Q19.149 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Seawater at the surface has a $\mathbf{\text{pH}}$ of about $\mathbf{\text{85}}$.

(a) Which of the following species has the highest concentration at this $\mathbf{pH}\mathbf{:}{\mathbf{H}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}\mathbf{;}\mathbf{HC}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{-}}\mathbf{;}\mathbf{C}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{2}\mathbf{-}}$ ? Explain.

(b) What are the concentration ratios $\left[C{O_3}^{2-}\right]\mathbf{/}\left[\mathrm{HC}{O_3}^{-}\right]$ and data-custom-editor="chemistry" $\left[\mathrm{HC}{O_3}^{-}\right]\mathbf{/}\left[H_2{\mathrm{CO}}_3\right]$ at this $\mathbf{\text{pH}}$?

(c) In the deep sea, light levels are low, and the $\mathbf{\text{pH}}$ is around $\mathbf{\text{7.5}}$. Suggest a reason for the lower $\mathbf{\text{pH}}$ at the greater ocean depth. (Hint: Consider the presence or absence of plant and animal life, and the effects on carbon dioxide concentrations.)

Step-by-Step Solution

Answer

1. Among all the three species,${\mathrm{HCO}}_3^{-}$ is more at $\text{pH=8.5}$.
2. The value of concentration ratios of $\frac{\left[{\mathrm{HCO}}_3^{-}\right]}{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]}$and $\frac{\left[{\mathrm{CO}}_3^2\right]}{\left[{\mathrm{HCO}}_3^{-}\right]}$are $1.4\times {10}^2$and $1.48\times {10}^{-2}$respectively.
3. Deep sea water has a higher acidity of $\text{pH=7.5}$

Step 1: Definition of Concept.

Capacity: The amount of heat energy required to raise the temperature of a body by a certain amount is known as heat capacity. The amount of heat in joules required to raise the temperature 1 Kelvin is known as heat capacity (symbol: C) in SI units.

Step 2: Find which of the following species has the highest concentration.

(a)

Considering the given information:

The equation considered as,

$\begin{array}{l}{\mathrm{H}}_2{\mathrm{CO}}_3\\ \rightleftarrows {\mathrm{H}}^{+}{+\mathrm{HCO}}_3^{-}\\ \left({\mathrm{K}}_{\mathrm{a}1}\right)\\ {\mathrm{HCO}}_3^{-}\rightleftarrows {\mathrm{H}}^{+}\\ {+\mathrm{CO}}_3^{2-}\\ \left({\mathrm{K}}_{\mathrm{a}2}\right)\\ {\mathrm{pK}}_{\mathrm{a}}=-{\mathrm{logK}}_{\mathrm{a}}\end{array}$

Now, for the 1st equation,

$\begin{array}{l}{\mathrm{H}}_2{\mathrm{CO}}_3\\ \rightleftarrows {\mathrm{H}}^{+}{+\mathrm{HCO}}_3^{-}\\ \left({\mathrm{K}}_{\mathrm{a}1}\right)\\ {\mathrm{K}}_{\mathrm{a}1}=4.5\times {10}^{-7}\\ {\mathrm{pK}}_{\mathrm{al}}=-\log \left(4.5\times {10}^{-7}\right)\end{array}$

${\mathrm{pK}}_{\mathrm{al}}=6.4368$

Now, for the 2nd equation,

$\begin{array}{l}{\mathrm{HCO}}_3^{-}\rightleftarrows {\mathrm{H}}^{+}\\ {+\mathrm{CO}}_3^{2-}\\ \left({\mathrm{K}}_{\mathrm{a}2}\right)\\ {\mathrm{K}}_{\mathrm{a}2}=4.7\times {10}^{-11}\\ {\mathrm{pK}}_{\mathrm{a}2}=-\log \left(4.7\times {10}^{-11}\right)\\ {\mathrm{pK}}_{\mathrm{a}2}=10.3279\end{array}$

From the above data it is shown that, ${\mathrm{pK}}_{\mathrm{a}2}>\mathrm{pH}$but ${\mathrm{pK}}_{\mathrm{al}}<\mathrm{pH}.(\mathrm{pH}=8.5$given)

If ${\mathrm{pK}}_{\mathrm{al}}$ is lower, the ${\mathrm{K}}_{\mathrm{al}}$ value is higher, implying that 1st dissociation occurs quickly and thus ${\mathrm{HCO}}_3^{-}$ is higher in the medium (because pH is basic, the conjugate base is higher in concentration).

Again, ${\mathrm{pK}}_{\mathrm{a}2}$is more that means ${\mathrm{K}}_{\mathrm{a}2}$ is less and hence dissociation is slower in rate. Hence in medium the reactant i.e. ${\mathrm{HCO}}_3^{-}$will be more.

Therefore, among all the three species ${\mathrm{HCO}}_3^{-}$ is more at $\text{pH=8.5}$.

Step 3: Find the concentration ratios.

