Na 2 S 2 O 3 solution was prepared with 8.7122 g of Na 2 S 2 O 3 * 5H 2 O and 0.

Question

Na2S2O3 solution was prepared with 8.7122 g of Na2S2O3 * 5H2O and 0.0256 g of Na2CO3 dissolved in 500 mL of Deionized water. Titrate was prepared with 3g of KI and 50 mL of KIO3, and 10 mL of H2SO4. 2 mL of startch indicator was added near ndpoint.

Standardization of sodium Thiosulfate Titrant 1.5054 Weight (g) of KIO3: Concentration of KIO3 0.01406909 Initial Burette Value (mL) Final Burette Value (mL) Volume of titrant (mL) Na2s203 Concentration (M) Trial 1 31.8 1.7 30.1 32.2 2.2 30 3 33.6 3.5 30.1

Explanation / Answer

the titration rxn-

6S2O32- +IO3-+6H+ --->3H2O+I-+3S4O62-

calculation of initial concentration of Na2S2O3

"Na2S2O3 solution was prepared with 8.7122 g of Na2S2O3 * 5H2O and 0.0256 g of Na2CO3 dissolved in 500 mL of Deionized water."

moles of Na2S2O3 in the solution=mass of Na2S2O3.5H2O/molar mass of Na2S2O3.5H2O=8.7122g/248.184g/mol=0.0351 moles

molarity of Na2S2O3=moles of Na2S2O3/total volume=0.0351 moles/0.5 L=0.0702 M

initial concentration=[Na2S2O3]o=0.0702M

initial concentration of KIO3=[KIO3]o=0.01406909M

trial 1; volume of KIO3 solution=50 ml KIO3+10mlacid=60ml

moles of KIO3=60ml*10^-3 L/ml *0.01406909M=0.000846 moles

moles of Na2S2O3 added=30.1ml*10^-3 L/ml*0.0702 mol/L=0.00211 moles

moles of Na2S2O3 neutralized=0.000846 moles,so remaining moles of Na2S2O3=0.00211-0.000846=0.00127 moles

final concentration of Na2S2O3=0.00127moles/total volume

total volume=30.1ml+60ml=90.1ml=0.0901 L

[Na2S2O3]=0.00127moles/0.0901L=0.0141M

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