Question

Can someone verify for me that I'm doing this correctly?

For data analysis, part b:

All S₂O₃^2- is consumed at the end of the reaction. Therefore, the moles of S₂O₃^2- consumed can be calculated using the equation below.
(moles S₂O₃^2-)_consumed = M_stock x (V_stock)

My instructor states "The stock sodium thiosulfate (S2O3) is 0.01M

0.01M x 0.010 L = 0.0001moles S2O3"

So I did the rest of this accordingly (below), and it differs a lot from what's on this website. Any explanations would be appreciated, thanks!

Chemical Kinetics

Lab Results

Record the following lab data in the table below for trial 4. (done)

(a) volume of potassium iodide solution added to Erlenmeyer flask A in mL	12.5 mL
(b) volume of potassium nitrate solution added to Erlenmeyer flask A in mL	12.5 mL
(c) volume of starch solution added to Erlenmeyer flask A in mL	5 mL
(d) volume of (NH4)2S2O8 solution added to Erlenmeyer flask B in mL	25 mL
(e) volume of (NH4)2SO4 solution added to Erlenmeyer flask B in mL	0 mL
(f) volume of Na2S2O3 solution added to Erlenmeyer flask B in mL	10 mL
(g) the initial color of the solution after adding flask B to flask A	grey/clear
(h) time for color change	41 s
(i) color of the final solution	black

Data Analysis

Fill in the following table. Look below the table for instructions on how to calculate the values for each row of the table.

a	total volume (L)	0.065 L
b	mol S2O32-consumed	0.0001 moles
c	mol I3-produced	0.00005 moles
d	[S2O32-] consumed (M)	0.001538 mol/L
e	[I3-] produced (M)	0.000769 mol/L

a) Calculate the final volume of the reaction mixture after the contents of beaker B are added to beaker A. Report your answer in liters.

b) All S₂O₃^2- is consumed at the end of the reaction. Therefore, the moles of S₂O₃^2- consumed can be calculated using the equation below.
(moles S₂O₃^2-)_consumed = M_stock x (V_stock)

c) The I₃^- produced in reaction 1 reacts with S₂O₃^2- in reaction 2 as shown.
I₃^- (aq) 2S₂O₃^2- (aq) → 3I^- (aq) S₄O₆^2- (aq)

Therefore, for every mole of I₃^- produced, 2 moles of S₂O₃^2- have reacted.
(moles I₃^-)_produced = (moles S₂O₃^2-)_consumed / 2

d) [S₂O₃^2-]_consumed = (moles S₂O₃^2-)_consumed / (V_total)

e) Calculation is the same as d, but using I₃^- instead of S₂O₃^2-.

Answer 1

A) The final volume is the addition of the volumes you used in the experiment

Total volume = 12.5 + 12.5 + 5 + 25+ 10 = 65 ml or 0.065 L, divide it by 1000

b) You say all the S2O3 is consumed, so lets find how many moles you have

V = 10 ml or 0.01 L

Molarity = 0.01

moles available = Molarity * Volume = 0.01 * 0.01 = 0.0001 moles

For moles of I-, in the stoichiometry you have that for every 2 moles of S2O3 3 moles of I- are produced

moles of I- = 0.0001 * 3 / 2 = 0.00015 moles of I^- produced

Molarity of I- produced = 0.00015 / 0.065 = 0.0023 M

These are my results the difference with you is that you took 1 mole of I- produced for every 2 moles of S2O3 and the reaction says that for every 2 moles of S2O3 you produce 3 moles of I-

Be carefull, I₃^- is triiodide on reactants side

I^-  is iodide ion on product side

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