How can I calculate the concentrations of hydrogen and hydroxide ions in a solution of sodium hydroxide?

Find the molar mass of NaOh and use stoichiometry to find moles. Then divide that total by the liters to get a molarity, which will also be your concentration value.

Lancenigo di Villorba (TV), Italy

You played about NaOH (e.g. Sodium Hydroxide) in aqueous solutions

NaOH(aq) ---> Na+(aq) + OH-(aq)

for complete dissociation by ionic way.
Once you used 2.5 g

2.5 / 40 = 6.25E-2 mol

these molar amount is dissoluted to give a 2.0 Liter of solutions

|Na+| = |OH-| = 6.25E-2 / 2.0 = 3.125E-2 M

So, YOU HAVE TO REMEMBER THAT THIS SOLUTE AFFECT
SELF-DISSOCIATION OF WATER

2 H2O(aq) ---> OH-(aq) + H3O+(aq)

A CHEMICAL PHENOMENON OBEYING TO AN EQUILIBRIUM

Kw = |H3O+| * |OH-|

WHICH BECOMES AT T=25°C=88°F

Kw = 1.0E-14 = |H3O+| * |OH-|

SO YOU MAY GIVE

|OH-| = 3.125E-2 M
|H3O+| = 1.0E-14 / 3.125E-2 = 3.2E-13 M
pH = -log(|H3O+|) = 12.5

I hope this helps you.

First determine how many mols of NaOH you have.
n=m/M, m= 2.50 g and M, the molecular mass of NaOH, is 40 g/mol. You have therefore n= 0.0625 mol NaOH
There are as many molecules(mol)  NaOH as are molecules (mol) of OH-. That means you have 0.0625 mol OH- and because there are as many atoms(mol) H as there are molecules (mol) OH- you have 0.0625 mol H.

Remember 1 mol consists of 6.023*10^23 atoms or molecules