2007-01-18 04:25:23 · 14 answers · asked by Lisa 1 in Science & Mathematics ➔ Chemistry
yes.
the bronsted-lowry definition of an acid is that it is a proton (H+) donor, meaning it gives off H+ in water. so when it's neutralized, the H+ given off neutralizes the base, forming water. Bases are proton acceptors, and therefore will add the H+ to create water as a product.
that's why the OH- will accept the H+ to create H20
2007-01-18 04:31:16 · answer #1 · answered by anniekel 2 · 0⤊ 0⤋
Yup, that's what happens, H+ combines with OH- to form water, then the other two (or more) elements combine to form a salt. One common example is:
HCl+NaOH --> HOH +NaCl
Hydrochloric Acid + Sodium Hydroxide --> Water + Sodium Chloride.
2007-01-18 04:46:41 · answer #2 · answered by Mr. Chemistry 2 · 0⤊ 0⤋
Yes you are right on track
We say neutralisation because the acid loses the H+ ion and the base loses its OH-.... and a neutral product ie H2O is obtained...
2007-01-18 04:31:48 · answer #3 · answered by Think 2 · 0⤊ 0⤋
yes! try any reaction!
HCl + KOH -----> KCl + H2O
see, teh hydroxl group of the base combines with the hydrogen in the acid to form water and a salt!
2007-01-18 04:30:04 · answer #4 · answered by uniciron 2 · 0⤊ 0⤋
Yes it's correct..
H+ and OH- combines to form water (H2O)
the two can neutralise each other by this reaction
2007-01-18 05:09:23 · answer #5 · answered by Plasma 1 · 0⤊ 0⤋
Yes. The hydroxide and hydronium ions combine to make water.
2007-01-18 04:30:54 · answer #6 · answered by borscht 6 · 0⤊ 0⤋
acid + base yields salt + water
2007-01-18 04:55:36 · answer #7 · answered by shiara_blade 6 · 0⤊ 0⤋
YES, plus the other components mix to form a "salt" (whether true table salt [NaCl] or otherwise, whatever is being mixed together).
2007-01-18 04:30:52 · answer #8 · answered by Anonymous · 0⤊ 0⤋
Absolutely correct.
2007-01-18 04:27:56 · answer #9 · answered by Keith P 7 · 0⤊ 0⤋
yes that's tru but not always
for example amonia solution doesn't have the 0h-
2007-01-18 04:31:19 · answer #10 · answered by T-bag 3 · 0⤊ 0⤋