Chapter 6, Problem 32E

a)

Interpretation Introduction

Interpretation: The concentration of ions in the solution has to be calculated.

Concept Introduction: Concentration can be defined in terms of molarity as moles of solute per volume of solution in litres. It can be given by the expression,

$\text{Molarity(in}\text{M)=}\frac{\text{mass}\text{of}\text{solute(in g)}}{\text{volume}\text{of}\text{solution(in}\text{L)}}$

Answer to Problem 32E

0.0200 mole of Sodium phosphate in 10.0 mL of solution.

${\text{M}}_{{\text{Na}}^{\text{+}}}\text{6}\text{.00}\text{M}{\text{M}}_{{\text{PO}}_{\text{4}}{}^{\text{3-}}}\text{2}\text{.00}\text{M}$

Explanation of Solution

Record the given data

Moles of Sodium phosphate= $\text{0}\text{.200}\text{moles}$

Volume of the solution= $\text{10}\text{.0}\text{mL}$

Calculation for the concentration of separate ions is as follows,

The balanced equation for dissolving ions,

${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4(s)}}\to {\text{3Na}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{+PO}}_{\text{4}}{{}^{\text{3-}}}_{\left(\text{aq}\right)}$

The molarity of ions can be calculated by the formula,

$\text{Molarity=}\frac{\text{mass}\text{of}\text{solute}}{\text{volume}\text{of}\text{solution}}$

Therefore, by substituting the given info in the formula, the concentrations of separate ions can be given as,

$\begin{array}{l}\text{Molarity=}\frac{\text{0}\text{.0200}}{\text{0}\text{.0100}}\\ \text{=2}\text{.00M}\end{array}$

In order, to find the concentration of the separate ion,

$\text{Concentration}\text{of}\text{individual}\text{ion=}\text{number}\text{of}\text{atoms}\times \text{the}\text{calculated}\text{molarity}$

Therefore, the concentrations of separate ions can be given by,

$\begin{array}{l}{\text{3Na}}^{\text{+}}\text{=3}\times \text{2}\text{.00=6}\text{.00}{\text{M}}_{{\text{Na}}^{\text{+}}}\\ {\text{PO}}_{\text{4}}{}^{\text{3-}}=\text{1}\times \text{3}\text{.00=}\text{2}\text{.00}{\text{M}}_{{\text{PO}}_{\text{4}}{}^{\text{3-}}}\end{array}$

The concentration of the individual ions is calculated by plugging in the values of molarity and number of atoms present in element. The concentration of individual ${\text{Na}}^{\text{+}}$ and ${\text{PO}}_{\text{4}}{}^{\text{3-}}$ are found to be $\text{6}\text{.00}{\text{M}}_{{\text{Na}}^{\text{+}}}$ and $\text{2}\text{.00}{\text{M}}_{{\text{PO}}_{\text{4}}{}^{\text{3-}}}$ respectively.

b)

Interpretation Introduction

Interpretation: The concentration of ions in the solution has to be calculated.

Concept Introduction: Concentration can be defined in terms of molarity as moles of solute per volume of solution in litres. It can be given by the expression,

$\text{Molarity(in}\text{M)=}\frac{\text{mass}\text{of}\text{solute(in g)}}{\text{volume}\text{of}\text{solution(in}\text{L)}}$

Answer to Problem 32E

0.300 mole of Barium nitrate in 600 mL of solution.

${\text{M}}_{{\text{Ba}}^{\text{2+}}}\text{0}\text{.500}\text{M}{\text{M}}_{{\text{NO}}_{\text{3}}{}^{\text{-}}}\text{1}\text{.00}\text{M}$

Explanation of Solution

Record the given data

Moles of Barium nitrate= $\text{0}\text{.300}\text{mole}$

Volume of the solution= $\text{600}\text{mL}$

Calculation for the concentration of separate ions is as follows,

The balanced equation for dissolving ions

${\text{Ba(NO}}_{\text{3}}{\text{)}}_{\text{2}\left(\text{s}\right)}\to {\text{Ba}}^{\text{2+}}{\text{+2NO}}_{\text{3}}{}^{\text{2-}}$

