6. The quantitative analysis for reduced glutathione in blood is complicated by
ID: 714636 • Letter: 6
Question
6. The quantitative analysis for reduced glutathione in blood is complicated by many potential interferents. In one study, when analyzing a solution of 10.0 ppb glutathione and 1.5 ppb ascorbic acid, the signal was 5.43 times greater than that obtained for the analysis of 10.0 ppb glutathione. What is the selectivity coefficient for this analysis? The same study found that analyzing a solution of 3.5x102 ppb methionine and 10.0 ppb glutathione gives a signal that is 0.906 times less than that obtained for the analysis of 10.0 ppb glutathione. What is the selectivity coefficient for this analysis? In what ways do these interferents behave differently? 7. Oungpipat and Alexander described a method for determining the concentration of glycolic acid (GA) in a variety of samples, including physiological fluids such as urine. In the presence of only GA, the signal is Ssample,1 kGACGA and in the presence of both glycolic acid and ascorbic acid (AA), the signal is Ssample.2 kGACGA+ KAACAA When the concentration of glycolic acid is 1.0 x 10-4 M and the concentration of ascorbic acid is 1.0 x 10-5 M, the ratio of their signals is sample1.44 Ssample 2 (a) Using the ratio of the two signals, determine the value of the selectivity ratio KGA,AA (b) Is the method more selective toward glycolic acid or ascorbic acid? (c) If the concentration of ascorbic acid is 1.0 x 10-5 M, what is the smallest concentration of glycolic acid that can be determined such that the error introduced by failing to account for the signal from ascorbic acid is less than 1%?Explanation / Answer
S(sample)=S(analyte)+S(interferent)=K(analyte)xC(analyte)+K(interferent)xC(interferent)
Where S is the signal of the sample, k is the interferent sensitity.
lets assume signal contribute due to only glutathione is 100.
k(g.t)=S(g.t)/C(g.T)=100/10=10.
S(sample)=(K(g.t)xC(g.t) )+ ( K(a.a)xC(a.a))
105.43=(10x10)+k(a.a)X1.5
105.43=100+1.5K(a.a)
5.43=1.5k(a.a)
k(a.a)=5.43/1.5=3.62.
Selectivity coefficiet=k(ascorbic acid)/k(glutathione)=3.62/10=0.362
2. The signal is 0.906 times lesser when interferent is present.
hence S(sample)=100-0.906=99.094
99.094=(10x10)+k(m.t)x350=100+350k(m.t)
350k(m.t)=99.094-100=-0.906
k(m, t)=-0.906/350=-0.002588
k(m.t, g.t)=k(m,t)/k(g.t)=-0.002588/10=-2.588x10-4
both the methods are selective only for interferenct and not for analyte.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.