Question

Serum taken from a patient being treated with lithium for manic-depressive illness was analyzed for lithium concentration. A reading of 322 units was obtained for the intensity of the 671 nm red emission line. Then 1.00 mL of a 10.2 mM Lithium standard was added to 9.00 mL of serum, and this spiked serum gave an intensity reading of 752 units. What is the original concentration of Li in the serum?

Answer 1

The formual required for this:

[X]_i / ([S]_f + [X]_f = I_x / (I_s+x)

where

[X]_i concentration of analyte in initial solution

[S]_f = concentration of stnadard in final solution

[X]_f concentration of analyte in final solution

I_x = signal from initial solution

(I_s+x) = signal form final solution

the data given:

[S]i = 10.2 mM

Recalculate [S]f

[S]f = [Si*] * Vinitial / Vfinal

[S]f = 10.2 * (1 mL)/(1 + 9 mL) = 1.02 mM

now...

[X]f = [X]i *  Vinitial / Vfinal

[X]f = [X]i *  (9 mL)/(1 + 9 mL)

[X]f = [X]i * 0.9

Now..

Ix = 322

Is+x = 752

Now, substitute all data

[X]_i / ([S]_f + [X]_f = I_x / (I_s+x)

[X]_i / (1.02+ 0.9*[Xi]) = 322/752

[X]_i / (1.02+ 0.9*[Xi]) =0.428

[X]_i = 0.428*1.02 + 0.428*0.9*[X]_i

(1-0.428*0.9)[X]_i = 0.43656

[X]_{i =} 0.43656/0.6148= 0.71

[X]_{i =} 0.71mM

then

Li sample --> 0.71 mM