EBK EXPLORING CHEMICAL ANALYSIS
EBK EXPLORING CHEMICAL ANALYSIS
5th Edition
ISBN: 9781319416942
Author: Harris
Publisher: VST
Question
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Chapter 17, Problem 17.26P

(a)

Interpretation Introduction

Interpretation:

pH range for predominant sulfurous acid forms has to be found.

Concept Introduction:

pH is calculated by Henderson-HasselBalch equation given as follows:

  pH=pKa+log([Salt][Acid])

(a)

Expert Solution
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Explanation of Solution

Equilibrium for dissociation of is given as follows:

  H2SO3HSO3+H+

Value of pH is calculated by Henderson-HasselBalch equation given as follows:

  pH=pKa+log[HSO3][H2SO3]        (1)

Substitute 1.857 for pKa in equation (1).

  pH=1.857+log[HSO3][H2SO3]        (2)

Equation (2) indicates there is large [H2SO3] relative to [HSO3] at pH of 1.857.

Value of pH is calculated by Henderson-HasselBalch equation given as follows:

  pH=pKa+log[SO32][HSO3]        (3)

Substitute 7.172 for pKa in equation (3).

  pH=7.172+log[SO32][HSO3]        (4)

Equation (4) indicates there is large [SO32] relative to [HSO3] at pH of 7.172. In between the two pH, the intermediate form [HSO3] is major species.

(b)

Interpretation Introduction

Interpretation:

Balanced cathodic and anodic half-reactions have to be found.

Concept Introduction:

Cell that is more negatively charge has better oxidation tendency and thus will readily lose electron. Electrons released from this electrode will reach the positive reduction potential electrode.

(b)

Expert Solution
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Explanation of Solution

Half cell reaction for reduction of water that occurs at cathode is written as follows:

  H2O(s)+e12H2(g)+OH

Half cell reaction for anodic compartment when iodide gets oxidized to triodide ion is written as follows:

  3I2I3+2e

(c)

Interpretation Introduction

Interpretation:

Balanced reactions amongst I3 -HSO3 and  I3 -S2O32 has to be written.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
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Explanation of Solution

Unbalanced reaction is formulated as follows:

  I3+HSO3I+SO42

Add 3 as coefficient of I to get equal I on both sides.

  I3+HSO33I+SO42

Since there are 4 O atoms on right, add one water molecule to left to have equal O atoms.

  I3+HSO3+H2O3I+SO42

Since there are 3 H atoms on right, add three H+ ions to right to have equal H atoms.

Finally, verify each side has equal charge. Thus balanced reaction amongst I3 and HSO3 is written as follows:

  I3+HSO3+H2O3I+SO42+3H+

Unbalanced reaction between I3 and S2O32 is formulated as follows:

  I3+S2O32I+S4O62

Add 3 as coefficient of I and 2 as coefficient of S2O32 to get equal  I , S and S4O on both sides.

  I3+2S2O323I+S4O62

(d)

Interpretation Introduction

Interpretation:

Concentration of total sulfite in undiluted wine has to be calculated.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
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Explanation of Solution

Balanced reactions amongst I3 and HSO3 is written as follows:

  I3+HSO3+H2O3I+SO42+3H+

Balanced reactions amongst I3 and S2O32 is written as follows:

  I3+2S2O323I+S4O62

Since 1 mol electrons carry a charge of 96485 C, so amount of iodide produced in 131 s is calculated as follows:

  Amount of iodide=(131 s)(1 mol96485 C)(10 mA1 s)(1 A1000 mA)(1 mol I32 mol e)=6.79×109 mol

Excess moles of thiosulfate are calculated as follows:

  Excess moles of  S2O32 =(6.79×109 mol I)(2 mol S2O321 mol I)=13.5×106 mol

Amount of S2O32 added initially is calculated as follows:

  Intial moles of S2O32 =(0.0507 M)(0.500 mL)(1 L1000 mL)=0.00002535 mol

Therefore amount of S2O32 consumed is calculated as follows:

  Consumed S2O32 =0.00002535 mol13.5×106 mol=11.77 μmol S2O32

Number of moles of SO32 after dilution is calculated as follows:

  Moles of SO32 =(6.55 μmol SO32)(10 mL20 mL)9 mL=3.64 μmol/mL SO32

Hence, in terms of molar concentration is calculated as follows:

  Molarity of SO32 =(3.64 μmol1 mL)(1000 mL1 L)(1 mol106 μmol)=3.64×103M SO323.64 mM

Thus, concentration of total sulfite in undiluted wine is 3.64 mM.

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