Question

Considering dilution, calculate the initial concentrations used for each trial. (It asks for [I-] and [S2O8 2-].)

Trial 1:

Water-3.00mL

0.10M KI-0.25mL

0.10M KCl-0.75mL

0.05M K2SO4-0.75mL

0.05M K2SO8-0.25mL

0.10M CuSO4-0mL

Answer 1

Concentration in M/L = # of moles / Volume in L

Volume of water added for dilution = 3.00 mL

a) [I^-] = ?

0.25 mL of 0.10 M

For KI, M = 0.10 M and v = 0.25 mL

M x v = 0.10 x 0.25 = 0.025 milimoles = 2.5 x 10^-5 moles.

On dilution final volume = 0.25 + 3 = 3.25 mL = 3.25 x 10^-3 L

Concentration of KI on dilution [KI] = # of moles / Volume = 2.5 x 10^-5 / 3.25 x 10^-3. = 7.7 x 10^-3 M/L

As KI is a strong electrolyte,

[I-] = [KI] = 7.7 x 10^-3 M/L

Initial concentration of I^- is 7.7 x 10^-3 M/L

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b) [S₂O₈^2-] = ?

Given 0.25mL of 0.05M K₂S₂O₈.

For K₂S₂O₈, M = 0.05 M and v = 0.25 mL

M x v = 0.05 x 0.25 = 0.125 milimoles = 1.25 x 10^-5 moles.

On dilution final volume = 0.25 + 3 = 3.25 mL = 3.25 x 10^-3 L

Concentration of S₂O₈^2- on dilution [K₂S₂O₈] = # of moles / Volume = 1.25 x 10^-5 / 3.25 x 10^-3. = 3.9 x 10^-3 M/L

As K₂S₂O₈ is a strong electrolyte,

[S₂O₈^2-] = [K₂S₂O₈] = 3.9 x 10^-3 M/L.

Initial concentration of S₂O₈^2- is 3.9 x 10^-3 M/L.

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For rest species if needed let me know.