Mass of KI = 1.75 grams Mass of (NH4)2S2O8 = 3.35 grams A) When KI is added to 2

Question

Mass of KI = 1.75 grams

Mass of (NH4)2S2O8 = 3.35 grams

A) When KI is added to 250mL of distilled water calculate the molarity of I-

B) When (NH4)2S2O8 is added to 250mL of distilled water calculate the molarity of S2O8^2-

C)

KI = 11.66mL

(NH4)2S2O8 = 20.05mL

S2O3^2- = 10mL

time (delta t) = 111 seconds

temp

Mass of KI = 1.75 grams Mass of (NH4)2S2O8 = 3.35 grams When KI is added to 250mL of distilled water calculate the molarity of I- When (NH4)2S2O8 is added to 250mL of distilled water calculate the molarity of S2O8^2- KI = 11.66mL (NH4)2S2O8 = 20.05mL S2O3^2- = 10mL time (delta t) = 111 seconds temp degree C = 22.25

Explanation / Answer

Because the standard and the unknown sample are treated the same way - 0.8 ml of each in 250 ml total volume - we can compare the set just by comparing the initial concentrations.

According to equation (1) one mol of iodate creates 3 moles of iodine. So standard iodine is 3 x 0,85 mM = 2.55 mM.

(1) KIO3+3H2SO4+5KI --> 3H2O +3I2+ 3K2SO4

This concentration shows 0.357 units of absorbance.
The absorbance of the unknown is 0.615 units indicating the iodione concentration to be c = 2.55 x 0.615/0.357 mM = 4.393 mM

According to equation (2) one mole of iodine is oxidized from iodid by one mole of H2O2

(2) H2O2 + 2KI -> 2KOH + I2

So H2O2 concentration of the tested 0.8 ml diluted H2O2 solution is 4.393 mM too.

Now we have to multiply this result by the dilution factor 1 ml was diluted to 250 ml, dilution factor 250.

And the final result is c = 250 x 4.393 mM = 1098.21 mM = 37.34 g/L.

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