Some analyses are best performed via the technique of \"back titration.\" For ex

Question

Some analyses are best performed via the technique of "back titration." For example, during the analysis of vitamin C with I_3^- a known excess of I_3^- can be added to a sample. Then the un-reacted I_3^- excess can be back titrated with thiosulfate (S_2O_3^2-). The important analysis of glucose (C_6H_12O_6) in nutritional and other biochemical processes can be performed in a similar I_3^- reaction. The stoichiometry is shown in the following balanced reaction of I_3^- and glucose: C_6H_12O_6 + I_3^- 3OH^- rightarrow C_6H_11O_7^- + 2H_2O + 3I^- To begin the analysis, 50.0 mL of 0.0526 M solution of I_3^- was added to a sample containing glucose. The excess (un-reacted) I_3^- required 23.7 mL of 0.0867 M S_2O_3^2- in a back titration. The stoichiometry of that is shown in the following balanced reaction of I_3^- reacting with S_2O_3^2. 2S_2O_3^2- + I_3^- rightarrow 3I^- + S_4O_6^2- How many moles of glucose were in the sample?

Explanation / Answer

As per the reaction 2S2O3-2 + I3- ---à3I- + S4O6-2

Moles of S2O3-2 required per moles of I3=2

Moles of S2O3-2 in 23.7ml of .0867M= Molarity* Volume(L)= 0.0867*23.7/1000=0.002055

Moles of I3 used = 0.002055/2=0.0010275

Moles of I3 supplied= 0.0526*50/1000 =0.00263

Moles of I3- consumed for reaction with glucose = 0.00263-0.002055= 0.000575

From the stoichiometry of the reaction C6H12O6+I3- + 3OH- --àC6H11O7- + 2H2O+3I-

Moles of glucose per moles of I3= 1, moles of glucose =0.000575

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