Construct a calibration curve by plotting the millivolt response (Y axis) versus
ID: 523822 • Letter: C
Question
Construct a calibration curve by plotting the millivolt response (Y axis) versus the log fluoride concentration (X axis). Using the calibration curve, calculate the average concentration of fluoride in the drinking water samples (don’t forget to take into account the sample dilution).
Preparation of Drinking Water Unknowns:
Pipet exactly 50.0 mL of (cold) tap water (do not use deionized water) into a 100 mL volumetric flask and dilute to 100 mL with the TISAB solution. Mix the solution thoroughly. Prepare another drinking water sample exactly like the first one.
The millivolt response was 250 mV for both samples.
Concentration was calculated using the log of F- in mg/L for the calibration curve
The slope of my calibration curve was y = -57.972x + 226.83
I know I need to plug in the millivolt response for y and then solve for x. Then take the inverse log of x to get a concentration, but I do not know how to “take into account the sample dilution” to give me the correct concentration.
Explanation / Answer
y = -57.972x + 226.83
y - millivolt response
x - log of fluoride concentration in ppm
We have y = 250mV
Then
250 mV = -57.972x + 226.83
-57.972x = 250 - 226.83
-57.972x = 23.17
x = 23.17 / -57.972
x = - 0.39967
log [F] = - 0.39967
[F] = 10-0.39967
[F] = 0.3984 ppm
1 ppm = 1 mg/L
So the concentration of Fluoride is 0.3984 mg/L
The total volume of the solution is 100mL
Mass of Fluoride in 100 mL of solution is
= 0.3984 (mg / L) x 100 x 10-3 L
= 0.03984 mg
This 0.03984 mg of Fluoride came from 50 mL of the tap water.
So the concentration of Fluoride in 50 mL of tap water is
= 0.03984 mg / 50 x10-3 L
= 0.7968 mg/L
The concentration of Fluoride in 50 mL of tap water is 0.7968 mg / L