Chapter 4, Problem 45PS

Identify the ions that exist in each aqueous solution, and specify the concentration of each ion.

(a) 0.25 M(NH₄)₂SO₄

(b) 0.123 M Na₂CO₃

(c) 0.056 M HNO₃

Interpretation Introduction

Interpretation:

The ions that exist in aqueous solution of $\text{0}\text{.25 M }{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$ has to be determined.  Concentration of each ion has to be determined.

Concept introduction:

• Concentration of solutions can be expressed in various terms; molarity is one such concentration expressing term.
• Molarity (M) of a solution is the number of gram moles of a solute present in one liter of the solution.

  $\text{Molarity}\text{=}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{element}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}$

• A solution containing one gram mole or $\text{0}\text{.1}$ gram of solute per litre of solution is called molar solution.
• Amount of substance (mol) can be determined by using the equation,

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.

Answer to Problem 45PS

Aqueous solution of $\text{0}\text{.25 M }{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}$ contains ${\text{NH}}_{\text{4}}{}^{\text{+}}$ and ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ ions.  The molar concentration of each ion that exists in each of the given aqueous solution is $\text{0}\text{.50}\text{M}$${\text{NH}}_4{}^{+}$ and $\text{0}\text{.25}\text{M}$${\text{SO}}_4{}^{\text{2-}}.$

Explanation of Solution

One mole of ${{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}$ compound contains two moles of ${\text{NH}}_{\text{4}}{}^{\text{+}}$ and $\text{1}\text{mol}\text{of}{\text{SO}}_4{}^{\text{2-}}$.  In solution the compound dissociates and the balanced chemical reaction equation for this can be written as,

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

So the ions that exist in the $\text{0}\text{.25}\text{M}{{\text{(NH}}_{\text{4}}\text{)}}_{\text{2}}{\text{SO}}_{\text{4}}$ aqueous solution are ${\text{NH}}_{\text{4}}{}^{\text{+}}$ and ${\text{SO}}_4{}^{\text{2-}}$

Molar concentration of ammonium ion ${\text{[NH}}_{\text{4}}{}^{\text{+}}]$= $\text{2×0}\text{.25}\text{M}{\text{NH}}_4{{}^{+}}_{\text{(aq)}}\text{=}\text{0}\text{.50}\text{M}$

Molar concentration of sulfate ion $\text{[}{\text{SO}}_4{}^{\text{2-}}]$ =$\text{0}\text{.25}\text{M}$

Interpretation Introduction

Interpretation:

The ions that exist in aqueous solution of $\text{0}{\text{.123 M Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ has to be determined.  Concentration of each ion has to be determined.

Concept introduction:

• Concentration of solutions can be expressed in various terms; molarity is one such concentration expressing term.
• Molarity (M) of a solution is the number of gram moles of a solute present in one liter of the solution.

  $\text{Molarity}\text{=}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{element}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}$

• A solution containing one gram mole or $\text{0}\text{.1}$ gram of solute per litre of solution is called molar solution.
• Amount of substance (mol) can be determined by using the equation,

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.

Answer to Problem 45PS

Aqueous solution of $\text{0}{\text{.123 M Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ contains ${\text{Na}}^{\text{+}}$ and ${\text{CO}}_3{}^{\text{2-}}$ ions.  The molar concentration of each ion that exists in each of the given aqueous solution is $\text{0}\text{.246}\text{M}{\text{Na}}^{\text{+}}$ and  $\text{0}\text{.123}\text{M}{\text{CO}}_{\text{3}}{}^{\text{2-}}$

Explanation of Solution

One mole of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ compound contains two moles of ${\text{Na}}^{\text{+}}$ and $\text{1}\text{mol}\text{of}{\text{CO}}_{\text{3}}{}^{\text{2-}}$.  The balanced chemical reaction equation for this can be written as,

  ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(s)}}\stackrel{}{\to }{\text{2Na}}^{\text{+}}{}_{\text{(aq)}}\text{+}{\text{CO}}_{\text{3}}{{}^{\text{2-}}}_{\text{(aq)}}$

So the ions that exist in the $\text{0}\text{.123}\text{M}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ aqueous solution are ${\text{Na}}^{\text{+}}$ and ${\text{CO}}_{\text{3}}{}^{\text{2-}}$

Molar concentration of sodium ion ${\text{[Na}}^{\text{+}}\text{]}$= $\text{2×0}\text{.123}\text{M}{\text{Na}}^{\text{+}}{}_{\text{(aq)}}\text{=}\text{0}\text{.246}\text{M}$

Molar concentration of carbonate ion ${\text{[CO}}_3{}^{2-}\text{]}$ =$\text{0}\text{.123}\text{M}$

Interpretation Introduction

Interpretation:

The ions that exist in aqueous solution of $\text{0}\text{.056}{\text{M HNO}}_{\text{3}}$ has to be determined.  Concentration of each ion has to be determined.

Concept introduction:

• Concentration of solutions can be expressed in various terms; molarity is one such concentration expressing term.
• Molarity (M) of a solution is the number of gram moles of a solute present in one liter of the solution.

  $\text{Molarity}\text{=}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{element}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}$

• A solution containing one gram mole or $\text{0}\text{.1}$ gram of solute per litre of solution is called molar solution.
• Amount of substance (mol) can be determined by using the equation,

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.

Answer to Problem 45PS

Aqueous solution of $\text{0}\text{.056}{\text{M HNO}}_{\text{3}}$ contains ${\text{H}}^{+}$ and ${\text{NO}}_3{}^{\text{-}}$ ions.  The molar concentration of each ion that exists in each of the given aqueous solution is $\text{0}\text{.056}\text{M}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}},\text{0}\text{.056}\text{M}{\text{NO}}_{\text{3}}{}^{\text{-}}.$

Explanation of Solution

One mole of ${\text{HNO}}_{\text{3}}$ compound contains one moles of ${\text{H}}^{\text{+}}$ and $\text{1}\text{mol}\text{of}{\text{NO}}_{\text{3}}{}^{\text{-}}$.  The ${\text{H}}^{\text{+}}$ ion combines ${\text{H}}_{\text{2}}\text{O}$ and forms ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}.$  The balanced chemical reaction equation for this can be written as,

  ${\text{HNO}}_{\text{3}}{}_{\text{(l)}}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}\text{+}{\text{NO}}_{\text{3}}{{}^{\text{-}}}_{\text{(aq)}}$

So the ions that exist in the $\text{0}\text{.056}\text{M}{\text{HNO}}_{\text{3}}$ aqueous solution are ${\text{H}}^{\text{+}}$ and ${\text{NO}}_{\text{3}}{}^{\text{-}}$

Molar concentration of hydronium ion ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$= $\text{1×0}\text{.056}\text{M}{\text{H}}^{\text{+}}{}_{\text{(aq)}}\text{=}\text{0}\text{.056}\text{M}$

Molar concentration of nitrate ion ${\text{[NO}}_{\text{3}}{}^{\text{2-}}\text{]}$ =$\text{1×0}\text{.056}\text{M}{\text{NO}}_{\text{3}}{{}^{\text{-}}}_{\text{(aq)}}\text{=}\text{0}\text{.056}\text{M}$