Chapter 11, Problem 70QP

(a)

Interpretation Introduction

Interpretation:

The concentration in parts per million $(\text{ppm})$ of nitrate ions from parts per billion $(\text{ppb})$ in the given sample of tap water is to be calculated.

Explanation of Solution

Given Information: The $(\text{ppb})$ concentration of nitrate ions in tap water is $95\text{ppb}$ .

Parts per million is the ratio of the mass of solute and the mass of solution multiplied by $1{\text{million (10}}^6\text{)}$ . The relationship may be expressed as follows:

$\text{Parts per million}=\frac{\text{Grams of solute}}{\text{Grams of solution}}\times {10}^6\text{ ppm}$ (1)

Parts per billion is the ratio of the mass of solute and the mass of solution multiplied by $1{\text{billion (10}}^9\text{)}$ . The relationship is expressed as follows:

$\text{Parts per million}=\frac{\text{Grams of solute}}{\text{Grams of solution}}\times {10}^9\text{ ppb}$ ……(2)

Divide equation (1) by equation (2), to get the following relationship:

$\text{ppm}=\text{ppb}\times {10}^{-3}$ ……(3)

Substitute the ppb value in equation (3) to obtain the $(\text{ppm})$ value as follows:

$\begin{array}{c}\text{Parts per million}=95\times {10}^{-}{}^3\text{ppm}\\ =0.095\text{ppm}\end{array}$

So, the $(\text{ppm})$ concentration of nitrate ions in tap water is $0.095\text{ppm}$ .

Since EPA has fixed the maximum permissible level of nitrate ions in water to be $10.0\text{ppm}$ and the given tap water contains $0.095\text{ppm}$ of nitrate ions, the tap water is safe to drink.

(b)

Interpretation Introduction

Interpretation:

The concentration of nitrate ions in ${\text{mg mL}}^{-1}$ from the concentration in parts per million in the given sample of tap water is to be calculated.

Explanation of Solution

Given Information: The $\left(\text{ppm}\right)$ concentration of nitrate ions in tap water is $0.095\text{ppm}$ and the density of solution is $1.0{\text{mg mL}}^{-1}$ .

Parts per million is the ratio of the mass of solute and the mass of solution multiplied by $1{\text{million (10}}^6\text{)}$ . For an aqueous solution with a density equal to $1.0{\text{mg mL}}^{-1}$ , parts per million may be taken equal to the milligrams of solute present per litre of solution. This relationship may be expressed as follows:

$\text{Parts per million}=\frac{\text{Mass of solute in milligrams}}{\text{Volume of solution in litres}}$

Since $1\text{litre}=1000\text{millilitres}$ , the above equation may be rearranged as follows:

$\begin{array}{c}\text{Parts per million}=\frac{\text{Mass of solute in milligrams}}{\text{Volume of solution in millilitres}\times {\text{10}}^3}\\ \text{Parts per million}\times {\text{10}}^3=\frac{\text{Mass of solute in milligrams}}{\text{Volume of solution in millilitres}}\\ \text{1000}\text{ppm}=1{\text{ mg mL}}^{-1}\mathrm{......}\left(4\right)\end{array}$

So, the above equation may be written in the following manner also:

$\text{1}\text{ppm}={10}^{-3}{\text{ mg mL}}^{-1}$

Given $\left(\text{ppm}\right)$ concentration of nitrate ions tap water is $0.095\text{ppm}$ . So, the ${\text{mg mL}}^{-1}$ concentration is as follows:

$\begin{array}{c}=0.095\times {10}^{-3}{\text{mg mL}}^{-1}\\ =\text{9}\text{.5}\times {\text{10}}^{\text{-5}}{\text{mg mL}}^{-1}\end{array}$

Hence the concentration of nitrate ions in ${\text{mg mL}}^{-1}$ is $\text{9}\text{.5}\times {\text{10}}^{\text{-5}}{\text{mg mL}}^{-1}$ .

(c)

Interpretation Introduction

Interpretation:

The mass of nitrate ions present in $100\text{mL}$ of $\text{9}\text{.5}\times {\text{10}}^{\text{-5}}{\text{mg mL}}^{-1}$ nitrate ion solution is to be determined.

Explanation of Solution

Given Information: $100\text{mL}$ of the solution has $\text{9}\text{.5}\times {\text{10}}^{\text{-5}}{\text{mg mL}}^{-1}$ nitrate ion concentration.

The mass of lead ions present in $100\text{mL}$ of the solution is as follows:

$\begin{array}{c}\text{Mass of nitrate ions}=\text{9}\text{.5}\times {\text{10}}^{\text{-5}}{\text{mg mL}}^{-1}\times 100\text{mL}\\ =9.5\times {10}^{-3}\text{mg}\end{array}$

So, the mass of nitrate ions present is $\text{9}\text{.5}\times {\text{10}}^{\text{-3}}\text{mg}$ . The amount of nitrate ions in grams is as follows:

$\begin{array}{c}\text{Mass of nitrate ions}=9.5\times {10}^{-3}\text{mg}\times \frac{1\text{g}}{1000\text{mg}}\\ =9.5\times {10}^{-6}\text{g}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The moles of nitrate ions present in $100\text{mL}$ of $\text{9}\text{.5}\times {\text{10}}^{\text{-5}}{\text{mg mL}}^{-1}$ nitrate ion solution is to be determined.

Explanation of Solution

Given Information: The mass of nitrate ions is $9.5\times {10}^{-6}\text{g}$ and the molar mass of nitrate ions is $62.004{\text{g mol}}^{-1}$ .

The number of moles is calculated as follows:

$\begin{array}{c}\text{Moles of nitrate ion}=\frac{\text{Mass in grams}}{\text{Molar mass in grams}}\\ =\frac{9.5\times {10}^{-6}\text{g}}{62.004{\text{g mole}}^{-1}}\\ =1.53\times {10}^{-7}\text{mole}\end{array}$

Hence, $100\text{mL}$ of the water contains $1.53\times {10}^{-7}\text{mole}$ of nitrate ions.