Chapter 18, Problem 18.32P

Interpretation Introduction

Interpretation:

The number of moles has to be calculated for $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{ or [OH}}^{\text{-}}\text{]}$ that is added to a liter of strong base solution to adjust its $\text{pH}$ from $\text{9}\text{.33 to 9}\text{.07}$.

Concept introduction:

A solution may be acidic and basic depends on relative strength of anions as an acid or base. Anions and cations can be acidic, basic and netural. Acidic have $\text{pH < 7}$, basic have $\text{pH > 7}$ and netural have $\text{pH=7}$.

The $\text{pH}$ is decreases when the solution is more acidic (less basic) similarly the $\text{pH}$ is increase when the solution is more basic (less acidic). Therefore ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is added to decrease the $\text{pH}$.

One mole of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ion reacts with one mole ${\text{OH}}^{\text{–}}$ ion, the difference in ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ will be equal to the ${\text{OH}}^{\text{–}}$ added.

The Relationships among $\text{pH, pOH, and pKw}$

  $\begin{array}{l}\text{pH = -}\text{log}{\text{(H}}_{\text{3}}{\text{O}}^{\text{+}}\text{)}\\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]={10}^{-\text{pH}}\\ \text{pOH = -}\text{log}{\text{(OH}}^{-}\text{)}\\ \left[{\text{OH}}^{-}\right]={10}^{-\text{pOH }}\end{array}$

  ${\text{pK}}_{\text{w}}\text{ = pH + pOH = 14}\text{.00 }$