3 · 0⤊ 0⤋
U WILL HAVE TO USE LOG TABLE
2006-08-22 14:00:24 · answer #5 · answered by aamir_azad30 1 · 0⤊ 2⤋