A standard iodate solution is prepared at 22.0 °C by dissolving 0.9123 g (uncorr

Question

A standard iodate solution is prepared at 22.0 °C by dissolving 0.9123 g (uncorrected mass) of KIO3 with deionized water in a 499.92 mL volumetric flask.To standardize the Na2S2O3 solution, a 10.005 mL aliquot of the KIO3 solution is acidified,treated with excess KI, and titrated with Na2S2O3, requiring 22.05 mL to reach the endpoint. Finally, a 10.005 mL aliquot of unknown Cu solution is reduced with excess KI. The liberated iodine requires 31.46 mL of Na2S2O3 solution to reach the end point. Calculate the unknown Cu concentration in g/L.

These are the correct answers below, the pathway to achieve these results is needed.

[KIO3 standard (corrected to 20 °C): 8.53255 mM]

[S2O3^2 titrant: 23.2294 mM]

[Unknown: 4.64160 g/L Cu]

(the answers may be different around the later decimal places but this is caused by buoyancy correction and density interpolation, which I understand how to adjust for I just need the basic work flow...I can adjust the density and buoyancy after that,thanks)

Mtrue/Mread=(1-0.0012/8)/(1-0.0012/density object @22.0'C)

Explanation / Answer

KIO3 + 6KI + 6HCl + 6Na2S2O3 ---> 6KCl + 3H2O + 3Na2S4O6 + 6NaI

moles of KIO3 = 0.9123 g/214 g/mol = 4.26 mmol

molarity of KIO3 solution = 4.26/0.49992 = 8.52 mM

Volume of KIO3 used for Na2S2O3 neutralization = 8.52 mM x 10.005 ml = 0.085 mmol

moles of Na2S2O3 reacted = 6 x 0.085 mmol = 0.511 mmol

molarity of Na2S2O3 solution = 0.511 mmol x 1000/22.05 ml = 23.2 mM

1 mole of Cu gives 1 mole of I2

2 moles of Na2S2O3 reacts with 1 mole of I2

moles of Na2S2O3 used for I2 from copper solution = 23.2 mM x 31.46 ml = 0.73 mmol

moles of I2 = moles of copper = 0.73/2 = 0.365 mmol

unknown copper concentration = 0.365 x 63.546/10.005 = 2.32 g/L

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