[H+] = Kw / [OH-] = 1.0*10-14 / 2.5*10-4 = 4.0*10-11 mol/L
solution with [OH-] = 2.5 x 10-9 , A solution with [H+] = 1.2 x 10-4, A solution with pH = 4.5
Na+ plus OH- plus H+ equals H2O plus Na+ plus Cl-
If -h = 14, then ... h = -14
(OH- is a base) (H+ is an acid) Therefore by adding water to HSO3, the OH- ion is produced therefore it is an Arrhenius base.
[H+] = Kw / [OH-] = 1.0*10-14 / 2.5*10-4 = 4.0*10-11 mol/L
solution with [OH-] = 2.5 x 10-9 , A solution with [H+] = 1.2 x 10-4, A solution with pH = 4.5
Na+ plus OH- plus H+ equals H2O plus Na+ plus Cl-
h = 0.538462 14 + 5h + 2h = 5h + 28h 14 + 7h = 33h 14 + 7h - 7h = 33h - 7h 14 = 26h 14/26 = 26/26h 0..538462 = h
pH= -log[H+] pH + pOH = 14 pOH = 14 - pH pOH= -log[OH], so the antilog of -pOH will give you the OH concentration.
If -h = 14, then ... h = -14
(OH- is a base) (H+ is an acid) Therefore by adding water to HSO3, the OH- ion is produced therefore it is an Arrhenius base.
H+ or a proton.
Think of water as HOH, which is basically a H+ ion and an OH- ion. So then, in solution, the reaction looks like this: Na+ + OH- + H+ + Cl- ----> Na+ + Cl- + H+ + OH- and then if we put the ions together, we get NaOH + HCl ----> NaCl + H2O.
That will be HOH
The concentration of OH- decreases as the concentration of H+ increases. This is beacause there is an equilibrium H2O <-> H+ + OH- and therefore the [H+][OH-] is a constant
pH = - log([H+]) , pOH = - log([OH-] , pH + pOH = 14 [X] = concentration of X