Water ionizes to a very small degree into hydronium ions, H3O+ ,
and hydroxide ions , OH-: 2H2O(l) reverse arrows H3O+(aq) + OH-(aq)
Kw = (H3O+) (OH-) = 1.00x10 -14
The very small value of the equilibrium constant , Kw, should
give you an appreciation for how few water molecules actually
ionize in pure water.
. In neutral solutions, (H3O+) = (OH-)
. In basin solutions, (H3O+) < (OH-)
. In acidic solutions , ( H3O+) > ( OH -)
But in all aqueous solutions, the product of the hydronium and
hydroxide concentrations is equal to Kw. This, Kw allows you to
calculate (H3O+) from (OH-) , or visa versa.
Part A- 5.00 x 10 -3 million of HBr are dissolved in water to
make 14.0 L of solution. What is the concentration of hydroxide
ions, (OH-) , in this solution?
(OH-) =________value, units
Part B- 6.00 g of NaOH are dissolved in cold water to make
5.00L of solution. What is the concentration of hydronium ions,
(H3O+) , in this solution?
(H3O+) =________value, units