Chapter 9, Problem 9.AE

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $\text{1}{\text{.0×10}}^{\text{-2}}\text{M}\text{NaOH}$ has to be calculated.

Concept Introduction:

The relationship between $\text{pH}\text{and}\text{pOH}$ is given as,

${\text{pH+pOH=-log K}}_{\text{w}}$

$\text{pH+pOH=14}$

The $\text{pH}$ of solution can be calculated as,

But,

$\begin{array}{l}{\text{A}}_{{\text{H}}^{\text{+}}}{\text{+A}}_{{\text{OH}}^{\text{-}}}{\text{=K}}_{\text{w}}\\ \\ {\text{A}}_{{\text{H}}^{\text{+}}}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{A}}_{{\text{OH}}^{\text{-}}}}\end{array}$

$\begin{array}{l}{\text{pH=-logA}}_{{\text{H}}^{\text{+}}}\\ \\ \text{pH=-log}{\left[{\text{H}}^{\text{+}}\right]}_{{\text{γH}}^{\text{+}}}\end{array}$

With the above equation, the negative logarithm of activity of Hydrogen ion can be measured.

To calculate the $\text{pH}$ of $\text{1}{\text{.0×10}}^{\text{-2}}\text{M}\text{NaOH}$

Answer to Problem 9.AE

The $\text{pH}$ of $\text{1}{\text{.0×10}}^{\text{-2}}\text{M}\text{NaOH}$ is found to be $\text{11}\text{.95}$ .

Explanation of Solution

Molarity of $\text{NaOH}$ = $\text{1}{\text{.0×10}}^{\text{-2}}\text{M}$

$\left[{\text{OH}}^{\text{-}}\right]$ = $\text{1}{\text{.0×10}}^{\text{-2}}\text{M}$

${\text{γ}}_{{\text{OH}}^{\text{-}}}\text{=0}\text{.900}$

$\text{μ=0}\text{.0100}\text{M}$

$\text{pH}$ can be calculated as,

$\begin{array}{l}{\text{A}}_{{\text{H}}^{\text{+}}}\text{=}\frac{{\text{K}}_{\text{w}}}{{\left[\text{OH}\right]}^{\text{-}}{\text{γ}}_{{\text{OH}}^{\text{-}}}}\\ \\ {\text{A}}_{{\text{H}}^{\text{+}}}\text{=}\frac{\text{1}{\text{.0×10}}^{\text{-14}}}{\left(\text{1}{\text{.0×10}}^{\text{-2}}\right)\left(\text{0}\text{.900}\right)}\\ \\ {\text{A}}_{{\text{H}}^{\text{+}}}\text{=1}{\text{.11×10}}^{\text{-12}}\\ \\ \text{pH=-log}\left[{\text{A}}_{{\text{H}}^{\text{+}}}\right]\\ \\ \text{pH=11}\text{.95}\end{array}$

The $\text{pH}$ of $\text{1}{\text{.0×10}}^{\text{-2}}\text{M}\text{NaOH}$= $\text{11}\text{.95}$

Conclusion

The $\text{pH}$ of $\text{1}{\text{.0×10}}^{\text{-2}}\text{M}\text{NaOH}$ was calculated using ${\text{A}}_{{\text{H}}^{\text{+}}}$ and the $\text{pH}$ of $\text{1}{\text{.0×10}}^{\text{-2}}\text{M}\text{NaOH}$ was found to be $\text{11}\text{.95}$ .