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Calibration curves and concentration

Syllabus HAVO: Chemical methods

Question 1 / 50 correct

A sample diluted fourfold contains 2.00 mg/L. What was the original concentration?

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More explanation and a worked example

A line from known samples

A calibration line connects a measured signal with known concentrations. Here A = 0.020 + 0.100 c, with c in mg/L. Signal A is dimensionless here. The data cover 0 to 5 mg/L.

Two separate steps

First subtract the intercept and divide by the slope. This gives the concentration of the measured sample. If it was diluted, multiply by the dilution factor afterwards. Outside the calibration range, further evidence is required.

A sample diluted fivefold has A = 0.370.

  1. c = (0.370 − 0.020) / 0.100 = 3.50 mg/L.
  2. The original solution is five times as concentrated.
Originally: 5 × 3.50 = 17.5 mg/L.
EXPLORE THE CALIBRATION

A = 0.02 + 0.1 c

012345600.20.40.6Ac (mg/L)

c = 3.5 mg/L

The same data in a table
c (mg/L)A
00.02
10.12
20.22
30.32
40.42
50.52
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