Question

Low concentrations of EDTA near the detection limit gave the following dimensionless instrument readings: 175, 104, 164, 193, 131, 189, 155, 133, 151, and 176. Ten blanks had a mean reading of 50.0. The slope of the calibration curve is 1.75 x 10^-9 M^-1.

a. Estimate the signal detection limit for EDTA.
b. What is the concentration detection limit?
c. What is the lower limit of quantification?

Answer 1

Answer:

Step-by-step explanation:

From the given information:

The mean of the readings is:$=\dfrac{175+104+164+193+131+189+155+133+151+ 176}{10}$

$= \dfrac{1571}{10}$

= 157.1

The standard deviation (SD) can be computed by using the expression:

$SD =\sqrt{ \dfrac{\sum_f(x_i - \bar x)^2}{n-1}}$

$SD =\sqrt{ \dfrac{(175-157.1)^2+(104-157.1)^2+(164-157.1)^2+...+(176-157.1)^2}{10-1}}$

Standard deviation = 28.195

∴

FOr the EDTA complexes;

The signal detection limit = (3*SD) + $y_{blanks}$

= (3*28.195) + 50

= 84.585 + 50

= 134.585

We need to point out that the value of the calibration curve given is too vague and it should be (1.75 x 10^9 M^-1) as oppose to  (1.75 x 10^-9 M^-1)

The concentration of detection limit is:

$=\dfrac{3 \times SD}{slope }$

$=\dfrac{3 \times 28.195}{1.75 \times 10^{9} \ M^{-1} }$

$\mathbf{= 4.833\times 10^{-8} \ M}$

The lower limit of quantification is:

$=\dfrac{10 \times SD}{slope }$

$=\dfrac{10 \times 28.195}{1.75 \times 10^{9} \ M^{-1} }$

$\mathbf{= 1.611 \times 10^{-7} \ M}$