The molarity of ions can be calculated by the formula,

$\text{Molarity=}\frac{\text{mass}\text{of}\text{solute}}{\text{volume}\text{of}\text{solution}}$

Therefore, by substituting the given info in the formula, molarity can be calculated by,

$\begin{array}{l}\text{Molarity=}\frac{\text{0}\text{.300}}{\text{0}\text{.600}}\\ \text{=0}\text{.500M}\end{array}$

In order, to find the concentration of the separate ion,

$\text{Concentration}\text{of}\text{individual}\text{ion=}\text{number}\text{of}\text{atoms}\times \text{the}\text{calculated}\text{molarity}$

Therefore, the concentrations of separate ions can be given by,

$\begin{array}{l}{\text{Ba}}^{\text{2+}}\text{=1}\times \text{0}\text{.5=0}\text{.5}{\text{M}}_{{\text{Ba}}^{\text{2+}}}\\ {\text{NO}}_{\text{3}}{}^{\text{-}}=\text{2}\times \text{0}\text{.5=}\text{1}\text{.00}{\text{M}}_{{\text{NO}}_{\text{3}}{}^{\text{-}}}\end{array}$

The concentration of the individual ions is calculated by plugging in the values of molarity and number of atoms present in element. The concentration of individual ${\text{Ba}}^{\text{2+}}$ and ${\text{NO}}_{\text{3}}{}^{\text{-}}$ are $\text{0}\text{.5}{\text{M}}_{{\text{Ba}}^{\text{2+}}}$ and $\text{1}\text{.00}{\text{M}}_{{\text{NO}}_{\text{3}}{}^{\text{-}}}$ respectively.

c)

Interpretation Introduction

Interpretation: The concentration of ions in the solution has to be calculated.

Concept Introduction: Concentration can be defined in terms of molarity as moles of solute per volume of solution in litres. It can be given by the expression,

$\text{Molarity(in}\text{M)=}\frac{\text{mass}\text{of}\text{solute(in g)}}{\text{volume}\text{of}\text{solution(in}\text{L)}}$

Answer to Problem 32E

1.00 g of Potassium chloride in 0.500 L of solution.

${\text{M}}_{{\text{K}}^{\text{+}}}\text{0}\text{.0268}\text{M}{\text{M}}_{{\text{Cl}}^{\text{-}}}\text{0}\text{.0268M}$

Explanation of Solution

Calculation

Record the given info

Mole of Potassium chloride=1.00 gram

Volume of solution= 0.500 L

for the concentration of individual ions is as follows,

The balanced equation can be given as,

$\text{KCl}\to {\text{K}}^{\text{+}}{\text{+Cl}}_{}{}^{\text{-}}$

In order to calculate molarity, grams are converted into moles by using the molar mass.

The molar mass can be calculated by sum of mass of individual elements.

Molar mass of Potassium chloride can be given as (1×40) + (1×35.5) =74.55 g/mol

Amount of Potassium chloride= $\frac{\text{Mass}}{\text{Molar}\text{Mass}}$

                                                  = $\begin{array}{l}\frac{\text{1}}{\text{74}\text{.55}}\\ \text{=0}\text{.01341}\text{mol}\end{array}$

$\begin{array}{l}\text{Concentration=}\frac{\text{amount}}{\text{volume}}\\ \end{array}$ $\begin{array}{l}\text{=}\frac{\text{0}\text{.01341}}{\text{0}\text{.500}}\\ \text{=0}\text{.0268}\text{M}\end{array}$

In order, to find the concentration of the separate ion,

$\text{Concentration}\text{of}\text{individual}\text{ion=}\text{number}\text{of}\text{atoms}\times \text{the}\text{calculated}\text{molarity}$

