i really need help with this pre lab. i will repost the question for another 300

Question

i really need help with this pre lab. i will repost the question for another 3000 pts if my website accepts your answer. i hope someone out there can help.

Oxidation of Iodide to Iodine; Spectroscopic measurement of TriIodide ion.

IO3^- + 5 I^- + 6 H^+ ===> 3 H20 + 3 I2 (standard)

H2O2 + 2 I^- + 2 H^+ ===> 2 H20 + 2 I2 (unknown)

I^- + I2 ===> I3^-

STANDARD:

1) Put into a clean 250 mL volumetric flask exactly ..... 0.95 ml of 0.85 mM iodate STANDARD.

2) Add an excess of 1M KI and 1M Sulfuric Acid and let stand for 5 minutes.

3) Dilute to volume.

UNKNOWN:

1) Pipet into a clean 250 mL volumetric flask 1.00 mL of your UNKNOWN and dilute

to volume with distilled water.

2) Put into a clean 250 mL volumetric flask exactly........ 0.95 mL of this diluted H2O2 UNKNOWN.

3) Add an excess of 1M KI and 1M Sulfuric Acid and let stand for 5 minutes.

4) Dilute to volume (250 mL).

Measure STANDARD and UNKNOWN spectroscopically at 360 nm using the same cuvet throughout.

% Transmission of BLANK.................................. 101.9 %

% Transmission of STANDARD................................ 48.3 %

% Transmission of UNKNOWN................................. 40.2 %

(g) Absorbance of STANDARD.................................... _____0.324_________

(h) Absorbance of UNKNOWN.................................... _____0.404__________

(i) Molarity of original UNKNOWN................................. ___________________ M

using Beer's law C1/A1 = C2/A2 calculate the molarity of the ORIGINAL unknown concentration of HOOH.

absorbance = log( %T blank / %T test )

part G) Absorbance of standard = log (101.9 / 48.3) = 0.324

part H) Absorbance of unknown = log (101.9 / 40.2) = 0.404

first i tried to find the number of moles of the standard

.00085M(standard)/.00095L = 0.89473 mol of standard

then divided that by .250L for the dilute to volume part

0.89473 mol (standard) / .250L = 3.5789 M Standard

then with beers law C1/A1=C2/A2

i found that i concentration of the unknown was

C1=[(C2)(A1)]/A2

C1=[(3.5789M std)(0.404)]/.324 = 4.462579 M unknown

when i go to enter this value on my school website i get this error:

A mole of Iodate ion produces 3 moles of iodine.

One mL of your UNKNOWN was diluted to 250 mL

Explanation / Answer

1) known

moles of Iodate = 0.95x10^ -3 x 0.85 x10^ -3 = 0.8075 x10^-6

since KI and H2SO4 is excess I2 moles = 3 x0.8075 x10^-6 = 2.4225 micro moles , vol = 250 ml , Conc = 2.4225 x10^-6/0.25 = 9.69 x10^-6 M

Absorbance = 0.324 = 9.69 x10^ -6 x const

Absorbance unknown = 0.404 = C x const

by above two eq
Conc Unknown/9.69x10^-6 = ( 0.404/0.324)

cocn unknown = 12.083 x10^-6 M

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