Cobalt was used as an internal standard to analyze a sample of titanium with ato

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 1.50 mL of a 13.7 g/mL solution of Co. The atomic absorbances were measured as 0.166 and 0.187 for Ti and Co, respectively. As a reference, a standard mixture containing 2.50 g Co/mL and 1.64 g Ti/mL was prepared and measured to have a signal-to-signal ratio of 1.35 Ti: 1.00 Co. Determine the concentration (moles/L) of titanium in the original unknown solution.

Explanation / Answer

The total volume of the test solution becomes, 4.00 mL + 1.50 mL = 5.50 mL

In this solution the concentration of Co is = 1.50 x 13.7/ 5.50 = 3.74 micro g/ mL

So, 3.74 micro g/ mL of Co gives a signal of 0.187

Now in the reference solution the concentration of Co is 2.50 micro g/ mL

So, it should give a signal = (0.187 x 2.50)/ 3.74 = 0.125

Now in the reference solution, the signal-to-signal ratio Ti : Co = 1.35 : 1.00

Therefore signal for Ti = 0.125 x 1.35 = 0.169

So, a signal of 0.169 corresponds to 1.64 g Ti/mL

Hence a signal of 0.166 in test solution corresponds to (1.64 x 0.166)/ 0.169 = 1.61 g Ti/mL

Now 1.61 g Ti/mL is the concentration of Ti in 5.50 mL of Test solution

So, the concentration of Ti in the 4.00 mL of original unknown solution = (1.61 x 5.50)/ 4.00 = 2.21 g Ti/mL

Again, 2.21 g Ti/mL = 2.21 x 103 g Ti/ L = 2.21 x 103 x 10-6 g Ti/ L

= (2.21 x 103 x 10-6)/ 47.87 moles of Ti/ L (as molar mass of Ti is 47.87)

= 4.62 x 10-5 moles/L

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