Question

Serum taken from a patient being treated with lithium for manic-depressive illness was analyzed for lithium concentration. A reading of 415 units was obtained for the intensity of the 671 nm red emission line. Then 1.00 mL of a 11.2 mM Lithium standard was added to 9.00 mL of serum, and this spiked serum gave an intensity reading of 879 units. What is the original concentration of Li in the serum?

Answer 1

[X]_i = initial concentration of serum.

V_o = original volume = 9mL & V_f  = final volume (after addition) = 9+1 = 10mL

[X]_f = [X]_i (V_o/V_f)  
= [X]_i (9mL/10 mL) = 0.9[X]_i

S = Lithium standard "spike" added = 11.2mM
[S]_f = [S]_i (V_s/V_f) = 11.2 mM (1 mL/10 mL)   
= 1.12 mM

[X]_i  / ([X]_{f +}  [S]_f ) = INITIAL reading / FINAL reading

Therefore: [X]_i  / ([X]_{f +}  [S]_f ) =   [X]_i / { 0.9 [X]_i + 1.12 mM} = 415 / 879 = 0.472
[X]_i = 0.919mM     ANSWER!