Chapter 23, Problem 23.8P

(a)

Interpretation Introduction

Interpretation:

$\text{pH}$ of the cloud water has to be calculated.

Concept Introduction:

$\text{pH}$ is scale to measure acidity and basicity of water based solution. $\text{pH}$ of a solution means negative logarithm of activity of ${\text{H}}^{+}$ ion with base 10. For dilute solution, $\text{pH}$ is mathematically represented as follows:

    $\text{pH}=-{\log }_{10}\left[{\text{H}}^{\text{+}}\right]$

Here, $\left[{\text{H}}^{\text{+}}\right]$ is the concentration of ${\text{H}}^{+}$ ion in solution.

Answer to Problem 23.8P

$\text{pH}$ of cloud water is $3.883$.

Explanation of Solution

$\text{pH}$ of cloud water is calculated as follows:

    $\begin{array}{c}\text{pH}=-{\log }_{10}\left[{\text{H}}^{+}\right]\\ =-{\log }_{10}\left(131\times {10}^{-6}\right)\\ =3.883\end{array}$

Hence, $\text{pH}$ of cloud water is $3.883$.

(b)

Interpretation Introduction

Interpretation:

Anion charges and cation charges present in rain water have to be calculated. Whether the magnitude of anion charges and cation charges are same, that has to be explained. Quality of analysis has to be predicted.

Concept Introduction:

Total anionic charge is the total charge carried by all anions and total cationic charge is the total charge carried by all cations.

Explanation of Solution

Total anionic charges are calculated as follows:

    $\begin{array}{c}\text{Total}\text{anionic}\text{charge}=\left(\text{Total charge}\text{carried}\text{by}{\text{Cl}}^{-}\right)+\left(\text{Total charge}\text{carried}\text{by}{\text{NO}}_3^{-}\right)\\ +\left(\text{Total charge}\text{carried}\text{by}{\text{SO}}_4^{-}\right)\\ =\left(-1\right)\left(\text{101}\text{μM}\right)+\left(-1\right)\left(360\text{μM}\right)+\left(-2\right)\left(156\text{μM}\right)\\ =-773\text{μM}\end{array}$

Total cationic charges are calculated as follows:

    $\begin{array}{c}\text{Total}\text{cationic}\text{charges}=\left(\text{Total}\text{charge}\text{carried}\text{by}{\text{H}}^{+}\right)+\left(\text{Total}\text{charge}\text{carried}\text{by}{\text{Na}}^{+}\right)\\ +\left(\text{Total}\text{charge}\text{carried}\text{by}{\text{NH}}_4^{+}\right)+\left(\text{Total}\text{charge}\text{carried}\text{by}{\text{K}}^{+}\right)\\ +\left(\text{Total}\text{charge}\text{carried}\text{by}{\text{Ca}}^{2+}\right)+\left(\text{Total}\text{charge}\text{carried}\text{by}{\text{Mg}}^{2+}\right)\\ =\left(+1\right)\left(131\text{μM}\right)+\left(+1\right)\left(100\text{μM}\right)+\left(+1\right)\left(472\text{μM}\right)\\ +\left(+1\right)\left(1.3\text{μM}\right)+\left(+2\right)\left(26\text{μM}\right)+\left(+2\right)\left(12\text{μM}\right)\\ =780.3\text{μM}\end{array}$

Hence magnitude of total anionic charge and total cationic charges are not exactly same.

The cloud water has slight excess of positive charge but practically it should be neutral. So the analysis was not $100\%$ accurate.

(c)

Interpretation Introduction

Interpretation:

Total mass of dissolved ion in cloud water has to be calculated.

Concept Introduction:

Total concentration of dissolved ion in a solution is calculated as follows:

    $\text{Concentration}\text{of}\text{dissolved ion}\left(\text{in}\text{mg}{\text{mL}}^{-1}\right)=\left[{\displaystyle \sum \begin{array}{l}\left(\begin{array}{l}\text{Concentraion}\text{of}\\ \text{ion in}\text{molarity}\end{array}\right)\\ \left(\text{Molar}\text{mass}\text{of}\text{ion}\right)\end{array}}\right]$

Answer to Problem 23.8P

Hence mass of ions present in $\text{1}\text{mL}$ cloud water sample is $\text{0}\text{.0512682}\text{mg}$.

Explanation of Solution

Total concentration of dissolved ion (in $\text{mg}{\text{mL}}^{-1}$)  in cloud water is calculated as follows:

    $\begin{array}{c}\text{Concentration}\text{of}\text{ion}=\left(\text{101}\times {10}^{-6}\text{M}\right)\left(35.5\text{g}{\text{mol}}^{-1}\right)+\left(360\times {10}^{-6}\text{M}\right)\left(62\text{g}{\text{mol}}^{-1}\right)\\ +\left(156\times {10}^{-6}\text{M}\right)\left(96\text{g}{\text{mol}}^{-1}\right)+\left(131\times {10}^{-6}\text{M}\right)\left(1\text{g}{\text{mol}}^{-1}\right)\\ +\left(100\times {10}^{-6}\text{M}\right)\left(23\text{g}{\text{mol}}^{-1}\right)+\left(472\times {10}^{-6}\text{M}\right)\left(14\text{g}{\text{mol}}^{-1}\right)\\ +\left(1.3\times {10}^{-6}\text{M}\right)\left(39\text{g}{\text{mol}}^{-1}\right)+\left(26\times {10}^{-6}\text{M}\right)\left(40\text{g}{\text{mol}}^{-1}\right)\\ +\left(12\times {10}^{-6}\text{M}\right)\left(24\text{g}{\text{mol}}^{-1}\right)\\ =51268.2\times {10}^{-6}\text{g}{\text{L}}^{-1}\\ =0.0512682\text{mg}{\text{mL}}^{-1}\end{array}$

Hence mass of ions present in $\text{1}\text{mL}$ cloud water sample is $\text{0}\text{.0512682}\text{mg}$.