Question

How would I use the Standard Addition method to calculate the ppm Standard with the following table?

Solution #	mL Sample	mL Standard	ppm Standard	Current (mA)
1	8.40	0.00	0.00	.0065614
2	8.40	0.200		.0089296
3	8,40	0.400		.0109682
4	8.40	0.600		.0130220
5	8.40	0.800		0.0159060

Answer 1

Ans. Preparation of aliquots using standard addition.

Example- 2^nd aliquot:

The concentration of standard solution is NOT mentioned. Let the concentration of standard solution of the analyte be, [X] = 100.0 ppm.

To 8.40 mL aliquot of original sample, 0.200 mL of 100 ppm standard is added.

Final volume of solution = 8.40 mL (original sample) + 0.200 mL (standard) = 8.60 mL

Now, Using                C1V1 = C2V2

C1= Concentration, and V1= volume of initial solution 1         ; Std. solution (ppm std.)

C2= Concentration, and V2 = Volume of final solution 2         ; Final spiked soln.

Putting the values in equation 1-

            100.0 ppm x 0.200 mL = C2 X 8.60 mL

            Or, C2 = (100.0 ppm x 0.200 mL) / 8.60 mL mL = 2.3256 ppm

* In above equation, C1V1 = ppm standard.

Similarly, calculations are done for all the aliquots in excel. The calculation table is shown in picture.

A graph is plotted using final [X] in aliquots on X-axis and respective current on Y-axis.

# Calculating [X] in un-spiked original sample.

Aliquot 1, to which no standard is added is called the un-spiked original sample. The concentration of [X] in un-spiked aliquot is thus equal to the [X] in original sample.

The linear equation obtained from graph “y = 0.001x + 0.0064” is in form of y = mx + c.

            Where, c = y-intercept         , m = slope

From the linear graph equation-

[X] in un-spiked original sample is given by = (y-intercept / slope) concentration units

= (0.0064 / 0.001) ppm

= 6.4

Therefore, [X] in original un-spiked sample= 6.4 ppm

Note: 1. Use the concentration of standard solution provided to you to calculate final [X] in finally diluted solution.

2. You can also plot ppm of X from standard solution vs respective current as needed. This graph would give you total amount of the desired molecule in 8.40 mL original sample, but not its concentration.