By equilibrium only in water:
Ionconcentration product = KW ,
meaning:
[H30+] * [OH-] = 1.0*10-14 (at 25oC)
H30+(aq) + OH-(aq) <==>> H2O(l)It is the aqueous solution of Acetic acid.
H3O is a strong acid.
H30+(aq) + OH-(aq) ----> H2O(l)
H3O= 0.9 mol/dm3 OH=1.2 mol/dm3
H3o+
If the PH of lemon juice at 298 k is found to be 2.32, the concentration of H3O plus ions in the solution would be 0.5 M.
In this reaction H3O+ is the conjugate acid. The original acid in this reaction is H3PO4
dissociation of acid in water: A + H2O <-> A- + H3O+ with dissociation constant Ka = [A-][H3O+]/[A][H2O] = [A-][H3O+]/[A]. dissociation of base in water: B + H2O <-> HB+ + OH- with dissociation constant Kb = [HB+][OH-]/[B][H2O] = [HB+][OH-]/[B] dissociation of water in itself: 2H2O <-> H3O+ + OH- with dissociation constant Kw = [H3O+][OH-]/[H2O]^2 = [H3O+][OH-] where [H2O] has been ommitted because it is a pure liquid. substituting relations for Ka and Kb into Kw gives: Kw = [H3O+][OH-] = (Ka[A]/[A-])(Kb[B]/[HB+]) = KaKb where [A] = [HB+] and [B] = [A-].
H3O= 0.9 mol/dm3 OH=1.2 mol/dm3
Cu+ H2O [OH + H3O= 2H2O]Copper plus more than one water = [CuOH + H3O]
Yes.
It must be 1 x 10-9
In neutral solutions, [H3O+] = [H2O].In bases, [OH-] = [H3O+].In bases, [OH-] is greater than [H3O+].In acids, [OH-] is greater than [H3O+].In bases, [OH-] is less than [H3O+].
H3o+
If the PH of lemon juice at 298 k is found to be 2.32, the concentration of H3O plus ions in the solution would be 0.5 M.
In this reaction H3O+ is the conjugate acid. The original acid in this reaction is H3PO4
H3O is a strong acid.
dissociation of acid in water: A + H2O <-> A- + H3O+ with dissociation constant Ka = [A-][H3O+]/[A][H2O] = [A-][H3O+]/[A]. dissociation of base in water: B + H2O <-> HB+ + OH- with dissociation constant Kb = [HB+][OH-]/[B][H2O] = [HB+][OH-]/[B] dissociation of water in itself: 2H2O <-> H3O+ + OH- with dissociation constant Kw = [H3O+][OH-]/[H2O]^2 = [H3O+][OH-] where [H2O] has been ommitted because it is a pure liquid. substituting relations for Ka and Kb into Kw gives: Kw = [H3O+][OH-] = (Ka[A]/[A-])(Kb[B]/[HB+]) = KaKb where [A] = [HB+] and [B] = [A-].
pH=10, means the concentration of OH- ions is 0.0001 M and concentration of H+ ions is 0.0000000001M
[H3O+][OH-] = Kw = 1x10^-14[OH-] = 1x10^-14/0.0034 = 1x10^-14/3.4x10^-3[OH-] = 2.9x10^-12 M