Chapter 12, Problem 12.119QP

(a)

Interpretation Introduction

Interpretation:

Given that limit for Barium in drinking water is $2\text{ ppm}$, Mass percent of Barium allowed in drinking water has to be calculated.

Concept Introduction:

Mass percent is one of parameters used to express concentration of a solution.  It is expressed as,

$\text{Mass percent of a solute = }\frac{mass\text{ of solute}}{mass\text{ of solution}}\times \text{ 100\%}$

Parts per million (ppm) is a term used to express the concentration of an element or any other substance in a material.  It is expressed as,

$\text{ppm = }\frac{mass\text{ of solute}}{mass\text{ of solution}}\times {\text{ 10}}^6$

Answer to Problem 12.119QP

Mass percentage of permissible level of Barium in drinking water is $2\times 10^{-4}\%.$

Explanation of Solution

Parts per million is expressed as,

$\text{ppm = }\frac{mass\text{ of solute}}{mass\text{ of solution}}\times {\text{ 10}}^6$

The term $\frac{mass\text{ of solute}}{mass\text{ of solution}}$ corresponds to mass percent of the solute.   By comparing and rewriting the two expressions, $2ppm$ Barium is written in mass percent as -

$\begin{array}{l}2ppm=\frac{mass\text{ of Ba}}{mass\text{ of Solution}}\times {10}^6\\ \frac{mass\text{ of solute}}{mass\text{ of solution}}=\text{ 2}\times {\text{10}}^{-6}\\ mass\text{ percent}\text{= 2}\times {\text{10}}^{-6}\times {10}^2\text{ g}\\ \text{= 2}\times {\text{10}}^{-4}\%\end{array}$

(b)

Interpretation Introduction

Interpretation:

Given that limit for Barium in drinking water is $2\text{ ppm}$, Molarity of solution of Barium salt of $2\text{ ppm}$ has to be calculated given the density of solution as $\text{1}\text{.0 g/mL}\text{.}$

Concept Introduction:

Mass percent is one of parameters used to express concentration of a solution.  It is expressed as,

$\text{Mass percent of a solute = }\frac{mass\text{ of solute}}{mass\text{ of solution}}\times \text{ 100\%}$

Parts per million (ppm) is a term used to express the concentration of an element or any other substance in a material.  It is expressed as,

$\text{ppm = }\frac{mass\text{ of solute}}{mass\text{ of solution}}\times {\text{ 10}}^6$

Answer to Problem 12.119QP

Molarity of solution of $2ppm$ Barium is calculated as $1.4\times 10^{-5}M.$

Explanation of Solution

Consider $1\text{ L}$ of the solution.  Density of the solution is $1.00\text{ g/mL}\text{.}$ then mass of the solution would be ${10}^3\text{ g}\text{.}$  calculate mass and number of moles of Barium as follows -

$\begin{array}{l}2ppm=\frac{mass\text{ of Ba}}{mass\text{ of Solution}}\times {10}^6\\ =\frac{2\text{ ppm}}{{10}^6}\times mass\text{ of solution}\\ \text{= }\frac{2\text{ ppm}}{{10}^6}\times {10}^3\text{ g}\\ \text{= 0}\text{.002 g Ba}\\ \\ \text{moles of Ba}\text{= 0}\text{.002 g Ba}\times \frac{1\text{ mol Ba}}{137.3\text{ g Ba}}\\ =\text{ 1}\text{.4 }\times {\text{10}}^{-5}\text{ mol Ba}\end{array}$

Molarity of solution containing $2\text{ ppm}$ Barium is thus calculated as,

$\begin{array}{l}molarity=\frac{mol\text{ Ba}}{liters\text{ of solution}}\\ =\frac{1.4\times {10}^{-5}mol\text{ Ba}}{1\text{ L}}\\ =\text{ 1}\text{.4}\times {\text{10}}^{-5}M\end{array}$

(c)

Interpretation Introduction

Interpretation:

Given that limit for Barium in drinking water is $2\text{ ppm}$, the concentration of Barium has to be expressed in mg per liter.

Concept Introduction:

Mass percent is one of parameters used to express concentration of a solution.  It is expressed as,

$\text{Mass percent of a solute = }\frac{mass\text{ of solute}}{mass\text{ of solution}}\times \text{ 100\%}$

Parts per million (ppm) is a term used to express the concentration of an element or any other substance in a material.  It is expressed as,

$\text{ppm = }\frac{mass\text{ of solute}}{mass\text{ of solution}}\times {\text{ 10}}^6$

Answer to Problem 12.119QP

Concentration of Ba in mg per liter is $2mg/L.$

Explanation of Solution

Previously we have calculated that $1\text{ L}$ of the solution contains $0.002\text{ g}$ of Barium.  It is equivalent to $2\text{ mg}$ of Barium.  Thus we could write that concentration of Barium in mg per liter is $2mg/L.$