Chapter 10, Problem 10.4P

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ and the concentration of ${\text{H}}_{\text{2}}\text{A}$ 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 10.4P

The $\text{pH}$ and the concentration of ${\text{H}}_{\text{2}}\text{A}$ is $\text{2}\text{.51}$ and $\text{1}{\text{.00×10}}^{\text{-8}}\text{M}$

Explanation of Solution

Given,

Dissociation of the given acid

$\begin{array}{l}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.100-x}}{\text{=K}}_{\text{1}}\text{x=}\text{3}{\text{.11×10}}^{\text{-3}}\\ \left[{\text{H}}^{\text{+}}\right]\text{=}\left[{\text{HA}}^{\text{-}}\right]\\ \text{pH=}\text{2}\text{.51}\\ \\ \left[{\text{HA}}^{\text{-}}\right]\text{= 0}\text{.100-x =}\text{0}\text{.0969M}\\ \\ \left[{\text{A}}^{\text{2-}}\right]\text{=}\frac{{\text{K}}_{\text{2}}\left[{\text{HA}}^{\text{-}}\right]}{\left[{\text{H}}^{\text{+}}\right]}\text{= 1}\text{.00 ×}{\text{10}}^{\text{-3}}\text{M}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ and the concentration of ions in ${\text{NaHSO}}_{\text{3}}$ 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 10.4P

The $\text{pH}$ and the concentration of ${\text{HA}}^{\text{-}}$ is $6.00$ and $\text{1}\text{.00}\times {\text{10}}^{\text{-3}}\text{M}$

Explanation of Solution

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

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

$\begin{array}{l}\text{=1}{\text{.00×10}}^{\text{-6}}\text{M=pH}\text{=}\text{6}\text{.00}\\ \left[{\text{H}}_{\text{2}}\text{A}\right]\approx \frac{\left[\text{H+}\right]\left[{\text{HA}}^{\text{-}}\right]}{{\text{K}}_{\text{1}}}\text{=1}\text{.00}\text{×}{\text{10}}^{\text{-3}}\text{M}\\ \left[{\text{A}}^{\text{2-}}\right]\text{=}\frac{{\text{K}}_{\text{2}}\left[{\text{HA}}^{\text{-}}\right]}{\left[{\text{H}}^{\text{+}}\right]}\text{= 1}{\text{.00×10}}^{\text{-3}}\text{M}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The $\text{pH}$ and the concentration of ions in ${\text{Na}}_{\text{2}}\text{A}$ 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:

The $\text{pH}$ and the concentration of ${\text{Na}}_{\text{2}}\text{A}$ is $\text{10}\text{.50}$ and $\text{1}\text{.00}\text{×}{\text{10}}^{\text{-10}}\text{M}$

Answer to Problem 10.4P

The $\text{pH}$ and the concentration of ${\text{Na}}_{\text{2}}\text{A}$ is $\text{10}\text{.50}$ and $\text{1}\text{.00}\text{×}{\text{10}}^{\text{-10}}\text{M}$

Explanation of Solution

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

$\begin{array}{l}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.100-x}}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{2}}}\text{=}\text{x}\text{=}\left[{\text{OH}}^{\text{-}}\right]\text{=}\left[{\text{HA}}^{\text{-}}\right]\text{=}\text{3}{\text{.16×10}}^{\text{-4}}\text{M}\\ \text{pH}\text{=10}\text{.50}\end{array}$

$\begin{array}{l}\left[{\text{A}}^{\text{2-}}\right]\text{=}\text{0}\text{.100-x=}\text{9}{\text{.97×10}}^{\text{-2}}\text{M}\\ \left[{\text{H}}_{\text{2}}\text{A}\right]\approx \frac{\left[\text{H+}\right]\left[{\text{HA}}^{\text{-}}\right]}{{\text{K}}_{\text{1}}}\text{=1}\text{.00}\text{×}{\text{10}}^{\text{-10}}\text{M}\end{array}$

$\begin{array}{l}\underset{\_}{\text{pH}\left[{\text{H}}_{\text{2}}\text{A}\right]\left[{\text{HA}}^{\text{-}}\right]\left[{\text{A}}^{\text{2-}}\right]}\\ \text{0}\text{.100M}{\text{H}}_{\text{2}}\text{A}\text{2}\text{.51}\text{9}{\text{.69×10}}^{\text{-2}}\text{3}{\text{.11×10}}^{\text{-3}}\text{1}{\text{.00×10}}^{\text{-8}}\\ \text{0}\text{.100M}\text{NaHA}\text{6}\text{.00}\text{1}{\text{.00×10}}^{\text{-3}}\text{1}{\text{.00×10}}^{\text{-1}}\text{1}{\text{.00×10}}^{\text{-3}}\\ \text{0}\text{.100M}{\text{Na}}_{\text{2}}\text{A}\text{10}\text{.50}\text{1}{\text{.00×10}}^{\text{-10}}\text{3}{\text{.16×10}}^{\text{-4}}\text{9}{\text{.97×10}}^{\text{-2}}\end{array}$

The concentration of solution is $\text{1}\text{.00}\text{×}{\text{10}}^{\text{-10}}\text{M}$