Question

Serum taken from a patient being treated with lithium for manic-depressive illness was analyzed for lithium concentration. A reading of 284 units was obtained for the intensity of the 671 nm red emission line. Then 1.00 mL of a 10.5 mM Lithium standard was added to 9.00 mL of serum, and this spiked serum gave an intensity reading of 7.60 × 102 units. What is the original concentration of Li in the serum?

Answer 1

Our first value of units is of the initial concentration.

I(x) = 284; 284 = k[Li+] ----- equation (1), where k is a proportionality constant.

The next equation is going to represent the initial concentration of the lithium ion plus the addition of the standard.

I(x+s) = 7.60 x 10²; 760 = k*{ [Li+ standard] * (V standard/ V total) } +k*{ [Li+ serum] * (V initial/ V total) }

Since we have known that 284 = k*[Li+ serum], we can plug that value into our equation as---

760 = k*{ [10.5] * (1 mL/ 10 mL) } + 284 * (9 mL/ 10 mL)

=> 760 =k*(1.05) + 255.6

=>k*(1.05) = 504.4

=> k =480.4

Now, putting the value of k in equation(1) we will get the original concentration of Li in the serum as---

equation (1) => 284 =k[Li⁺]

=> [Li⁺] = 284/480.4

=> [Li⁺] = 0.59 mM

{* note that intensity reading is taken as 7.60 × 10² units instead of 7.60 × 102 units}