9. A 0.2056 g chocolate sample was digested (completely dissolved) in nitric aci
ID: 3303325 • Letter: 9
Question
9. 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 475 ppb chromium standard. A standard addition calibration was obtained by measuring the constant volume solutions using ICP-MS to determine the chromium concentrations (results shown below). *See page 16 in textbook for ppb definition Vs ful) LINEST Output 1888.881 9950.456 63.61053 360.0394 0.995484 495.6791 S+X 30000 y 1888 3x + 9950.5 R2 =0.9955 9508 14062 17488 20037 24003 27661 25000 20 20000 15000 60 80 100 10000 DEVSQ= 60.72158 5000 0 0 6 [Crl (ppb) 8 A. what is the chromium concentration (in ppb) ± ,for the diluted (5 mL) sample? B. what is the chromium concentration (in ppb) ± 95% confidence interval for the diluted (5 mL) sample? C. What is the chromium concentration (in ppb) for the original (solid) chocolate sample?Explanation / Answer
(a)
In the data given for the diluted 5mL sample, making use of the best fit equation, we get the following conc. values for the six solutions in series:
0, 0.573, 1.05, 1.405, 1.956, 2.46
The 95% C.I. is then : -0.52 to 3 ppb
(b)
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
(C)
For the original chocolate sample, we put y = 0 in the best fit equation.
Putting y=0, we get:
x = -1.386
The -ve sign is because of extrapolation into the -ve x axis.
So, [Cr] in chocolate sample = 1.386 ppb
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.