[BilloRani2012]

6.39 mL of a solution of CaCl2 is placed in a 100 mL volumetric flask. A small amount of an indicator is added that forms a coloured complex with Ca2+ and the volume of the solution is made up to 100 mL. A set of standards was made similarly, with the final concentrations (in micromolar) and absorbances at 570 nm shown.

Conc. (μM)       Abs at 570 nm
0                 0.001
100                 0.121
200                 0.239
300                 0.362
400                 0.481
500                 0.602
600                 0.718
700                 0.756
800                 0.762
900                 0.770

If the A570 of the solution in the volumetric flask was 0.518, what was the original concentration of Ca2+ in mM? Answer to three significant figures.

[BilloRani2012]

okay....i don't really know where to start, but do i have to use the equation:

A = ebc?

im kinda confused

[enahs]

A = ebc is correct. Your task is to determine what e is.

When given a set of data like this, you want to plot the data and do a trend line. The slop of the line is you e value.

absorbance on y-axis, concentration on x.

[BilloRani2012]

okay....oh and i will have to convert micromolar to molar for the concent. right??

[BilloRani2012]

i have tired it:

I plotted the graph: and to find of the value for e, i did this:

m = y1 - y2 / x1 - x2

so i used to points on the line: (i converted micro molar to molar for the CONCENTRATION)

so, m = 0.239 - 0.481 / 0.0002 - 0.0004
     m = 1210

does that sound right??

so you said that this is my e value, so wen i put it in the equation i get:

A = 1210*b*c
0.518 = 1210*b*c

so i have 2 unkowns??  Im kinda stuck..

[Borek]

You can combine eb into one constant.

Even if you don't, b cancels out in the final calculations.

[BilloRani2012]

okay so would the original concent. just be:

0.518 = 1210*b*c

c = 0.518/1210
c = 4.28 x 10^-4

is that right??
       

[BilloRani2012]

i really need help....im kinda stuck

[Borek]

Assume A=k*c. k combines both path length and specific absorbance, as long as you are using same substance and same cuvette, that's perfectly valid approach.

Best way of dealing with the problem is to draw a plot, find the linear part, then use linear regression to find out k, or to draw a calibration curve and use graphic method to read the concentration. However, just by looking at the table it is obvious that given value of absorbance (0.518) is between 0.481 and 0.602, so the concentration must be between 400 and 500 μM.

[BilloRani2012]

I have tried what you said

I drew the graph on excel and got a straight line.
I then found the slope by doing: y2-y1 / x2-x1
I chose 2 points of the line. I chose (200,0.239) and (400, 0.481)
So, my e value (the slope) turned out to be: 1.21 x 10^-3 or 0.00121

Then, i put that into the equation: Abs = eC
so, 0.518 = 0.00121*C
so i got, C = 0.518/0.0012
             C= 431.7
             C= 432
does that sound that right??

[Borek]

So far so good, now just take dilution into account.

[BilloRani2012]

do you mean use the equation:

C1V1=C2V2 and find C1??

[Borek]

Yes.

[BilloRani2012]

so would this be right then:

C1V1=C2V2
C1*6.39 = 432*100
C1*6.39 = 43200
         C1= 6760 ( Divide by 10 000)
          C = 0.676

IS THAT RIGHT??

[DrCMS]

so would this be right then:

C1V1=C2V2
C1*6.39 = 432*100
C1*6.39 = 43200
         C1= 6760 ( Divide by 10 000)
          C = 0.676

IS THAT RIGHT??

   No.  Why did you divide by 10000?

What are the units for the answer 6760?  What units should the answer be in?  How do you convert from one to the other.