Question

Calculate the moles of I₃^-reduced by S₂O₃^2- using the following data:

• Sample volume (mL) = 200.0
• Buret Reading, initial (mL) = 3.85
• Buret Reading, final (mL) = 18.25
• Volume of Na₂S₂O₃ dispensed (mL) = 14.4 mL
• Average molar concentration of Na₂S₂O₃ (mol/L) = 0.02077
• Moles of Na₂S₂O₃ dispensed (mol) = __________
• Moles of I₃^- reduced by S₂O₃^2- (mol) = __________

Hint: Compute for moles of Na₂S₂O₃ first then use Equation 31.3 of your lab manual to convert it to moles I₃^- .

Answer 1

Answer:

This can be solved using simple volumetric law,

Which states,

V_thio× M_thio = V_I3- × M_I3-

Volume of thio sulphate(V_thio) dispensed is given as 14.4 mL.

Molarity of thio sulphate (M_thio) is given as 0.02077 mol L^-1.

From this molarity value, we can calculate the moles of thio sulphate dispensed, by the relation;

Molarity = moles / volume and

Moles = molarity × volume.

Volume of thio sulphate is 0.014 L (14.4 mL).

So, moles = 0.02077 mol L^-1 × 0.014 L

Moles of sodium thio sulphate = 0.00030 moles.

Now come back to the volumetric equation;

Volume of iodine sample(V_i3-)used is 200.0 mL.

Rearranging the volumetric equation for M_i3- gives,

M_i3- = (Vthio × Mthio) / Vi3-

Mi3- = ( 14.4 mL × 0.02077 M) / 200 mL.

Mi3- = (0.2991 / 200 ) M

Mi3- = 0.001496 M.

Moles of I3- = molarity × volume. Molarity was calculated in above step. Volume = 0.2L

Moles of i3- = 0.001496 mol L^-1 × 0.2 L

Moles of i3- = 0.00030 moles.

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