A 0.2056 g chocolate sample was digested (completely dissolved) in nitric acid a
ID: 486910 • Letter: A
Question
A 0.2056 g chocolate sample was digested (completely dissolved) in nitric acid and the resulting solution diluted to a total volume of 10.00 mL. A series of solutions were then prepared with total constant volumes of 5.00 ml, each containing 1.00 ml of the chocolate solution plus variable volumes of a 125 ppb chromium standard. The constant volume solutions were analyzed using ICP-MS to determine the chromium concentrations (results shown below). *See page 16 in textbook for ppb definition. A. What is the chromium concentration (in ppb) plusminus 95% confidence interval for the diluted (5 ml) sample? _____________Explanation / Answer
8. From the given data
concentration of standard (Cs) in the dilute sample when,
V(ml) Cs(ppb)
0 0
0.02 125 x 0.02/5 = 0.5
0.04 125 x 0.04/5 = 1.0
0.06 125 x 0.06/5 = 1.5
0.08 125 x 0.08/5 = 2.0
0.1 125 x 0.1/5 = 2.5
Concentration of chromium in dilute unknown sample Cx
with 0.5 ppb standard
Cx/(Cx + 0.5) = 9508/14062
Cx = 1.044 ppb
with 1.0 ppb standard
Cx/(Cx + 1.0) = 9508/17488
Cx = 1.191 ppb
with 1.5 ppb standard
Cx/(Cx + 1.5) = 9508/20037
Cx = 1.354 ppb
with 2.0 ppb standard
Cx/(Cx + 2.0) = 9508/24003
Cx = 1.312 ppb
with 2.5 ppb standard
Cx/(Cx + 2.5) = 9508/27661
Cx = 1.310 ppb
mean concentration = 1.242
standard deviation = sq.rt.(sum(x-X)/5)
= sq.rt.((0.0392 + 0.0026 + 0.0125 + 0.0049 + 0.0046)/5)
= 0.113
at 95% confidence level the range would be 1.102 to 1.382 ppb
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