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 unknowns have a mean reading of 60.0. The slope of the calibration curve is 1.75×109M−1.
a) Estimate the signal detection limit for EDTA.
b) What is the concentration detection limit?
c) What is the lower limit of quantitation?

Answer 1

Answer:

Following are the solution to these question:

Explanation:

Calculating the mean:

$\bar{x}=\frac{175+104+164+193+131+189+155+133+151+176}{10}$

  $=\frac{1571}{10}\\\\=157.1$

Calculating the standardn:

$\sigma=\sqrt{\frac{\Sigma(x_i-\bar{x})^2}{n-1}}$

Please find the correct equation in the attached file.

$=28.195$

For point a:

$=3s+yblank \\\\=3 \times 28.195+50\\\\=84.585+50\\\\=134.585$

For point b:

$=3 \ \frac{s}{m}\\\\ = \frac{(3 \times 28.195)}{1.75 \times 10^9 \ M^{-1}}\\\\= 4.833 \times 10^{-8} \ M$

For point c:

$= 10 \frac{s}{m} \\\\= \frac{(10 \times 28.195)}{1.75 x 10^9 \ M^{-1}}\\\\ = 1.611 \times 10^{-7}\ M$

It is calculated by using the slope value that is $1.75 \times 10^9 M^{-1}$. The slope value $1.75 \times 10^9 M^{-1}$is ambiguous.