Therefore, the concentrations of separate ions can be given by,

$\begin{array}{l}{\text{K}}^{\text{+}}=\text{1}\times \text{0}\text{.0268=0}\text{.0268}{\text{M}}_{{\text{K}}^{\text{+}}}\\ {\text{Cl}}^{\text{-}}=\text{1}\times \text{0}\text{.0268=}\text{0}\text{.0268}{\text{M}}_{{\text{Cl}}^{\text{-}}}\end{array}$

The concentration of the individual ions is calculated by plugging in the values of molarity and number of atoms present in element. The concentration of individual ${\text{K}}^{\text{+}}$ and ${\text{Cl}}^{\text{-}}$ are $\text{0}\text{.0268}{\text{M}}_{{\text{K}}^{\text{+}}}$ and $\text{0}\text{.0268}{\text{M}}_{{\text{Cl}}^{\text{-}}}$

d)

Interpretation Introduction

Interpretation: The concentration of ions in the solution has to be calculated.

Concept Introduction: Concentration can be defined in terms of molarity as moles of solute per volume of solution in litres. It can be given by the expression,

$\text{Molarity(in}\text{M)=}\frac{\text{mass}\text{of}\text{solute(in g)}}{\text{volume}\text{of}\text{solution(in}\text{L)}}$

Answer to Problem 32E

132 g of Ammonium sulphate in 1.50 L of solution.

${\text{M}}_{{\text{NH}}_{\text{4}}{}^{\text{+}}}\text{0}\text{.66}\text{M}{\text{M}}_{{\text{SO}}_{\text{4}}{}^{\text{2-}}}\text{1}\text{.32}\text{M}$

Explanation of Solution

Record the given info

Mole of Ammonium sulphate=132 gram

Volume of solution=1.50L

Calculation for the concentration of individual ions is as follows,

The balanced equation for dissolving ions,

${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{2NH}}_{\text{4}}{}^{\text{2+}}{\text{+SO}}_{\text{4}}{}^{\text{2-}}$

In order to calculate molarity, grams are converted into moles by using the molar mass.

The molar mass can be calculated by sum of mass of individual elements.

Molar mass of Ammonium sulphate is 132 g/mol

$\text{Amount}\text{of}\text{Ammonium}\text{sulphate}$= $\frac{\text{Mass}}{\text{Molar}\text{Mass}}$

$\text{Amount}\text{of}\text{Ammonium}\text{sulphate}\text{=}\frac{\text{132}}{\text{132}}\text{=1}\text{.00}\text{mol}$

Therefore, from the amount of Ammonium sulphate, the molarity can be calculated as,

$\begin{array}{l}\text{Concentration=}\frac{\text{amount}}{\text{volume}}\\ \end{array}$

$\begin{array}{l}\text{Concentration=}\frac{\text{1}}{\text{1}\text{.5}}\\ \text{=0}\text{.666}\text{M}\end{array}$

In order, to find the concentration of the separate ion,

$\text{Concentration}\text{of}\text{individual}\text{ion=}\text{number}\text{of}\text{atoms}\times \text{the}\text{calculated}\text{molarity}$

Therefore, the concentrations of separate ions can be given by,

$\begin{array}{l}{\text{NH}}_{\text{4}}{}^{\text{+}}\text{=2×0}\text{.666=}\text{1}\text{.32}{\text{M}}_{{\text{NH}}_{\text{4}}{}^{\text{+}}}\\ {\text{SO}}_{\text{4}}{}^{\text{2-}}\text{=1×0}\text{.666=}\text{0}\text{.666}{\text{M}}_{{\text{SO}}_{\text{4}}{}^{\text{2-}}}\end{array}$

The concentration of the individual ions is calculated by plugging in the values of molarity and number of atoms present in element. The concentration of individual ${\text{NH}}_{\text{4}}{}^{\text{+}}$ and ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ are

$\text{1}\text{.32}{\text{M}}_{{\text{NH}}_{\text{4}}{}^{\text{+}}}$ and $\text{0}\text{.666}{\text{M}}_{{\text{SO}}_{\text{4}}{}^{\text{2-}}}$.