(b)

Considering the given information:

The equation considered as,

$\begin{array}{l}{\mathrm{H}}_2{\mathrm{CO}}_3\\ \rightleftarrows {\mathrm{H}}^{+}{+\mathrm{HCO}}_3^{-}\\ \left({\mathrm{K}}_{\mathrm{a}1}\right)\\ {\mathrm{HCO}}_3^{-}\rightleftarrows {\mathrm{H}}^{+}\\ {+\mathrm{CO}}_3^{2-}\\ \left({\mathrm{K}}_{\mathrm{a}2}\right)\\ {\mathrm{pK}}_{\mathrm{a}}=-{\mathrm{logK}}_{\mathrm{a}}\end{array}$

For the first equation,

$\begin{array}{l}{\mathrm{H}}_2{\mathrm{CO}}_3\\ \rightleftarrows {\mathrm{H}}^{+}{+\mathrm{HCO}}_3^{-}\\ \left({\mathrm{K}}_{\mathrm{a}1}\right)\end{array}$

Thus the ${\mathrm{pK}}_{\mathrm{a}}$ is calculated as,

$\begin{array}{c}{\mathrm{K}}_{\mathrm{al}}=4.5\times {10}^{-7}\\ {\mathrm{pK}}_{\mathrm{al}}=-\log \left(4.5\times {10}^{-7}\right)\\ {\mathrm{pK}}_{\mathrm{al}}=6.4368\end{array}$

Thus the pH is calculated as,

$\begin{array}{c}\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[\mathrm{salt}\right]}{\left[\mathrm{acid}\right]}\\ \mathrm{pH}={\mathrm{pK}}_{\mathrm{al}}+\log \frac{\left[{\mathrm{HCO}}_3^{-}\right]}{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]}\\ 8.5=6.35+\log \frac{\left[{\mathrm{HCO}}_3^{-}\right]}{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]}\end{array}$

The value of concentration ratios of $\frac{\left[{\mathrm{HCO}}_3^{-}\right]}{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\mid \right.}=1.4\times {10}^2$

For the second equation,

$\begin{array}{l}{\mathrm{HCO}}_3^{-}\rightleftarrows {\mathrm{H}}^{+}\\ {+\mathrm{CO}}_3^{2-}\\ \left({\mathrm{K}}_{\mathrm{a}2}\right)\end{array}$

Thus the ${\mathrm{pK}}_{\mathrm{a}}$ is calculated as,

$\begin{array}{c}{\mathrm{K}}_{\mathrm{a}2}=4.7\times {10}^{-11}\\ {\mathrm{pK}}_{\mathrm{a}2}=-\log \left(4.7\times {10}^{-11}\right)\\ {\mathrm{pK}}_{\mathrm{a}2}=10.33\end{array}$

Thus the pH is calculated as,

$\begin{array}{c}\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}2}+\log \frac{\left[{\mathrm{CO}}_3^2\right]}{\left[{\mathrm{HCO}}_3\right]}\\ 8.5=10.33+\log \frac{\left[{\mathrm{CO}}_3^2\right]}{\left[{\mathrm{HCO}}_3^{-}\right]}\\ \frac{\left[{\mathrm{CO}}_3^2\right]}{\left[{\mathrm{HCO}}_3^{-}\right]}=1.48\times {10}^{-2}\end{array}$

The value of concentration ratios of $\frac{\left[{\mathrm{CO}}_3^2\right]}{\left[{\mathrm{HCO}}_3\right]}=1.48\times {10}^{-2}$.

Thus the value of concentration ratios of $\frac{\left[{\mathrm{HCO}}_3\right]}{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]}$ and $\frac{\left[{\mathrm{CO}}_3^2\right]}{\left[{\mathrm{HCO}}_3\right]}$are $1.4\times {10}^2$and $1.48\times {10}^{-2}$respectively.

 

Step 4: Suggest a reason for the lower pH at the greater ocean depth.

(c)

The presence of dissolved carbon dioxide in the water causes the acidity of sea water.

There are plants on the upper sea level, and sunlight can also reach the upper sea level. As a result, photosynthesis occurs on the upper level of the sea, and plants absorb carbon dioxide through the photosynthesis process. 

As a result, the concentration of dissolved carbon dioxide in the upper sea level decreases.

However, because of the respiration of sea animals and plants at the upper level, there will be some dissolved carbon dioxide in the upper level. As a result, sea water at the upper levels is acidic, but in smaller amounts.

However, at deep sea level, $\text{pH=7.5}$, which indicates that the water is more acidic. Because the pressure in the deep sea is so high that sunlight cannot penetrate, plants cannot survive there, but some animals can. As a result, only respiration takes place there, rather than photosynthesis. As a result, the deep-sea level has a higher concentration of dissolved carbon dioxide than the upper level. As a result, the acidity is higher here.

Therefore, deep sea water has a higher acidity of $\text{pH=7.5}$.