Chapter 19, Problem 19.142P

(a)

Interpretation Introduction

Interpretation:

Species among ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$, ${\text{HCO}}_{\text{3}}^{\text{-}}$, ${\text{CO}}_{\text{3}}^{\text{2-}}$ having highest concentration at $\text{pH}\text{=}\text{8}\text{.5}$ has to be calculated.

Concept introduction:

$pH$:

$\text{pH}$ is a scale used to specify the acidity or basicity a solution.  It ranges from $0-14$.  $\text{pH}$ $7.0$ is considered as neutral solution, $\text{pH}$ more than $7.0$ is taken as basic solution whereas $\text{pH}$ less than $7.0$ is considered as acidic solution (at ${\text{25}}^{\text{ο}}\text{C}$).  It is the measurement of activity of free ${\text{H}}^{\text{+}}$ and ${\text{OH}}^{\text{-}}$ in solution.

$p{K}_a$:

$p{K}_a$ is the negative log of acid dissociation constant that determines the strength of the acid. Lower the $p{K}_a$ value stronger is the acid.

${\text{pK}}_{\text{a1}}$ is the 1^st dissociation constant for the acid and ${\text{pK}}_{\text{a2}}$ is the 2^nd dissociation constant of the acid.

Answer to Problem 19.142P

Among all the three species, ${\text{HCO}}_{\text{3}}^{\text{-}}$ is more at $\text{pH}\text{=}\text{8}\text{.5}$.

Explanation of Solution

The equation considered as,

  ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\stackrel{}{\rightleftarrows }{\text{H}}^{\text{+}}\text{+}{\text{HCO}}_{\text{3}}^{\text{-}}\mathrm{...}({\text{K}}_{\text{a1}})$

  ${\text{HCO}}_{\text{3}}^{\text{-}}\stackrel{}{\rightleftarrows }{\text{H}}^{\text{+}}+{\text{CO}}_{\text{3}}^{\text{2-}}\mathrm{...}({\text{K}}_{\text{a2}})$

  ${\text{pK}}_{\text{a}}=-\log {\text{K}}_{\text{a}}$

Now, for the 1^st equation,

  ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\stackrel{}{\rightleftarrows }{\text{H}}^{\text{+}}\text{+}{\text{HCO}}_{\text{3}}^{\text{-}}\mathrm{...}({\text{K}}_{\text{a1}})$

  ${\text{K}}_{\text{a1}}=4.5\times {10}^{-7}$

  ${\text{pK}}_{\text{a1}}=-\log (4.5\times {10}^{-7})$

  ${\text{pK}}_{\text{a1}}=6.4368$

Now, for the 2^nd equation,

  ${\text{HCO}}_{\text{3}}^{\text{-}}\stackrel{}{\rightleftarrows }{\text{H}}^{\text{+}}+{\text{CO}}_{\text{3}}^{\text{2-}}\mathrm{...}({\text{K}}_{\text{a2}})$

  ${\text{K}}_{\text{a2}}=4.7\times {10}^{-11}$

  ${\text{pK}}_{\text{a2}}=-\log (4.7\times {10}^{-11})$

  ${\text{pK}}_{\text{a2}}=10.3279$

From the above data it is shown that, ${\text{pK}}_{\text{a2}}\text{>}\text{pH}$ but ${\text{pK}}_{\text{a1}}<\text{pH}$. ($\text{pH}\text{=}\text{8}\text{.5}$ given)

If ${\text{pK}}_{\text{a1}}$ is less that means ${\text{K}}_{\text{a1}}$ value is more which means the 1^st dissociation occurs rapidly and hence ${\text{HCO}}_{\text{3}}^{\text{-}}$ will be more in medium (as the pH is basic so the conjugate base is more in concentration.)

Again, ${\text{pK}}_{\text{a2}}$ is more that means ${\text{K}}_{\text{a2}}$ is less and hence 2^nd dissociation is slower in rate. Hence in medium the reactant i.e. ${\text{HCO}}_{\text{3}}^{\text{-}}$ will be more.

Hence among all the three species ${\text{HCO}}_{\text{3}}^{\text{-}}$ is more at $\text{pH}\text{=}\text{8}\text{.5}$.

