Please please please answer this question fully and please answer it only if you

Question

Please please please answer this question fully and please answer it only if you're 100% of the correct answer! Thank you! 0/12 points My Notes Some people have an allergic reaction to the food preservative sulfite (so32 Sulfite in wine was measured by the following procedure: To 50.0 mL of wine were added 5.00 mL of solution containing (0.8043 g KIO3 5 g KI)/100 mL. Acidification with 1.0 mL of 6.0 M H2SO4 quantitatively converted IO3 nto 13 The 13 reacted with SO3 to generate SO4 leaving excess I3 in solution. The excess I3 required 12.86 mL of 0.04818 M Na2s203 to reach a starch end point. (a) Write the reaction that occurs between KIO3 and KI when the H2SO4 is added. (Use the lowest possible coefficients. Omit spectator ions and states-of-matter. Help chem Pad Greek (b) Write a balanced reaction between 13 and sulfite. Use the lowest possible coefficients. Omit spectator ions and states-of-matter.) chem Pad Help (c) Find the concentration of sulfite in the wine. Express your answer in mol/L and in mg so32 per liter. x mol/L HINTS: Don't use scientific notation, do put zero before decimal point, carefully evaluate significant figures. 406.7 mg/L.

Explanation / Answer

1) 2IO3^- +12H+10e ------->I2+6H2O

2I- --->I2+2e]*5

...............................................................

12H+ +2IO3^- +10I- ---------->6I2+6H2O

or 6H+ +IO3^- +5I- ------>3I2 +3H2O

or 3H2SO4 +KIO3+5KI ----->3I2+3H2O

1mole KIO3 gives 3 moles I2 or I3-

so 1 mole I3- is given by 1/3 moles KIO3-

2)SO3^2- +H2O --------->SO4^2- +2H+ +2e

I3- ---->I2 +I-

I2+2e---------->2I-

.......................................................................................

)SO3^2- +H2O +I3- ------------>SO4^2- +2H+ +3I-

3)2S2O3^2- +I2-->2S4O32- +2I-

2 mole S2O32-=1 moles I2 or I3-

I2+I---->I3-

molarity of KIO3=0.8043g/214.001g/mol=0.00376 moles

[KIO3]=0.00376 moles/100ml=0.0376moles/1000ml=3.76*10^-2 mol/L

moles of KIO3 added=5ml*3.76*10^-2 M=0.005L*3.76*10^-2 M=18.8*10^-7=1.9*10^-4 moles

moles of I3- produced by KIO3=3*1.9*10^-4 moles=5.7 *10^-4 moles

moles of S2O32-)=0.04818 moles/L*0.01286L=6.19*10^-4 moles

moles of excess I3-=1/2*moles of S2O3=1/2*6.19*10^-4 moles=3.09*10^-4 moles

moles of I3- produced =moles of sulfite+excess I3-

or,5.7 *10^-4 moles moles=moles of sulfite +3.09*10^-4 moles

moles of sulfite=2.6*10^-3 moles

[SO32-]=moles/total volume=2.6*10^-3 moles/50ml=2.6*10^-3 moles/0.05L= 0.052 M

in mg/L, [SO32-]=0.052 mol/L*80.066 g/mol=4.1634 g/L=4163.4 mg/L

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