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 end point. 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 what the correct step answers should be, just need to show how to do work:

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

[S2O3 2− titrant: 23.2294 mM]

FINAL ANSWER: [Unknown: 4.64160 g/L Cu]

Answer 1

KIO3: Volume = 499.92 ml

Mass = 0.9123 g

Molar mass of KIO3 = 214 g/mol

Concentration of KIO3 = (mass*1000)/(molar mass*volume)

= (0.9123*1000)/(214*499.92) = 0.00852753 M = 8.52753 mM (at 22 oC)

Concentration of KIO3 = 8.52753 mM (at 22 oC)

When Na2S2O3 is titrated with KIO3 in excess of KI, in net balanced equation, 1 mol of IO3- reacts with 6 mles of S2O32-.

Concentration of KIO3 = M1 = 8.52753 mM

volume = V1 = 10.005 ml

Volume of Na2S2O3 = V2 = 22.05 ml

6(M1V1) = (M2V2)

6(8.52753*10.005) = (M2) (22.05)

M2 = 23.2158 mM

Concentration of S2O32- titrant = 23.2158 mM

In the reaction of Na2S2O3 with I2, volume of S2O32- used = 31.46 ml = 31.46*10^-3 L

concentration = 23.2158 mM = 23.2158*10^-3 M

Number of moles = concentration*volume = (23.2158*10^-3 M)(31.46*10^-3 L) = 730.3691*10^-6 mol

2 moles of Na2S2O3 react with 1 mole of I2.

So, 730.3691*10^-6 mol of Na2S2O3 react with mole of I2 = 730.3691*10^-6 mol/2 = 365.184*10^-6 mol

On reaction of Cu2+ with KI, 2 moles of Cu2+ react to give 1 mole of I2.

So, moles of Cu2+ reacting to give 365.184*10^-6 mol of I2 = 365.184*10^-6 mol*2 = 730.3691*10^-6 mol

Atomic mass of Cu = 63.5 g/mol

Mass of Cu = moles*mass = 730.3691*10^-6 mol*63.5 = 0.04638 g

Volume = 10.005 ml = 0.010005 L

Concentration = mass/volume = 0.04638 g/0.010005 L = 4.6355 g/L

concentration of Cu = 4.6355 g/L