How To Find Concentration From Calibration Curve

Solution 1:

I assume your teacher explained the HPLC separation. If injected a mixture of four ions you lot will get iv peaks. Each peak has an area, which is proportional to the concentration of the substance.

Imagine yous wanted to determine the concentration of chloride ions in your tap water. What you lot would exercise is that you will prepare several known concentrations of chloride ions using $\ce{NaCl}$ solutions, say $0$ to $\pu{350 ppm}$ (look at your calibration chart). Inject them one past 1, and measure the height area, let united states of america say the 2d acme is that of chloride ion. The pinnacle shape is Gaussian in real work just yous can "estimate" them as triangles. Recall triangle's expanse is very easy to calculate.

First step: Plot the peak area for concentration. What you become is a calibration curve.

The adjacent step is to find a mathematical equation that fits this data.

Fortunately it is linear in your example of the form $y=mC$ where $y= \text{peak area}$, $k=\text{slope}$, $C=\text{concentration}$ in $\pu{ppm}$.

Now yous would inject your tap water in the HPLC, you lot do non know the concentration ($C_1$) but you practise know the peak expanse.

The last pace is called interpolation. You go from the known area, and read the corresponding concentration centrality. Viola! Now you know the chloride concentration in your tap water!

Alternatively, you use the equation, $y=0.1312C$, so I know the surface area of the tap water chloride injection, say, information technology was $\pu{40.0 units}$, so $C_1 = 40/0.1312 = \pu{304.9 ppm}$.

In typical tap h2o information technology is around $\pu{100 ppm}$, I just made up those numbers.

Bonus: Rationalize the units of the summit area on the $y$-axis! Why it is conductivity times minutes?

Solution 2:

Suppose it is a calibration curve of high functioning liquid chromatography (HPLC) to determine concentration of chloride ion in unknown solutions. The detector is sensitive to charge ions, which gave the peaks, area of which is proportional to concentration of ions. I assume each height area is highly specific for ion in interest. First you make a calibration bend using standard $\ce{Cl-}$ solutions with known concentrations:

The bend is a straight line, which goes through $(0,0)$ origin ($y = 0.1312x$) with an excellent correlation with $R^two = 0.9995$ ($R^2 = 1.0$ is the best correlation). I insert an instance of chromatography trace to explicate how concentration of ion is related to expanse under the relevant superlative. When running the standard solution, you'd find the right retention time ($t_R$) for the chloride ion. Once you lot prepare your scale curve, you can run your unknown sample (suppose it contains $\ce{Cl-}$ and $\ce{NO3-}$ ions) and find the area under the chloride peak (Nowadays, all meridian areas of the chromatogram are automatically given by the software). Suppose it was $\pu{22.five Absorbance/min}$ as indicated in the paradigm. Since the equation of the best fitted direct line is given automatically ($y = 0.1312x$), y'all tin can submit the $y$-value to information technology and notice the $10$-value, which is the concentration in $\pu{ppm}$:

$$22.five = 0.1312x \ \Rightarrow \ x = \frac{22.5}{0.1312} \approx \pu{172 ppm}$$

Modernistic technalogy made these calculations like shooting fish in a barrel by giving the equation for the calibration curve and the areas under the curve. In sometime-way, for example, you have to cut each summit from each printed chromatogram, and weigh them by belittling balance to correlate them for known concentration. Farther, you plot the values in a graph paper and brand the best fitted directly line. Once, yous got the peak surface area value for your unknown solution, you mark that value in $y$-axis and draw a directly line parallel line to $x$-axis until it meets the curve. From that crossing betoken, you again depict a directly line parallel line to $y$-axis until it meets $x$-centrality (meet maroon lines in the image). The point second parallel line meet $x$-centrality is the chloride ion concentration in your unknown sample.

Note that you lot cannot find the $\ce{NO3-}$ ion concentration in your unknown sample (see the insert) using the same scale curve. The detector been used might bear witness different response to the nitrate ion compared to chloride ions. Therefore, you must make dissimilar calibration curve for the nitrate ions using standard nitrate solutions.

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Solution 3:

The line goes according to : $y = 0.1312~x$ ; In other words the area of the peak is equal to $0.1312$ times the concentration of chloride in ppm. This result is extremely useful. If for example, in a later measurement you obtain a peak whose surface area is $y = 35$, information technology means that the concentration $x$ of chloride ion responsible of this peak is : $x = 35/0.1312 = 266.eight$ ppm

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Solution iv:

If we take 1st and final points on your graph we'd see that:

• At concentration $\pu{350ppm}$ your detector shows number $45$
• At concentration effectually $0$ your detector shows number effectually $0$
• And so what if detector shows $175$ (which is $0.5*(350-0))$? You lot may conclude that the concentration is $\pu{22.5ppm}$ (which $0.5*(45-0)$)

Calibration curve gives y'all understanding of the relationship between the number that detector shows and the concentration. So that when you see some reading you could calculate the concentration.

Note that ii points usually isn't enough because you don't necessarily have a directly line. So you need more measurements to see if information technology's true. If information technology's non true - you'll have to come with more complicated equations than $y=0.1312x$ and your results will be less precise.

Source: https://newbedev.com/how-to-read-a-chromatography-calibration-curve

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