It must be 1 x 10-9
Yes.
Dissolving in water = splitting in ionsCH3COONH4 --> CH3COO- + NH4+CH3COO-, acetate is a weak base: CH3COO- + H2O CH3COOH + OH-NH4+, ammonium is a weak acid: NH4+ + H2O NH3 + H3O+Totally in water: CH3COO- + NH4+ CH3COOH + NH3 and 2H2O H3O+ + OH-
The pH is a measure of the concentration of H3O+ in a solution. The lower the pH, the higher the concentration of H3O+. This is because of the way it is defined:pH = - log10 [H3O+]or in other words, the pH is the negative logarithm (in base 10) of the concentration of H3O+.Water, and water-based solutions have a special property: if you multiply the concentration of H3O+ and the concentration of OH-, you always get a constant number, no matter what. Mathematically, that is:[H3O+] * [OH-] = 1 x 10-14This also says the the two concentrations are inverselyproportional. So when one is high, the other has to be low.So, getting back to your question, we know the pH of each solution. From that we know the concentration of H3O+. Again, lower the pH, the higher the concentration of H3O+. And since the concentrations of H3O+ and OH- are inversely proportional, when H3O+ is high, the OH- concentration is low. So which solution has the lowest amount of H3O+? That's the one that has the highest pH, and that will also have the highest concentration of OH-.See the Related Questions for more information about pH, acids and bases.
OH- and H+
it is the concentration of H3O+ times the concentration of OH-
H3O= 0.9 mol/dm3 OH=1.2 mol/dm3
Cu+ H2O [OH + H3O= 2H2O]Copper plus more than one water = [CuOH + H3O]
Yes.
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+].
By equilibrium only in water:Ionconcentration product = KW ,meaning:[H30+] * [OH-] = 1.0*10-14 (at 25oC)H30+(aq) + OH-(aq) > H2O(l)
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