Question

What concentrations of H⁺ and OH^- are produced by H₂O dissociation in 0.01 M NaOH?

Short answer

The concentrations of H⁺ and OH^- are produced by H₂O dissociation in 0.01 M NaOH are 10^-12M and 0.01 M respectively.

Step-by-step solution

Introduction

NaOH is a strong base which is highly soluble in water. Dissociation of NaOH in water gives the following reaction

$NaOH\stackrel{H_{2}O}{\to }N{a}^{+}+O{H}^{-}$

Here it is asked to findconcentrations of H⁺ and OH^- are produced by H₂O dissociation in 0.01 M NaOH.

Determine the concentration

As NaOH is a strong base therefore it can be said that the concentration of OH^- in the solution will be 0.01 M.

Now the concentration of H⁺ will be

$\begin{array}{rcl}\left[H^{+}\right]\left[O{H}^{-}\right]& =& K_w\\ K_w& =& {10}^{-14}\\ \left[H^{+}\right]\left[O{H}^{-}\right]& =& {10}^{-14}\\ \left[H^{+}\right]\left[0.01\right]& =& {10}^{-14}\\ \left[H^{+}\right]& =& \frac{{10}^{-14}}{0.01}\\ \left[H^{+}\right]& =& {10}^{-12}M\\ & & \end{array}$