Chapter 8, Problem 8.17QA

Interpretation Introduction

To find:

Total mass of ions in $2.75 L$ of Mackenzie River water, where it contains $0.820 mM {Ca}^{2+}, 0.430 mM {Mg}^{2+}, 0.300 mM {Na}^{+}, 0.0200 M K^{+}, 0.250 mM {Cl}^{-},$

$0.380 mM S{O}_4^{2-}, 1.82 mM HC{O}_3^{-}$.

Answer to Problem 8.17QA

Solution:

Total mass of ions in $2.75 L$ of Mackenzie River water is $2.72 g$.

Explanation of Solution

1) Concept

To determine the mass of the solute needed to prepare a solution, we need to calculate the moles of the solute. We can calculate the moles of the solute from the molarity of the solution and volume.

A mole is the SI unit of an amount of chemical substance. When writing in units, it is written as “mol.”

2) Formula

i) $Molarity\left(M\right)= \frac{Moles of solute}{Volume in L}$

ii) $Molar mass = \frac{mass of substance}{mol}$

3) Given

i) $0.820 mM {Ca}^{2+}$

ii) $0.430 mM {Mg}^{2+}$

iii) $0.300 mM {Na}^{+}$

iv) $0.0200 M K^{+}$

v) $0.250 mM {Cl}^{-}$

vi) $0.380 mM S{O}_4^{2-}$

vii) $1.82 mM HC{O}_3^{-}$

4) Calculation

Calculate the mass of each ion from molarity and volume:

a) $0.820 mM {Ca}^{2+}$

Convert molarity from $mM$ to $M$.

$1 M=1000 mM$

$0.820 mM {Ca}^{2+}\times \frac{1 M}{1000 mM}=0.000820 M{ Ca}^{2+}$

Calculate the moles of ions.

$2.75 L\times \frac{0.000820 mol}{L{ Ca}^{2+}}=0.002255 mol {Ca}^{2+}$

Convert moles to mass using molar mass.

$0.002255 {Ca}^{2+}\times \frac{40.078 g}{1 mol}=0.09038 g {Ca}^{2+}$

b) $0.430 mM {Mg}^{2+}$

Convert molarity from $mM$ to $M.$

$0.430 mM {Mg}^{2+}\times \frac{1 M}{1000 mM}=0.000430 M$

Calculate moles of ${Mg}^{2+}$ in $2.75 L$.

$2.75 L \times \frac{0.000430 mol}{L {Mg}^{2+}}=0.0011825 mol{ Mg}^{2+}$

Convert moles to mass.

$0.0011825 {Mg}^{2+}\times \frac{24.0305 g}{1 mol}= 0.02874 g {Mg}^{2+}$

c) $0.300 mM {Na}^{+}$

Convert molarity from $mM$ to $M.$

$0.300 mM {Na}^{+}\times \frac{1 M}{1000 mM}=0.00030 M {Na}^{+}$

Calculate moles of ${Na}^{+}.$

$2.75 L \times \frac{0.00030 mol}{L {Na}^{+}}= 0.000825 mol {Na}^{+}$

Convert moles to mass.

$0.000825 mol {Na}^{+}\times \frac{22.99 g}{1 mol}=0.0190 g {Na}^{+}$

d) $0.0200 M K^{+}$

Calculate the moles of $K^{+}$.

$2.75 L \times \frac{0.0200 mol}{L K^{+}}=0.0550 mol K^{+}$

Convert moles to mass.

$0.0550 mol K^{+}\times \frac{39.098 g}{1 mol }=2.1504 g K^{+}$

e) $0.250 mM {Cl}^{-}$

Convert molarity from $mM$ to $M.$

$0.250 mM {Cl}^{-}\times \frac{1 M}{1000 mM}=0.000250 M {Cl}^{-}$

Calculate the moles of ${Cl}^{-}$.

$2.75 L \times \frac{0.000250 mol}{L {Cl}^{-}}= 0.0006875 mol {Cl}^{-}$

Convert moles to mass.

$0.0006875 mol {Cl}^{-}\times \frac{35.453 g}{1 mol}=0.02437 g {Cl}^{-}$

f) $0.380 mM S{O}_4^{2-}$

Convert molarity from $mM$ to $M.$

$0.380 mM S{O}_4^{2-}\times \frac{1 M}{1000 mM}=0.000380 M S{O}_4^{2-}$

Calculate the moles of $S{O}_4^{2-}$.

$2.75 L \times \frac{0.000380 mol}{L S{O}_4^{2-}}=0.001045 mol S{O}_4^{2-}$

Convert moles to mass using molar mass.

$0.001045 mol S{O}_4^{2-}\times \frac{96.061 g}{1 mol}=0.1004 g S{O}_4^{2-}$

g) $1.82 mM HC{O}_3^{-}$

Convert molarity from $mM$ to $M.$

$1.82 mM HC{O}_3^{-}\times \frac{1 M}{1000 mM}=0.00182 M HC{O}_3^{-}$

Calculate the moles of $HC{O}_3^{-}$.

$2.75 L \times \frac{0.00182 mol}{L HC{O}_3^{-}}= 0.00501 mol HC{O}_3^{-}$

Convert moles to mass using molar mass.

$0.00501 mol HC{O}_3^{-}\times \frac{61.016 g}{1 mol}=0.3057 g HC{O}_3^{-}$

Calculate the total mass.

$Total mass=0.09038 g+0.02874 g+0.0190 g+2.1504 g+0.02437 g+0.1004 g +0.3057 g$

$Total mass=2.72 g$

Total mass of ions in $2.75 L$ of Mackenzie River water is $2.72 g$.

Conclusion:

Using the molarity and volume in liter, calculate the moles of each ion, convert to mass in grams, and then calculate the total mass.