Chapter 8, Problem 109P

Interpretation Introduction

(a)

Interpretation:

The concentrations of $[{\text{H}}_{\text{3}}{\text{O}}^{+}]$ and $[{\text{OH}}^{-}]$ at pH levels of 8.2 and 8.1 should to be calculated.

Concept Introduction:

pH can be defined as the negative logarithm of hydrogen ion concentration or hydronium ion concentration. It can be represented in the form of an equation below.

$p\text{H}=-\log [{\text{H}}_{\text{3}}{\text{O}}^{+}]$

Also, product of concentration of hydronium ion and hydroxide ion is ionization constant of water.

$\begin{array}{c}{K}_w=\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]\\ ={10}^{-14}\end{array}$.

Answer to Problem 109P

Thus, at pH 8.2, the concentrations are.

$[{\text{H}}_{\text{3}}{\text{O}}^{+}]=6.31\times {10}^{-9}$

$[{\text{OH}}^{-}]=1.585\times {10}^{-6}$

At pH 8.1, the concentrations are.

$[{\text{H}}_{\text{3}}{\text{O}}^{+}]=7.943\times {10}^{-9}$

$[{\mathrm{OH}}^{-}]=1.259\times {10}^{-6}$.

Explanation of Solution

pH can be defined as the negative logarithm of hydrogen ion concentration or hydronium ion concentration. It can be represented in the form of an equation below.

$\text{pH}=-\log [{\text{H}}_{\text{3}}{\text{O}}^{+}]$

From the above equation, the hydronium ion concentration can be calculated.

$[{\text{H}}_{\text{3}}{\text{O}}^{+}]={10}^{-\text{pH}}$

The hydronium ion and hydroxide ion concentrations are related to each other. Their product is equal to the ionic product of water. This is represented in the form of equation below.

$K_w=[{\text{H}}_{\text{3}}{\text{O}}^{+}][{\text{OH}}^{-}]=1\times {10}^{-14}$

Step 1: At pH 8.2.

Calculation of hydronium and hydroxide concentration at pH 8.2.

$\begin{array}{c}[{\text{H}}_{\text{3}}{\text{O}}^{+}]={10}^{-\text{pH}}\\ ={10}^{(-8.20)}\\ =6.31\times {10}^{-9}\end{array}$

$\begin{array}{c}{K}_w=[{\text{H}}_{\text{3}}{\text{O}}^{+}][{\text{OH}}^{-}]=1\times {10}^{-14}\\ =(6.31\times {10}^{-9})[{\text{OH}}^{-}]=1\times {10}^{-14}\\ =[{\text{OH}}^{-}]=\frac{1\times {10}^{-14}}{(6.31\times {10}^{-9})}\\ =[{\text{OH}}^{-}]=1.585\times {10}^{-6}\end{array}$

Step 2: At pH 8.1.

Calculation of hydronium and hydroxide concentration at pH 8.1.

$\begin{array}{c}[{\text{H}}_{\text{3}}{\text{O}}^{+}]={10}^{-\text{pH}}\\ ={10}^{(-8.10)}\\ =7.943\times {10}^{-9}\end{array}$

$\begin{array}{c}{K}_w=[{\text{H}}_{\text{3}}{\text{O}}^{+}][{\text{OH}}^{-}]=1\times {10}^{-14}\\ =(7.943\times {10}^{-9})[{\text{OH}}^{-}]=1\times {10}^{-14}\\ =[{\text{OH}}^{-}]=\frac{1\times {10}^{-14}}{(7.943\times {10}^{-9})}\\ =[{\text{OH}}^{-}]=1.259\times {10}^{-6}\end{array}$.

Interpretation Introduction

(b)

Interpretation:

The decrease in pH by 0.1 unit is equal to an increase in acidity of about 30% should be demonstrated.

Concept Introduction:

We know that a change in 1 unit of pH makes a lot of change in acidity. This can be calculated by using the difference in the pH.

Answer to Problem 109P

Thus, a change in 0.1 units in pH of the solution causes an increase in acidity by 26%. This is rounded to 30% which is a significant figure.

Explanation of Solution

The pH initially is 8.2 and the final pH is 8.1. It is clear that there is a decrease in pH making the ocean acidic. But, let us calculate whether this 0.1 pH change is really significant.

This is calculated by the equation,

Fractional change = ${10}^{(-\Delta \text{pH})}-1$

$\begin{array}{c}\text{Fractional}\text{change}={10}^{(-\Delta \text{pH})}-1\\ ={10}^{(-(8.1-8.2)}-1\\ ={10}^{(0.1)}-1\\ =0.26\end{array}$

This 0.26 on a 100 scale makes 26%. To make it a round figure we make it 30%. Thus, the decrease in pH by 0.1corresponds to an increase in acidity of about 30%.

Interpretation Introduction

(c)

Interpretation:

The relationship between the pH of seawater and the bioavailability of carbonate ion should be determined.

Concept Introduction:

The carbon dioxide present in the environment has most of the effect in the acidity of the ocean water. The relative amounts of carbon dioxide are increasing in the environment causing its negative effect to the environment.

Answer to Problem 109P

As the pH of the sea water decreases, the availability of carbonate ion decreases.

Explanation of Solution

The carbon dioxide present in the environment reacts with the water in the sea water and immediately forms carbonic acid.

${\text{H}}_{\text{2}}\text{O}+{\text{CO}}_{\text{2}}\underset{}{\to }{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$

The carbonic acid is unstable. It gets immediately dissociated into bicarbonate ion and hydrogen ion.

${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\underset{}{\rightleftharpoons }{\text{HCO}}_{\text{3}}{}^{-}+{\text{H}}^{+}$

So, when the carbon dioxide levels increase in the water and according to the first reaction the amount of carbonic acid increases. The increased amount of carbonic acid because of its instability it dissociates to different ions. This results in increase in hydrogen ions that result in increase in acidity of the sea water. At the same time the availability of carbonate ion decreases.

Thus, increase in pH of the seawater results in the decrease in the availability of carbonate ion.