(b)

Interpretation Introduction

Interpretation:

At $\text{pH}\text{=}\text{8}\text{.5}$ the concentration ratio of $\frac{\left[{\text{HCO}}_{\text{3}}^{\text{-}}\right]}{\left[{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\right]}$ and $\frac{\left[{\text{CO}}_{\text{3}}^{\text{2-}}\right]}{\left[{\text{HCO}}_{\text{3}}^{\text{-}}\right]}$ has to be calculated.

Concept introduction:

$pH$:

$\text{pH}$ is a scale used to specify the acidity or basicity a solution. It ranges from $0-14$. $\text{pH}$ $7.0$ is considered as neutral solution, $\text{pH}$ more than $7.0$ is taken as basic solution whereas $\text{pH}$ less than $7.0$ is considered as acidic solution (at ${\text{25}}^{\text{ο}}\text{C}$). It is the measurement of activity of free ${\text{H}}^{\text{+}}$ and ${\text{OH}}^{\text{-}}$ in solution.

$p{K}_a$:

$p{K}_a$ is the negative log of acid dissociation constant that determines the strength of the acid. Lower the $p{K}_a$ value stronger is the acid.

${\text{pK}}_{\text{a1}}$ is the 1^st dissociation constant for the acid and ${\text{pK}}_{\text{a2}}$ is the 2^nd dissociation constant of the acid.

Henderson-Hasselbalch Equation:

This equation is used to determine the $\text{pH}$ of a buffer solution.  ${\text{K}}_{\text{a}}$ is the given quantity which is acid dissociation constant. It is done for a given concentration of acid and its conjugate base.

  $\text{pH=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\left[\text{salt}\right]}{\left[\text{acid}\right]}$

Answer to Problem 19.142P

The value of concentration ratios of $\frac{\left[{\text{HCO}}_{\text{3}}^{\text{-}}\right]}{\left[{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\right]}$ and $\frac{\left[{\text{CO}}_{\text{3}}^{\text{2-}}\right]}{\left[{\text{HCO}}_{\text{3}}^{\text{-}}\right]}$ are respectively $1.4\times {10}^2$ and $1.48\times {10}^{-2}.$

Explanation of Solution

The equation considered as,

  ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\stackrel{}{\rightleftarrows }{\text{H}}^{\text{+}}\text{+}{\text{HCO}}_{\text{3}}^{\text{-}}\mathrm{...}({\text{K}}_{\text{a1}})$

  ${\text{HCO}}_{\text{3}}^{\text{-}}\stackrel{}{\rightleftarrows }{\text{H}}^{\text{+}}+{\text{CO}}_{\text{3}}^{\text{2-}}\mathrm{...}({\text{K}}_{\text{a2}})$

  ${\text{pK}}_{\text{a}}=-\log {\text{K}}_{\text{a}}$

For the 1^st equation,

  ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\stackrel{}{\rightleftarrows }{\text{H}}^{\text{+}}\text{+}{\text{HCO}}_{\text{3}}^{\text{-}}\mathrm{...}({\text{K}}_{\text{a1}})$

Thus the $p{K}_a$ is calculated as,

  ${\text{K}}_{\text{a1}}=4.5\times {10}^{-7}$

  ${\text{pK}}_{\text{a1}}=-\log (4.5\times {10}^{-7})$

  ${\text{pK}}_{\text{a1}}=6.4368$

Thus the $\text{pH}$ is calculated as,

  $\text{pH=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{\left[\text{salt}\right]}{\left[\text{acid}\right]}$

  $\begin{array}{l}\text{pH=}{\text{pK}}_{\text{a1}}\text{+}\text{log}\frac{\left[{\text{HCO}}_{\text{3}}^{\text{-}}\right]}{\left[{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\right]}\\ \\ 8.5=6.35+\text{log}\frac{\left[{\text{HCO}}_{\text{3}}^{\text{-}}\right]}{\left[{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\right]}\end{array}$

The value of concentration ratios of $\frac{\left[{\text{HCO}}_{\text{3}}^{\text{-}}\right]}{\left[{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\right]}=1.4\times {10}^2$

For the 2^nd equation,

  ${\text{HCO}}_{\text{3}}^{\text{-}}\stackrel{}{\rightleftarrows }{\text{H}}^{\text{+}}+{\text{CO}}_{\text{3}}^{\text{2-}}\mathrm{...}({\text{K}}_{\text{a2}})$

