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 4.00 mL Ti solution of unknown concentration with 3.50 mL of a 12.3 μg/mL solution of Co. The atomic absorbances were measured as 0.103 and 0.212 for Ti and Co, respectively. As a reference, a standard mixture containing 2.20 μg Co/mL and 2.48 μg Ti/mL was prepared and measured to have a signal-to-signal ratio of 1.79 Ti: 1.00 Co. Determine the concentration (moles/L) of titanium in the original unknown solution.

Answer 1

A mixture was prepared by combining a 4.00 mL Ti solution of unknown concentration with 3.50 mL of a 12.3 μg/mL solution of Co.

total Co in the sample is 43.05 μg

The atomic absorbances were measured as 0.103 and 0.212 for Ti and Co, respectively.

As a reference, a standard mixture containing 2.20 μg Co/mL and 2.48 μg Ti/mL was prepared and measured to have a signal-to-signal ratio of 1.79 Ti: 1.00 Co. Assuming 1 mL each was mixed the amount of cobalt is 2.2 μg and titanium is 2.48 μg

2.20 μg Co gave a signal of 1.00;

2.48 μg Ti gave a signal of 1.79;

ratio of signal to metal quantity is

Co/Ti = 1/1.79 = 2.2/2.48

0.55 signal = 0.887 metal ratio

signal ratio in the unknown mixture is

Co/Ti = 0.212/0.103 =2.058 signal = 3.319 metal ratio

since total Co in the unknown was 43.05 μg Titanium quantity should be 43.05/3.319 = 12.97 μg in 4.0 mL

so concentration of titanium is 3.24 μg/mL in the unknown sample