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
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.
5 answers
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.
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.
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.
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
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
1 answer