[earthnation112]

Hi, just had a question which states:

“The following solutions are required:
1.   10 ml of a 1/250th diluted seawater sample + 5 ml 4 ppm magnesium solution diluted to 100 ml.
2.   10 ml of a 1/250th diluted seawater sample + 10 ml 4 ppm magnesium solution diluted to 100 ml.
3.   10 ml of a 1/250th diluted seawater sample + 15 ml 4 ppm magnesium solution diluted to 100 ml.
4.   10 ml of a 1/250th diluted seawater sample + 20 ml 4 ppm magnesium solution diluted to 100 ml.”


Information given regarding this experiment is:

Apparatus and Solutions Required
1. Master stock solution of containing 1000 ppm magnesium is provided (containing 10.139 g of A.R. MgSO4.7H2O dissolved in 1 litre of 0.5 M HCl)
2. Seawater sample
3. All other apparatus is available in the tray near the equipment


When it’s says for the first solution you need “5 ml 4 ppm magnesium solution diluted to 100 ml.”

My attempt to make the solution is:

Using formula: (req ppm x req vol)/stock = no of ml for req vol
(4 ppm x 5 ml)/ 1000 ppm = 0.02 ml

Would this mean extracting 0.02ml and adding this to 99.88ml of deionised would make a "5ml 4ppm magnesium solution diluted to 100 ml"? Thank You

[Borek]

And not

(10 ml of a 1/250th diluted seawater sample + 5 ml 4 ppm magnesium solution) diluted to 100 ml

?

[earthnation112]

And not

(10 ml of a 1/250th diluted seawater sample + 5 ml 4 ppm magnesium solution) diluted to 100 ml

?

So to make a "5 ml of 4 ppm magnsium solution" from the stock solution(1L) which is 1000 pm all I do is extract 5 ml and add 95 ml of deionsed water to make a 100 ml solution of 4 ppm?