Question

Cobalt was used as an internal standard to analyze a sample of titanium with atomic absorption spectroscopy. A mixture was prepared by combining a 6.00 mL Ti solution of unknown concentration with 2.50 mL of a 13.9 μg/mL solution of Co. The atomic absorbances were measured as 0.155 and 0.208 for Ti and Co, respectively. As a reference, a standard mixture containing 1.78 μg Co/mL and 1.09 μg Ti/mL was prepared and measured to have a signal-to-signal ratio of 2.06 Ti: 1.00 Co. Determine the concentration (moles/L) of titanium in the original unknown solution.

Answer 1

Standard concentration of Co = 1.78 g/mL

Standard concentration of Ti = 1.09 g/mL

signal-to-signal ratio = 2.06 Ti: 1.00 Co.

A_{Ti} / A_{Co} = 2.06 / 1.0

We have,

A_{Ti} / [Ti] = F * A_{Co} / [Co]

A_{Ti} / A_{Co} = F * [Ti] / [Co]

2.06 / 1.0 = F * (1.09 / 1.78)

2.06 = F * 0.61

F = 3.37

Now,

Atomic absorbance of Ti = 0.155

Atomic absorbance of Co = 0.208

[Ti] = 13.9 μg/mL

Again using:

A_{Ti} / [Ti] = F * A_{Co} / [Co]

(0.155 / 13.9) = 3.37 * (0.208 / [Co])

0.011 = 0.701 / [Co]

[Co] = 0.701 / 0.011

[Co] = 63.72 g/mL