Thus the $p{K}_a$ is calculated as,

  ${\text{K}}_{\text{a2}}=4.7\times {10}^{-11}$

  ${\text{pK}}_{\text{a2}}=-\log (4.7\times {10}^{-11})$

  ${\text{pK}}_{\text{a2}}=\text{ 10}\text{.33}$

Thus the $\text{pH}$ is calculated as,

  $\begin{array}{l}\text{pH=}{\text{pK}}_{\text{a2}}\text{+}\text{log}\frac{\left[{\text{CO}}_{\text{3}}^{\text{2-}}\right]}{\left[{\text{HCO}}_{\text{3}}^{\text{-}}\right]}\\ \\ 8.5=10.33+\text{log}\frac{\left[{\text{CO}}_{\text{3}}^{\text{2-}}\right]}{\left[{\text{HCO}}_{\text{3}}^{\text{-}}\right]}\\ \\ \frac{\left[{\text{CO}}_{\text{3}}^{\text{2-}}\right]}{\left[{\text{HCO}}_{\text{3}}^{\text{-}}\right]}=1.48\times {10}^{-2}\end{array}$

The value of concentration ratios of $\frac{\left[{\text{CO}}_{\text{3}}^{\text{2-}}\right]}{\left[{\text{HCO}}_{\text{3}}^{\text{-}}\right]}=1.48\times {10}^{-2}$

Thus the value of concentration ratios of $\frac{\left[{\text{HCO}}_{\text{3}}^{\text{-}}\right]}{\left[{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\right]}$ and $\frac{\left[{\text{CO}}_{\text{3}}^{\text{2-}}\right]}{\left[{\text{HCO}}_{\text{3}}^{\text{-}}\right]}$ are respectively $1.4\times {10}^2$ and $1.48\times {10}^{-2}.$

(c)

Interpretation Introduction

Interpretation:

Reason for lower $\text{pH}$ at deep sea level has to be interpreted.

Concept introduction:

$pH$:

$\text{pH}$ is a scale used to specify the acidity or basicity a solution. It ranges from $0-14$. $\text{pH}$ $7.0$ is considered as neutral solution, $\text{pH}$ more than $7.0$ is taken as basic solution whereas $\text{pH}$ less than $7.0$ is considered as acidic solution (at ${\text{25}}^{\text{ο}}\text{C}$). It is the measurement of activity of free ${\text{H}}^{\text{+}}$ and ${\text{OH}}^{\text{-}}$ in solution.

Photosynthesis:

The process of photosynthesis occurs in plants for their survival.  In this process carbon dioxide is absorbed from nature and oxygen is released to nature by plants.  Sunlight is the main important component for photosynthesis to occur.  Here glucose is formed.

  $\text{6}{\text{CO}}_{\text{2}}\text{+}\text{6}{\text{H}}_{\text{2}}\text{O}\stackrel{\text{sunlight}}{\rightleftarrows }{\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{+}{\text{6O}}_{\text{2}}$

Respiration:

Respiration is a biochemical process in which in the cells of organisms energy is evolved due to reaction between oxygen and glucose and by this process carbon dioxide is released along with water and ATP

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{+}{\text{6O}}_{\text{2}}\underset{}{\rightleftarrows }\text{6}{\text{CO}}_{\text{2}}\text{+}\text{6}{\text{H}}_{\text{2}}\text{O}\text{+}\text{ATP}$

Explanation of Solution

The acidity of the sea water is due to presence of dissolved carbon dioxide in the water.

On the upper sea level there are plants and also on upper level of sea sunlight can enter. As a result of that on the upper level of sea photosynthesis occurs and according to the process of photosynthesis carbon dioxide is taken up by plants.  This results a decrease in the dissolved carbon dioxide concentration in the upper sea level.

But still due to the respiration by the sea animals and the plants at upper level there will be some amount of dissolved carbon dioxide in the upper level. So sea water at upper level is acidic but less in amount.

However, in deep sea level, $\text{pH}\text{=}7.\text{5}$ that means water is more acidic here.  In deep sea level pressure is very high and sunlight cannot enter and thus plants cannot survive there but some animals only can survive there.  Hence photosynthesis does not occur there only respiration occurs.  As a result of that the concentration of dissolved carbon dioxide is more in the deep-sea level than in the upper level.  So the acidity here is more.

Hence, deep sea water has a higher acidity the value of $\text{pH}\text{=}7.\text{5}$.