3.12 g of bubble gum that contains Al is burnt in a high temperature, then is di

Question

3.12 g of bubble gum that contains Al is burnt in a high temperature, then is dissolved in 5 ml of nitrate acid, filtered, then diluted until the volume reached 100 ml. 5.00 ml of this solution is added into a 25.0 ml volumetric flask and into it added 300.0 ppm Al ion, diluted until 25.0 ml, then sprayed to AAS (Atomic Absorption Spectroscopy) flame and analyzed for Al in 309 nm. This is the data of the measurement of volume added to sample and the AAS signal. Calculate the concentration of Al in sample, using double internal standard method.

Explanation / Answer

This is standard addition method

When 0.1 ml of 300 ppm Al ion standard added

Concentration of Al ion standard in 25 ml solution (Cs) = 300 ppm x 0.1 ml/25 ml = 1.2 ppm

concentration of Al in unknown solution = Cx

AAS signal for sample without Al standard (Ax) = 0.214

AAS signal for solution with 0.1 ml Al standard added (Ax+As) = 0.386

Using,

Cx/(Cx + Cs) = Ax/(Ax + As)

Cx/(Cx + 1.2) = 0.214/0.386

0.386Cx = 0.214Cx + 0.2568

Cx = 0.2568/0.172 = 1.493 ppm

So concentration of Al in 25 ml solution = 1.493 ppm

concentration in 5 ml aliquot = 1.493 ppm x 25/5 = 7.465 ppm

concentration of Al in original 100 ml solution = 7.465 ppm x 100/5 = 149.3 ppm

So,

concentration of Al in sample = 149.3 ppm

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.