Worked example: Calculating concentration using the Beer–Lambert law | AP Chemistry | Khan Academy

So I have a question here from the Cots, Trickle, and Townsend Chemistry and Chemical Reactivity book, and I got their permission to do this. It says a solution of potassium permanganate has an absorbance of 0.53 when measured at 540 nanometers in a 1 centimer cell. What is the concentration? What is the concentration of the potassium permanganate?

Prior to determining the absorbance for the unknown solution, the following calibration data were collected for the spectrophotometer. The way that we would tackle this is we know that there is a linear relationship between absorbance and concentration. We could describe it something like this: that absorbance is going to be equal to some slope times our concentration, and you could say some y-intercept.

If we're purist about it, then the y-intercept should be zero because at a zero concentration, you should have a zero absorbance. But the way that chemists would typically do it is that they would put these points into a computer and then have the computer do a linear regression. You could also do that by hand, but that's a little bit out of the scope of this video.

I did that; I went to Desmos and I typed in the numbers that they gave, and this is what I got. So I just typed in these numbers, and then it fit a linear regression line to it, and it got these parameters: m is equal to this, and b is equal to this.

Now we could say significant figures; it seems like the small significant figures here we have are three, but we could just view the m and the b as intermediate numbers in our calculations. So what I'm going to do is I'm going to use this m and b, and then my final answer I'm going to round to three significant figures.

So what this tells us is that our absorbance is going to be 5.65333 times our concentration minus 0.008. Now they've given us what a is. Let me get rid of all of this stuff here. They told us that our absorbance is 0.539. So we know that 0.539 is equal to 5.65333c minus 0.0086.

And then if you want to solve for c, let's see. We could add this to both sides first, so you get 0.539 plus 0.0086 is equal to 5.65333c. Then divide both sides by this, and you would get c is equal to, or is going to be approximately equal to—be a little careful; all of these would really be approximates.

c is going to be approximately equal to 0.539 plus 0.0086 divided by 5.65333. Of course, we want to round to three significant figures. All right. 0.539 plus 0.0086 is equal to that divided by 5.65333 is equal to this.

So if we go three significant figures, this is going to be 0.0969. So I would write the concentration is approximately 0.0969 molar.