Intensity of serum signal Question

Question

The lithium concentration in serum taken from a patient being treated with lithium for manic-depressive illness was analyzed using flame emission spectroscopy. A sample of the serum gave a reading of 305 units for the intensity of the 671 nm red emission line. Then, 1.00 mL of a 11.31 mM lithium standard was added to 9.00 mL of serum. This spiked serum gave an intensity reading of 789 units at the 671 nm emission line.

What is the original concentration of Li+ in the serum?

Li+ =. units mM

𝐼X=𝑘[Li+serum]

how would I solve for the concentration?

Answer 1

The 1.00 ml of 11.31 mM that is added to the sample is considered as an "internal standard". We can calculate the [Li⁺] in the sample as follows:

1.00 ml x 1 L / 1000 ml x 11.31 mmol / L = 0.01131 mmoles added

Total volume is now 10.00 ml (1.00 ml + 9.00 ml)

[Li⁺] = 0.01131 mmols / 10.00 ml = 1.131 mM

Intensity = 789 units

Intensity due to the internal standard = 789 units - 305 units = 484 units

484 units / 1.131 mM = 427.9 units / mM

Unknown intensity = 305

[Li⁺] = 305 units x 1 mM / 427.9 units = 0.713 mM