Chapter 10, Problem 10.AE

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ and the concentration of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}$ has to be fined.

Concept introduction:

$\left[{\text{H}}^{\text{+}}\right]\text{=}\sqrt{\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}\text{(0}{\text{.050)+K}}_{\text{1}}{\text{K}}_{\text{w}}}{{\text{K}}_{\text{1}}\text{+(0}\text{.050)}}}$

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.AE

The $\text{pH}$ and the concentration of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}$ is $\text{2}{\text{.03×10}}^{\text{-2}}$ and $\text{6}{\text{.7×10}}^{\text{-8}}\text{M}$

Explanation of Solution

Given,

Dissociation of the given acid

$\begin{array}{l}{\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}\underset{}{\rightleftharpoons }{\text{HSO}}_{\text{3}}^{\text{-}}{\text{+H}}^{\text{+}}\\ \text{0}\text{.050-x}\text{x}\text{x}\end{array}$

$\begin{array}{l}\frac{\text{x2}}{\text{0}\text{.050-x}}{\text{=K}}_{\text{1}}\text{=}\text{1}{\text{.39×10}}^{\text{-2}}\\ \text{x=2}{\text{.03×10}}^{\text{-2}}\\ \left[{\text{HSO}}_{\text{3}}^{\text{-}}\right]\text{=}\left[{\text{H}}^{\text{+}}\right]\text{=2}{\text{.03×10}}^{\text{-2}}\text{M}\\ \text{pH=1}\text{.69}\\ \left[{\text{HSO}}_{\text{3}}^{}\right]\text{=0}\text{.050-x=0}\text{.030M}\end{array}$

$\left[{\text{SO}}_{\text{3}}^{\text{2-}}\right]\text{=}\frac{{\text{K}}_{\text{2}}\left[{\text{HSO}}_{\text{3}}^{\text{-}}\right]}{\left[{\text{H}}^{\text{+}}\right]}{\text{=K}}_{\text{2}}\text{=}\text{6}{\text{.7×10}}^{\text{-8}}\text{M}$

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ and the concentration of ions in ${\text{NaHSO}}_{\text{3}}$ has to be fined.

Concept introduction:

$\left[{\text{H}}^{\text{+}}\right]\text{=}\sqrt{\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}\text{(0}{\text{.050)+K}}_{\text{1}}{\text{K}}_{\text{w}}}{{\text{K}}_{\text{1}}\text{+(0}\text{.050)}}}$

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.AE

The $\text{pH}$ and the concentration of ${\text{NaHSO}}_{\text{3}}$ is $\text{4}\text{.57}$ and $\text{0}\text{.050M}$

Explanation of Solution

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

$\left[{\text{H}}^{\text{+}}\right]\text{=}\sqrt{\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}\text{(0}{\text{.050)+K}}_{\text{1}}{\text{K}}_{\text{w}}}{{\text{K}}_{\text{1}}\text{+(0}\text{.050)}}}$

$\begin{array}{l}\text{=2}{\text{.71×10}}^{\text{-5}}\text{M=pH=4}\text{.57}\\ \left[{\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}\right]\text{=}\frac{\left[\text{H+}\right]\left[{\text{HSO}}_{\text{3}}^{\text{-}}\right]}{{\text{K}}_{\text{1}}}\text{=}\frac{\text{(2}{\text{.71×10}}^{\text{-5}}\text{)(0}\text{.050)}}{\text{1}{\text{.39×10}}^{\text{-2}}}\text{=9}{\text{.7×10}}^{\text{-5}}\text{M}\end{array}$

$\begin{array}{l}\left[{\text{SO}}_{\text{3}}^{\text{2-}}\right]\text{=}\frac{{\text{K}}_{\text{2}}\left[{\text{HSO}}_{\text{3}}^{\text{-}}\right]}{\left[{\text{H}}^{\text{+}}\right]}\text{=1}{\text{.2×10}}^{\text{-4}}\text{M}\\ \left[{\text{HSO}}_{\text{3}}^{\text{-}}\right]\text{=0}\text{.050M}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The $\text{pH}$ and the concentration of ions in ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{3}}$ has to be fined.

Concept introduction:

$\left[{\text{H}}^{\text{+}}\right]\text{=}\sqrt{\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}\text{(0}{\text{.050)+K}}_{\text{1}}{\text{K}}_{\text{w}}}{{\text{K}}_{\text{1}}\text{+(0}\text{.050)}}}$

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 ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{3}}$ is $\text{9}\text{.94}$ and $\text{7}{\text{.2×10}}^{\text{-13}}\text{M}$

Answer to Problem 10.AE

The $\text{pH}$ and the concentration of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{3}}$ is $\text{9}\text{.94}$ and $\text{7}{\text{.2×10}}^{\text{-13}}\text{M}$

Explanation of Solution

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

$\begin{array}{l}\left[{\text{SO}}_{\text{3}}^{\text{2-}}\right]{\text{+H}}_{\text{2}}\text{O}\underset{}{\rightleftharpoons }{\text{HSO}}_{\text{3}}^{\text{-}}{\text{+OH}}^{\text{-}}\\ \text{0}\text{.050-x}\text{x}\text{x}\end{array}$

$\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.050-x}}{\text{=K}}_{\text{b1}}\text{=}\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a2}}}=1.49\times {10}^{-7}$

$\left[{\text{HSO}}_{\text{3}}^{\text{-}}\right]\text{=}\text{x}\text{=}\text{8}{\text{.6×10}}^{\text{-5}}\text{M}$

$\left[{\text{H}}^{\text{+}}\right]\text{=}\frac{{\text{K}}_{\text{w}}}{\text{x}}\text{=1}{\text{.16×10}}^{\text{-10}}\text{M=pH=9}\text{.94}$

$\begin{array}{l}\left[{\text{SO}}_{\text{3}}^{\text{2-}}\right]\text{=0}\text{.050-x =0}\text{.050M}\\ \left[{\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}^{}\right]\text{ = }\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{HSO}}_{\text{3}}^{\text{-}}\right]}{{\text{K}}_{\text{1}}}\text{=}\text{7}{\text{.2×10}}^{\text{-13}}\text{M}\end{array}$

The concentration of solution is $\text{7}{\text{.2×10}}^{\text{-13}}\text{M}$