Standard Addition Question:
Serum containing Na+ gave a signal of 4.27 mV in an atomic emission analysis. Then 5.00 mL of 2.08 M NaCl were added to 95.0 mL of serum. This spiked serum gave a signal of 7.98 mV. Find the original concentration of Na+ in the serum.
I can't figure out the 0.113 M answer using the method poorly explained in the textbook.
Okay, I just figured out the answer and where my confusion came from. Correct me if I am wrong.
x/0.104+0.950x = 4.27/7.98
7.98x = 0.44408+4.0565x
7.98x-4.0565x = 0.44408
3.9235x = 0.44408
x = 0.44408/3.9235
x = 0.1131846566
We have two signals let say s1 and s2 where s1=4.27mV and s2=7.98mV
Vol of the sample (Vs)=5.00mL
Vol of dilution (Vx)=95.0mL
Total vol(Vt)=100mL
Conc of sample (Cs)=2.08m
Consider the equation
S1/s2=(CxVt)/(VxCx +VsCs)
Where Cx is conc of Na
By making Cx the subject then substitute the values to the equation you will get 0.113m
You know, of course, that I have no idea how your book explains it, poorly or otherwise. Here is how you do it but I don't know it's the same method your book uses. Personally, when I was in grad school we did it all by graphing it. Makes a lot more sense to me. You can do it mathematically but you need a graph to see what is going on. Do this and you can just sketch it on a sheet of paper.
Draw the Y axis and X axis and mark the origin as zero concn. At the zero concn (which is what you're calling the sample), place a dot ON the Y axis and label it 4.056 mv.(That 4.27 was obtained in the pure sample; i.e., in the 95.00 mL sample before dilution. When you dilute it to 100 it will be 4.27 x (95/100) = 4.056) That's the zero point for the sample. How much Na^+ did you add? That's 2.08 M x 5.00/100 = 0.104M
On the sketch, go out on the x axis some distance and place a dot at 0.104 M and move up on the Y axis some distance to 7.98 and place a dot there. Then you can sketch a straight line from the 4.056 mv mark at the zero point to the 7.98 mv point for 0.104M. That's your graph. You can see that 0.104 M added to the sample gave you 7.98-4.056 = 3.92 mv. So stick that into A = abc (and since the cell length is constant we can dispense with b--of course there is no cell in AES but it is a constant length) so 3.92 = ac.
a = absorptivity constant = 3.92/0.104 = 37.72. Use that constant for a in the sample calculation.
A = ac. 4.27 = 37.72c and
c = about 0.113 M.
If you do this on a sheet of graph paper and do it accurately, You extrapolate that straight line TO THE LEFT so it crosses the x axis to the left of the origin and you read the concn of the sample at that point.
If this is not the same method as your book wants it done, you can write up the procedure from your book and I will interpret it and try to put it in words you can understand better than it is explained to you.
OK. I see I didn't do it the same way but you can also see that my numbers are the same. That 3.92 is the absorptivity constant and the 4.056 actually was 4.0565 but I just dropped the last 5. The book method uses ratio/proportion while I tried to reason it out from the beginning. Hope this helped a little.
Unfortunately, there is often more than one way to answer a textbook example. In class, it's one way (the quickest way), and in the textbook, it's another.
The "quick" way is fine for tests and exams... but it is useful to me that I learn alternate methods as well.
A step by step approach was lacking from the textbook explanation (my only complaint).
Anyway; thanks for your suggestion. I will give your method a shot.
To find the original concentration of Na+ in the serum, we can use the method of standard addition. This method involves adding a known quantity of a standard solution to a sample and measuring the change in signal to determine the concentration of the analyte in the sample.
Let's break down the problem step by step:
Step 1: Calculate the change in signal caused by adding the standard solution.
In this case, we added 5.00 mL of 2.08 M NaCl to 95.0 mL of serum. The initial signal of the serum was 4.27 mV, and the signal after adding the NaCl was 7.98 mV. Therefore, the change in signal is:
Change in signal = Signal after adding NaCl - Initial signal
= 7.98 mV - 4.27 mV
= 3.71 mV
Step 2: Calculate the concentration of the added NaCl solution.
To do this, we can use the formula:
Concentration = (moles of solute)/(volume of solution in liters)
First, let's convert the volume of NaCl solution into liters:
Volume = 5.00 mL = 5.00 x 10^-3 L
Next, calculate the moles of NaCl using the molarity and volume:
Moles of NaCl = Molarity x Volume
= 2.08 M x 5.00 x 10^-3 L
= 0.0104 moles
Step 3: Calculate the dilution factor caused by adding the NaCl solution.
The dilution factor can be calculated using the volumes of the sample and the added solution. In this case, the serum volume is 95.0 mL, and the NaCl volume is 5.00 mL.
Dilution factor = (Total volume of sample + volume of added solution)/(Volume of sample)
= (95.0 mL + 5.00 mL)/(95.0 mL)
= 1.053
Step 4: Calculate the concentration of Na+ in the serum.
This can be calculated using the formula:
Concentration of Na+ in serum = (Change in signal)/(Slope x Dilution factor)
The problem states that the concentration of Na+ in the serum after adding NaCl is 0.113 M, so we can use this value as the initial concentration of Na+ in the serum. Rearranging the formula, we get:
(Change in signal)/(Slope x Dilution factor) = Concentration of Na+ in serum
Solving for the concentration of Na+ in serum, we have:
Concentration of Na+ in serum = (Change in signal)/(Slope x Dilution factor)
= 3.71 mV/(Slope x 1.053)
= 0.113 M
Therefore, the original concentration of Na+ in the serum is 0.113 M.