STANDARD: Put into a clean 250 mL volumetric flask exactly ..... 0.90 ml

Question

of 1.00 mM iodate STANDARD.
Add an excess of 1M KI and 1M Sulfuric Acid and let stand for 5 minutes. Dilute to volume.
UNKNOWN: Pipet into a clean 250 mL volumetric flask 1.00 mL of your UNKNOWN and dilute to volume with distilled water.
Put into a clean 250 mL volumetric flask exactly ........ 0.90 mL
of this diluted H2O2 UNKNOWN.Add an excess of 1M KI and 1M Sulfuric Acid and let stand for 5 minutes. Dilute to volume (250 mL). Measure STANDARD and UNKNOWN spectroscopically at 360 nm using the same cuvet throughout.
% Transmission of BLANK.................................. 98.0 %
% Transmission of STANDARD................................ 52.4 %
% Transmission of UNKNOWN................................. 32.4 %
(g) Absorbance of STANDARD.................................... ___________________
(h) Absorbance of UNKNOWN.................................... ___________________
(i) Molarity of original UNKNOWN................................. ___________________ M

Answer 1

I have a bone to pick with this problem. The TRANSMITTANCE (this is not a 1948 Plymouth transmission) is what was measured.

A blank = log(100/98) = about0.009 but you be more accurate.
A std = log(100/52.4) = about 0.28
A unknown = log(100/32.4) = about 0.49

Subtract blank from BOTH std and unknwn absorbances to arrive at net absorbance.
A std = about 0.28-0.009 = about 0.27
A unknown = about 0.49-0.009 = about 0.48. Remember to redo these.

Std = 1 mM x (0.90/250) = ?

Use A = ebc for standard and solve for e. You have A, b, and c.

Use A = ebc for unknown. You have A, e3 (from the first calculation) and solve for c (in mM). This is concn of unknown in the measured solution. The concn of the unknown in the original solution is 250 times that.

Answer 2

Thanks DrBob222, i just need steps to solve for question i. the steps you gave is confusing

Answer 3

What Dr. Bob222 did was give you the setups to calculate the molarity. Transmittance is related to absorbance, so you have to calculate the absorbance of the blank and subtract it from your sample to to acquire the correct absorbance value for your sample.

A unknown- A blank= actual absorbance value for unknown sample=log(100/32.4)-log(100/98)=actual absorbance value for unknown sample

Now, for your standard, you started off with 1mM you took a 0.90mL sample, and then diluted that to 250ML, what is the molarity after the dilution? Can you use MV=MV? Yes.

1 mM x (0.90/250)= Molarity of standard.

You need to know the molar absorptivity, can you calculate it if you know the absorbance, length of the cuvette, and the molarity? Yes.

A=ebc or better yet, A/bc=e

****You know the absorbance of the standard from the second equation that Dr. Bob222 gave you.

A std = log(100/52.4)

Once you calculate the molar absorptivity, you can use that to calculate the concentration

A/be=c

You know e from what you calculated above, you know b, which is always just 1, and you know A from the first step: A unknown - A blank= actual absorbance value for unknown sample=log(100/32.4)-log(100/98)=actual absorbance value for unknown sample.

SOLVE

Answer 4

Since I don't know what figure you are referring to in your later post, I do not want to try and help you solve that one. Maybe Dr. Bob222 will assist you with that one.

Answer 5

Hi Faruk--I thought I gave you the steps (and all but worked the problem for you). Perhaps Devron's post will help.