The question requires us to calculate the concentration, in mol/L, of H3O+ and OH- ions for a 0.00016 M NaOH solution.
Since NaOH is a strong base, we can consider that it completely dissociates in water:
[tex]NaOH_{(aq)}\to Na^+_{(aq)}+OH^-_{(aq)}[/tex]From the dissociation reaction shown above, we can say that the concentration of OH- ions in a NaOH solution will be the same as the concentration of NaOH.
Thus: [OH-] = 0.00016 mol/L
Now, we can apply the ion-product constant of liquid water (Kw) to calculate the concentration of H3O+ ions:
[tex]\begin{gathered} K_{w_{}}=\lbrack H_3O^+\rbrack\times\lbrack OH^-\rbrack \\ \lbrack H_3O^+\rbrack=\frac{K_w}{\lbrack OH^-\rbrack} \end{gathered}[/tex]Since the value of [OH-] is 0.00016 M and considering the value of Kw as 1.00 x 10^-14, we calcuate [H3O+]:
[tex]\lbrack H_3O^+^{}\rbrack=\frac{1.00\times10^{-14}}{0.00016}=6.3\times10^{-11}mol/L[/tex]Therefore, the concentration of [OH-] and [H3O+] ions are 1.6 x 10^-4 and 6.3 x 10^-11 M, respectively.
