Chapter 4, Problem 125IL

You wish to determine the weight percent of copper in a copper-containing alloy. After dissolving a 0.251 -g sample of the alloy in acid, an excess of KI is added, and the Cu²⁺ and I⁻ ions undergo the reaction

2 Cu²⁺(aq) + 5 I⁻(aq) → 2 CuI(s) + I₃⁻(aq)

The liberated I₃⁻ is titrated with sodium thiosulfate according to the equation

I₃⁻ (aq) + 2 S₂O₃²⁻(aq) → S₄O₆²⁻(aq) + 3 I⁻ (aq)

1. (a) Designate the oxidizing and reducing agents in the two reactions above.
2. (b) If 26.32 mL of 0.101 M Na₂S₂O₃ is required for titration to the equivalence point, what is the weight percent of Cu in the alloy?

Interpretation Introduction

Interpretation:

The weight percent of $\text{Cu}$ in the given sample of alloy, oxidizing and reducing agent in the given reactions should be determined.

Concept introduction:

• The relation between the number of moles and mass of the substance is ,

     $Numberofmole=\frac{Massingram}{Molarmass}$

     $Massingramofthesubs\tan ce=Numberofmole\times Molarmass$

• 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.
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.
• $\text{Amount}\text{of substance}=\text{Concentration}\text{of}\text{substance}\times \text{volume}\text{of}\text{the}\text{substance}$
• Weight percent of elements of a compound is the ratio of weight of element to the weight of whole compound and multiplied with hundred.
• Oxidizing agent is the reduced (gains electrons) atom to cause another atom to be oxidized
• Reducing agent is the oxidized (loses electron) atom to cause another atom to be reduced.

Answer to Problem 125IL

In the first reaction  $C{u}^{2+}$ acts as the oxidizing agent and $I^{-}$ acts as the reducing agent. And in second reaction $I_3^{-}$ is the oxidizing agent and $S_2{O}_3^{2-}$ is the reducing agent.

Explanation of Solution

The balanced equation for the first reaction is,

    $2C{u}_{(aq)}^{2+}+5I_{(aq)}^{-}\to 2Cu{I}_{(s)}+I_{(aq)}^{-}$

In this reaction $C{u}^{2+}$ acts as the oxidizing agent and $I^{-}$ acts as the reducing agent.

The balanced equation for the second given reaction is,

            $I_{3(aq)}^{-}+2S_2{O}_{3(aq)}^{2-}\to S_4{O}_{6(aq)}^{2-}+3I_{(aq)}^{-}$

In this reaction $I_3^{-}$ is the oxidizing agent and $S_2{O}_3^{2-}$ is the reducing agent.

Interpretation Introduction

Interpretation:

The weight percent of $\text{Cu}$ in the given sample of alloy, oxidizing and reducing agent in the given reactions should be determined.

Concept introduction:

• The relation between the number of moles and mass of the substance is ,

     $Numberofmole=\frac{Massingram}{Molarmass}$

     $Massingramofthesubs\tan ce=Numberofmole\times Molarmass$

• 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.
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.
• $\text{Amount}\text{of substance}=\text{Concentration}\text{of}\text{substance}\times \text{volume}\text{of}\text{the}\text{substance}$
• Weight percent of elements of a compound is the ratio of weight of element to the weight of whole compound and multiplied with hundred.
• Oxidizing agent is the reduced (gains electrons) atom to cause another atom to be oxidized
• Reducing agent is the oxidized (loses electron) atom to cause another atom to be reduced.

Answer to Problem 125IL

The weight percent of $\text{Cu}$ in the given sample of alloy is $\text{67}\text{.3}\text{\%}$.

Explanation of Solution

Herein $0.251g$ sample of alloy that contains copper is dissolved in acid, an excess of $KI$ is added. And the following reaction will take place.

                  $2C{u}_{(aq)}^{2+}+5I_{(aq)}^{-}\to 2Cu{I}_{(s)}+I_{(aq)}^{-}$

The liberated $I_3^{-}$ is titrated against sodium thiosulfate according to the equation,

             $I_{3(aq)}^{-}+2S_2{O}_{3(aq)}^{2-}\to S_4{O}_{6(aq)}^{2-}+3I_{(aq)}^{-}$

The amount of $N{a}_{2}S{2}_{}{O}_3$ in $26.32mLof0.101MN{a}_2{S}_2{O}_3$ can be calculated as follows,

$\text{Amount}{\text{of Na}}_{\text{2}}\text{S}{\text{2}}_{}{\text{O}}_{\text{3}}=\text{Concentration}{}_{{\text{Na}}_{\text{2}}\text{S}{\text{2}}_{}{\text{O}}_{\text{3}}}\text{×volume}{}_{{\text{Na}}_{\text{2}}\text{S}{\text{2}}_{}{\text{O}}_{\text{3}}}$

                             $\frac{0.101\text{mol}}{\text{L}}\text{×0}\text{.02632}\text{L}\text{=}\text{0}\text{.002658}\text{moles }$

From the balanced equation it is clear that $N{a}_{2}S{2}_{}{O}_3$ and $Na{I}_3$ are reacts in $2:1$ ratio. Therefore the amount of $Na{I}_3$ is,

        $0.002658N{a}_{2}S{2}_{}{O}_3\times \frac{1molNa{I}_3}{2molN{a}_{2}S{2}_{}{O}_3}=0.001329mol$

Similarly the amount of Copper alloy can be determined from the amount of $Na{I}_3$

Therefore,

   $0.001329Na{I}_3\times \frac{2molCu}{1molNa{I}_3}=0.002658mole$

Mass of copper involved in this reaction is $0.169g$. And it is obtained by multiplying the amount of copper with its molar mass $63.54g/mol$.

The weight percent of $Cu$ = $\frac{0.169g}{0.251g}\times 100$

                    =    $\text{67}\text{.3}\text{\%}$