Question

I need to calculate the concentration of magnesium in wastewater. A 10 ml sample is taken and diluted to 50ml.

To 2.00 ml of this solution is added 3ml 0.0 ppm magnesium standard. The absorbance is 0.336.
For the second one, 3ml 1.0 ppm magnesium standard is added to 2.00 ml of the solution. The absorbance is 0.478.

I have to calculate the concentration magnesium in wastewater.

Answer 1

hmmm

Answer 2

I use the following formula:

\(\Large C_A=\dfrac{S_1*C_s*V_s}{(S_2-S_1)*V_0}\)
Cs= concentration standard
CA= concentration analyte
S1= signal sample (absorbance)
S2= signal sample + standard
VS= volume added standard
V0= volume sample

\(\Large C_A=\dfrac{0.336*1*3}{(0.478-0.336)*2}=3.549\ ppm\)
For the last part i multiply the ppm with 5 (as the sample is 5x diluted)

Is this the correct way?
I'm worried about \(\large V_0\), would it be 2 or 5 (because 3 ml 0.0ppm Mg is added)

Answer 3

@aaronq do you know?

Answer 4

...is added 3ml 0.0 ppm magnesium standard...

is that zero?

Answer 5

the mass of the magnesium as i calculated is 7 mgr.
you can find the concentration in any volume. if divided by 2. your answer is correct.

Answer 6

Oke thank you @ardalan