Chapter 11, Problem 69QP

Drinking water may contain a low concentration of lead ion, ${\text{Pb}}^{\text{2+}},$ due to corrosion of old lead pipes. The EPA has determine that the maximum safe level ion in water is 15 ppb. Suppose a sample of tap water determine to have a lead ion concentration of 0.0090 ppm. Assume the density of the solution is 1.00 g/mL.

(a) What is the concentration of lead ion in the tap water in units of ppb? Is it safe to drink?

(b) What is the concentration of lead ion in units of units of mg/mL?

(c) What mass of lead ion is in 100.0 mL of this drinking water?

(d) How many moles of lead ion are in 100.0 mL of the water?

Interpretation Introduction

Interpretation:

The concentration of ${\text{Pb}}^{2+}$ ions in parts per billion $\left(\text{ppb}\right)$ and parts per million $(\text{ppm})$ in s sample of tap water is to be calculated.

Explanation of Solution

Explanation:

Given Information: The concentration of lead ions in tap water, $0.0090\text{ppm}$ .

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 $1{\text{billion (10}}^9\text{)}$ . Their relationship may be expressed as follows:

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

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

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

Substitute the ppm value in equation (3), to obtain the ppb value as follows:

$\begin{array}{c}\text{Parts per billion}=0.0090\times {10}^3\text{ppb}\\ =9.0\text{ppb}\end{array}$

Thus, the $\left(\text{ppb}\right)$ concentration of ${\text{Pb}}^{2+}$ in tap water is $9.0\text{ppb}$ .

Since EPA has determined the maximum permissible level of ${\text{Pb}}^{2+}$ ions in water to be $15.0\text{ppb}$ and the given sample of tap water contains only $9.0\text{ppb}$ of ${\text{Pb}}^{2+}$ ions, the given tap water is safe to drink.

Interpretation Introduction

Interpretation:

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

Explanation of Solution

Given Information: The $\left(\text{ppb}\right)$ concentration of lead ion in the tap water. It is equal to $0.0090\text{ppm}$ and density of the solution is $1.00{\text{g mL}}^{-1}$ .

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 the given aqueous solution, the density is $1.00{\text{g mL}}^{-1}$ . Parts per million may be taken equal to the milligrams of solute present per litre of solution. This relationship is expressed as follows:

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

Since $1\text{litre}=1000\text{millilitre}$ , the above equation is rearranged as follows:

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

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

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

It is given that the $\left(\text{ppb}\right)$ concentration of lead ions in tap water sample is $0.0090\text{ppm}$ . So, the ${\text{mg mL}}^{-1}$ concentration of lead ions is as follows:

$\begin{array}{c}{\text{Lead ion concentration in mg mL}}^{-1}=0.0090\times {10}^{-3}{\text{mg mL}}^{-1}\\ =\text{9}\text{.0}\times {\text{10}}^{\text{-6}}{\text{mg mL}}^{-1}\end{array}$

Hence the concentration of lead ions in ${\text{mg mL}}^{-1}$ in the given tap water sample is $\text{9}\text{.0}\times {\text{10}}^{\text{-6}}{\text{mg mL}}^{-1}$ .

Interpretation Introduction

Interpretation:

The mass of lead ions present in $100\text{mL}$ of $\text{9}\text{.0}\times {\text{10}}^{\text{-6}}{\text{mg mL}}^{-1}$ solution is to be determined.

Explanation of Solution

Given Information: $100\text{mL}$ of the solution has $\text{9}\text{.0}\times {\text{10}}^{\text{-6}}{\text{mg mL}}^{-1}$ concentration.

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

$\begin{array}{c}{\text{Mass of Pb}}^{\text{2+}}\text{ions}=\left(9.0\times {10}^{-6}{\text{mg mL}}^{-1}\right)\times 100\text{mL}\\ =9.0\times {10}^{-4}\text{mg}\end{array}$

So, the mass of lead ions present in $100\text{mL}$ of the solution is $9.0\times {10}^{-4}\text{mg}$ . The amount of lead ions in the unit of grams is calculated as follows:

$\begin{array}{c}{\text{Mass of Pb}}^{\text{2+}}\text{ion}=9.0\times {10}^{-4}\text{mg}\times \frac{1\text{g}}{1000\text{mg}}\\ =9.0\times {10}^{-7}\text{g}\end{array}$

Interpretation Introduction

Interpretation:

The moles of lead ions present in $100\text{mL}$ of $\text{9}\text{.0}\times {\text{10}}^{\text{-6}}{\text{mg mL}}^{-1}$ concentration is to be determined.

Explanation of Solution

Given Information: The mass of lead ions is $9.0\times {10}^{-7}\text{g}$ and the molar mass of lead is $207.2{\text{g mol}}^{-1}$ .

The number of moles is calculated as follows:

$\begin{array}{c}\text{Moles of lead ions}=\frac{\text{Mass in grams}}{\text{Molar mass in grams}}\\ =\frac{9.0\times {10}^{-7}\text{g}}{207.2{\text{g mol}}^{-1}}\\ =4.34\times {10}^{-9}\text{mol}\end{array}$

Hence, $100\text{mL}$ of the water contains $4.34\times {10}^{-9}\text{mol}$ of lead ions.