When NaOH is dissolved in water, it dissociates to form Na+ and OH- ions. Since NaOH is a strong base, it dissociates completely in water. The balanced chemical equation for the dissociation of NaOH in water is:
NaOH(s) → Na+(aq) + OH-(aq)
Since the concentration of NaOH given in the problem is 3.0 x 10^-2 M, we know that the concentration of OH- ions in the solution is also 3.0 x 10^-2 M. This is because for every one molecule of NaOH that dissolves, one molecule of OH- is produced.
To find the hydronium ion (H3O+) concentration in the solution, we need to use the fact that water undergoes autoprotolysis to produce equal concentrations of H3O+ and OH- ions in any aqueous solution. The balanced chemical equation for autoprotolysis of water is:
2H2O(l) ⇌ H3O+(aq) + OH-(aq)
At equilibrium, the product of the H3O+ and OH- ion concentrations equals the ion product constant (Kw) for water at the given temperature. At 25°C, Kw = 1.0 x 10^-14.
Since the solution is basic, the concentration of OH- ions is greater than the concentration of H3O+ ions. Thus, we can use the equation:
Kw = [H3O+][OH-]
Substituting the given values, we get:
1.0 x 10^-14 = [H3O+][3.0 x 10^-2]
Solving for [H3O+], we get:
[H3O+] = 1.0 x 10^-14 / 3.0 x 10^-2
[H3O+] = 3.3 x 10^-13 M
Therefore, the hydronium ion concentration in the solution is 3.3 x 10^-13 M.