Question 10 of 14 Sapling Learning Map d Atomic emission spectroscopy and the me

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Question 10 of 14 Sapling Learning Map d Atomic emission spectroscopy and the method of standard addition are used to determine the Na concentration and its uncertainty in an unknown sample. Increasing amounts of a 1.75 g/mL standard Na. solution are added to a series of volumetric flasks containing 15.00 mL of the unknown solution. The volumetric flasks are then diluted to a final volume of 200.0 mL. Atomic emission spectroscopy is then used to measure the emission intensity of each solution. Unknown Volume (mL) Standard Volume (mL) Final Volume (ml) Emission Intenslity 15.00 15.00 15.00 15.00 15.00 0.00 5.00 10.00 15.00 20.00 200.0 200.0 200.0 200.0 200.0 327.0 470.0 618.0 783.0 924.0 Based on the data in the table above, determine the original concentration of Na as well as its uncertainty in the unknown sample. Report the uncertainty with three significant figures Concentration Uncertainty Number Number 1.413 / mL / ml, O Previous G.roUp&View; Saadion Check Answer Next Exit.- Hint about us careeprivacy policyterms of uscontactva help

Explanation / Answer

Consider the sample with 5 ml of 1.75 ug/ml standard added

final concentration of added standard in solution = 1.75 ug/ml x 5 ml/200 ml = 0.04375 ug/ml

final concentration of Na+ in this sample solution = [Na+]15/200 = 0.075[Na+]

So,

[Na+]/0.04375 + 0.075[Na+] = 327/470

[Na+] concentration in diluted solution = 0.0538 ug/ml

[Na+] in original solution = 0.0538 x 200/15 = 0.717 ug/ml

Now for uncertainty we calculate Na+ concentration in rest of the solutions and then find standard deviation

sample with 10 ml standard added, Na+ concentration = 0.555 ug/ml

sample with 15 ml standard added, Na+ concentration = 0.441 ug/ml

sample with 20 ml standard added, Na+ concentration = 0.377 ug/ml

mean (X) = 0.522

uncertainty = sq.rt.[(x-X)^2/4)] = sq.rt.((0.038 + 0.0011 + 0.0066 + 0.0210)/4) = 0.129 ug/ml

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