Suppose a student needs to standardize a sodium thiosulfate, Na2S2O3,Na2S2O3, solution for a titration experiment. To do so, he or she will react it with a solution of iodine. The student adds a 1.00 mL1.00 mL aliquot of 0.0200 M KIO30.0200 M KIO3 solution to a flask, followed by 3 mL3 mL of distilled water, 0.2 g0.2 g of solid KI, KI, and 1 mL H2SO4.1 mL H2SO4. The student then titrates the solution with sodium thiosulfate solution in order to determine the exact concentration of Na2S2O3.Na2S2O3. The end point of the titration is reached after 0.90 mL0.90 mL of Na2S2O3Na2S2O3 is dispensed from a microburet. What is the concentration of the standard sodium thiosulfate solution?

The concentration of the standard sodium thiosulfate solution is ; 0.133

The net chemical reaction equation

IO₃⁻  +  6H⁺  +   6S₂O₃³⁻ >  I⁻  +  3S₄O₆²⁻  + 3H₂O

First step : Determine the moles of  KIO₃

number of moles = molarity * volume

                            = 0.02 mol / L * 0.01 L

                            = 0.00002 mol

From the net chemical reaction equation

one ( 1 ) mole of KIO₃ reacts with 6 moles of S₂O₃²⁻

∴ number of moles in the reaction = 6 * 0.00002 = 0.00012 mol.

Final step : Determine the concentration of sodium thiosulfate solution

Given Volume = 0.90 ml = 0.0009 L

∴ concentration of sodium thiosulfate solution

   =  0.00012 / 0.0009  =  0.133

Hence we can conclude that The concentration of the standard sodium thiosulfate solution is ; 0.133

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0.133

Explanation:

reaction between KIO3 and KI in acidic medium

IO3⁻ +5I⁻ +6h⁺ → 3I₂ + 3H₂O

I₂ reacts with thiosulphate

NaS₂O₃  → 2Na⁺ + S₂O₃²⁻

net reaction

IO⁻₃ + 6H⁺ + 6S₂O₃³⁻ → I⁻ + 3S₄O₆²⁻ + 3H₂O

mole of KIO₃

= molarity x volume

= 0.00002mol

a mole of KIO₃ has reaction with 6 mol of S₂O₃²⁻

= 2x6x10⁻⁵

= 0.00012 mol

volume = 0.90 ml

1 ml = 0.001L

0.90ML  = 0.0009L

to get concentration,

molarity/volume

= 0.00012/0.0009

= 0.133m

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.