Chapter 10, Problem 10.8P

(a)

Interpretation Introduction

Interpretation:

The equation 10-11 for potassium hydrogen phthalate has to be derived.

Concept introduction:

$\left[{\text{H}}^{\text{+}}\right]\text{=}\sqrt{\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}{\text{F+K}}_{\text{1}}{\text{K}}_{\text{w}}}{{\text{K}}_{\text{1}}\text{+F}}}$

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

$\text{pH=}\text{-log}{\left[{\text{H}}^{\text{+}}\right]}_{\text{γ}}{}_{{\text{H}}^{\text{-}}}$

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

Explanation of Solution

$\begin{array}{l}\text{Charge}\text{balance:}\left[{\text{K}}^{\text{+}}\right]\text{+}\left[{\text{H}}^{\text{+}}\right]\text{=}\left[{\text{OH}}^{\text{-}}\right]\text{+}\left[{\text{HP}}^{\text{-}}\right]\text{+2}\left[{\text{P}}^{\text{2-}}\right]\text{(1)}\\ \text{Mass}\text{balance:}\left[{\text{K}}^{\text{+}}\right]\text{=}\left[{\text{H}}_{\text{2}}\text{P}\right]\text{+}\left[{\text{HP}}^{\text{-}}\right]\text{+}\left[{\text{P}}^{\text{2-}}\right]\text{(2)}\\ \\ \text{Equilibria:}{\text{K}}_{\text{1}}\text{=}\underset{\_}{\left[{\text{H}}^{\text{+}}\right]{\text{γ}}_{{\text{H}}^{\text{+}}}\left[{\text{HP}}^{\text{-}}\right]{\text{γ}}_{{\text{HP}}^{\text{-}}}}(3)\\ \left[{\text{H}}_{\text{2}}\text{P}\right]{\text{γ}}_{{\text{H}}_{\text{2}}\text{P}}\\ \\ {\text{K}}_{\text{2}}\text{=}\underset{\_}{\left[{\text{H}}^{\text{+}}\right]{\text{γ}}_{{\text{H}}^{\text{+}}}\left[{\text{P}}^{\text{2-}}\right]{\text{γ}}_{{\text{P}}^{\text{2-}}}}(4)\\ \left[{\text{H}}_{\text{2}}\text{P}\right]{\text{γ}}_{{\text{H}}_{\text{2}}\text{P}}\\ \text{Kw=}\left[{\text{H}}^{\text{+}}\right]{\text{γ}}_{{\text{H}}^{\text{+}}}\left[{\text{OH}}^{\text{-}}\right]{\text{γ}}_{{\text{OH}}^{\text{-}}}(5)\end{array}$

$\left[{\text{K}}^{\text{+}}\right]$ Value is substituted in equation (1) and (2)

$\left[{\text{H}}_{\text{2}}\text{P}\right]\text{+}\left[{\text{H}}^{\text{+}}\right]\text{-}\left[{\text{P}}^{\text{2-}}\right]\text{-}\left[{\text{OH}}^{\text{-}}\right]\text{=0}$

Given values apply to equation (3), (4), and (5) we get

$\begin{array}{l}\\ \frac{\left[{\text{H}}^{\text{+}}\right]{\text{γ}}_{{\text{H}}^{\text{+}}}\left[{\text{HP}}^{\text{-}}\right]{\text{γ}}_{{\text{HP}}^{\text{-}}}}{{\text{K}}_{\text{1}}{\text{γ}}_{{\text{H}}_{\text{2}}\text{P}}}\text{+}\left[{\text{H}}^{\text{+}}\right]\text{-}\frac{{\text{K}}_{\text{2}}\left[{\text{HP}}^{\text{-}}\right]{\text{γ}}_{{\text{HP}}^{\text{-}}}}{\left[{\text{H}}^{\text{+}}\right]{\text{γ}}_{{\text{H}}^{\text{+}}}{\text{γ}}_{{\text{P}}^{\text{2-}}}}\text{-}\frac{{\text{K}}_{\text{w}}}{\left[{\text{H}}^{\text{+}}\right]{\text{γ}}_{{\text{H}}^{\text{+}}}{\text{γ}}_{{\text{OH}}^{\text{-}}}}\end{array}$

That can be rearranged the equation we get

$\left[{\text{H}}^{\text{+}}\right]\sqrt{\frac{\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}\left[{\text{HP}}^{\text{-}}\right]{\text{γ}}_{{\text{HP}}^{\text{-}}}{\text{γ}}_{{\text{H}}_{\text{2}}\text{P}}}{{\text{γ}}_{{\text{H}}^{\text{+}}}{\text{γ}}_{{\text{P}}^{\text{2-}}}}\text{+}\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{w}}{\text{γ}}_{{\text{H}}_{\text{2}}\text{P}}}{{\text{γ}}_{{\text{H}}^{\text{+}}}{\text{γ}}_{{\text{OH}}^{\text{-}}}}}{{\text{K}}_{\text{1}}{\text{γ}}_{{\text{H}}_{\text{2}}\text{P}}\text{+}\left[{\text{HP}}^{\text{-}}\right]{\text{γ}}_{{\text{H}}^{\text{+}}}{\text{γ}}_{{\text{HP}}^{\text{-}}}}}$

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of given solution has to be found.

Concept introduction:

$\left[{\text{H}}^{\text{+}}\right]\text{=}\sqrt{\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}{\text{F+K}}_{\text{1}}{\text{K}}_{\text{w}}}{{\text{K}}_{\text{1}}\text{+F}}}$

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

$\text{pH=}\text{-log}{\left[{\text{H}}^{\text{+}}\right]}_{\text{γ}}{}_{{\text{H}}^{\text{-}}}$

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

Answer to Problem 10.8P

The $\text{pH}$ of given solution is $\text{4}\text{.03}$

Explanation of Solution

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

The strength of the ions is

$\begin{array}{l}\left[{\text{HP}}^{\text{-}}\right]\approx \text{0}\text{.050}\text{M,}{\text{γ}}_{{\text{HP}}^{\text{-}}}\text{=0}\text{.835,}{\text{γ}}_{{\text{P}}^{\text{2-}}}\text{=0}\text{.485,}{\text{γ}}_{{\text{H}}_{\text{2}}\text{P}}\approx \text{1}\text{.00}\\ {\text{γ}}_{{\text{H}}^{\text{+}}}\text{=0}\text{.86,}{\text{γ}}_{{\text{OH}}^{\text{-}}}\text{=0}\text{.81}\end{array}$

$\left[{\text{H}}^{\text{+}}\right]\text{=1}{\text{.09×10}}^{\text{-4}}\text{=}\text{pH}\text{=}\text{-log}\left[{\text{H}}^{\text{+}}\right]{\text{γ}}_{{\text{H}}^{\text{+}}}\text{=4}\text{.03}$