1) You need the salts of the metals of your interest; such as CuCl2.2H2O for copper, Pb(NO3)2 for lead and etc. 

2) Then you have to prepare the stock solution (up to 1L, usually) of the metals with your desired concentration (ppm or mg/l, both are the same unit). For this, please follow these calculations (i will take the example of copper salt mentioned above): if you prepare the stock solution of 500 mg/l, it is 500 mg of Cu/1000 ml of deionized water.

3) Now, how to calculate 500 mg of Cu from the salt? You must find the molecular weight of the salt which is 170.5 g for CuCl2.2H2O, and the molar mass of the metal which is 63.5 for Cu. Therefore, 1g(1000 mg) of Cu from the salt = (170.5/63.5) g of salt. That goes to 2.685 g from the salt bottle. But, that one is for 1g (1000 mg) and you need 500 mg. So, divide it by 2. The result would be 1.34 g.

4) You are almost there. Take that 1.34 g of salt powder, a 1L of volumetric flask, a funnel, and deionized water.

5) With the help of the funnel, drop the powder into the flask, pour a little deionized water and shake it in order to have a homogenous solution.

6) When it's done, fill the flask up to 1000 ml or 1L and you have 500 ppm (or mg/l) of Cu2+ solution.

It applies to all metals or adsorbates solution preparation.

Now, take convenient amounts of adsorbate solution (5ml or 10 ml) from the flask and dilute it to make various concentrations for your adsorption study. Please follow the calculations for dilution:

1) Suppose you want to do experiment with 50 ppm of Cu. Then make this equation: C1 V1 = C2 V2.

C1=initial concentration, V1=volume you need to take from the stock solution you prepared (from the flask), C2= your desired concentration (50 ppm in this case), V2= the final volume with which you'll do your experiment.

So, for our example, C1= 500, V1= ?, C2= 50, V2= 250 (it'll depend on your flexibility).

Therefore, V1=25 ml.

Then, take 25 ml from the flask and mix deionized water (225 ml) to make the solution up to 250 ml. And you have 50 ppm of Cu. 

You can make various concentrations from your stock solution using this calculation. 

Be very precise during the measurements. 

Wish you all the best.

Salehin.

1) You need the salts of the metals of your interest; such as CuCl2.2H2O for copper, Pb(NO3)2 for lead and etc. 

2) Then you have to prepare the stock solution (up to 1L, usually) of the metals with your desired concentration (ppm or mg/l, both are the same unit). For this, please follow these calculations (i will take the example of copper salt mentioned above): if you prepare the stock solution of 500 mg/l, it is 500 mg of Cu/1000 ml of deionized water.

3) Now, how to calculate 500 mg of Cu from the salt? You must find the molecular weight of the salt which is 170.5 g for CuCl2.2H2O, and the molar mass of the metal which is 63.5 for Cu. Therefore, 1g(1000 mg) of Cu from the salt = (170.5/63.5) g of salt. That goes to 2.685 g from the salt bottle. But, that one is for 1g (1000 mg) and you need 500 mg. So, divide it by 2. The result would be 1.34 g.

4) You are almost there. Take that 1.34 g of salt powder, a 1L of volumetric flask, a funnel, and deionized water.

5) With the help of the funnel, drop the powder into the flask, pour a little deionized water and shake it in order to have a homogenous solution.

6) When it's done, fill the flask up to 1000 ml or 1L and you have 500 ppm (or mg/l) of Cu2+ solution.

It applies to all metals or adsorbates solution preparation.

Now, take convenient amounts of adsorbate solution (5ml or 10 ml) from the flask and dilute it to make various concentrations for your adsorption study. Please follow the calculations for dilution:

1) Suppose you want to do experiment with 50 ppm of Cu. Then make this equation: C1 V1 = C2 V2.

C1=initial concentration, V1=volume you need to take from the stock solution you prepared (from the flask), C2= your desired concentration (50 ppm in this case), V2= the final volume with which you'll do your experiment.

So, for our example, C1= 500, V1= ?, C2= 50, V2= 250 (it'll depend on your flexibility).

Therefore, V1=25 ml.

Then, take 25 ml from the flask and mix deionized water (225 ml) to make the solution up to 250 ml. And you have 50 ppm of Cu. 

You can make various concentrations from your stock solution using this calculation. 

Be very precise during the measurements. 

Wish you all the best.

Salehin.

Thanks a lot Salehin. Extremely helpful. If you have any papers give link of that in heavy metal adsorption by various adsorbates. I will meet u in further discussions.

Dear Dr. Sundarrajan, I am glad that it was of your use. Please find the attached papers. 

Thank you Salehin. Great job!

Nothing to add, except the choice of the salt, which should be soluble, of course, but not too hygroscopic for a good accuracy. For example, nitrates of many transition metals (chromium, nickel, iron, etc.) are EXTREMELY hygroscopic. Therefore, once open, the flasks containing these compounds absorb huge amounts of water, and the salts become deliquescent. So there is no chance to get the required concentrations by weighing the salts, which can never be dry. Better using chlorides or sulfates, far less hygroscopic, providing that their solublity is good enough, of course.

Alain

I used soluble salts of the metal needed such as chloride or nitrate of metals to prepare 1000 ppm solution and then dilute to the desired conc. in deionized water, making a calibration curve and then find the conc. of the unknown sample.

Thanks Mr. Alain. I prefer salts as you said to avoid hygroscopic nature of others.

Yenial, Will nitric acid interfere in the adsorption behaviour of the adsorbate?

One important point is missing...that is the purity of the salt :).. although most of the times analytical grade salts are used... but its always important to check the % purity to get accurate results. Never forget to divide the required value(mass) by its % purity to get the actual value...regarding adding nitric acid to the salt solution.... then i wont recommend it... as while peroforming adsorption studies, pH plays a vital role and may alter the mechanism of the uptake...therefore, nitric acid should be added only if low pH aduustment is required.... . to be sure of precipitation try to consult the speciation diagram of the metals .. regards

Elaborate answer provided by Sirajus Salehin. A great effort.

Dr. kazmi, can you please explain your point by giving an example. Are you saying that we should not nitric acid to the stoch solution for preventing precipitation. because, in adsorption studies, we always set the pH in the range  of 2 to 10 almost.

Do you mean the preparation of aqueous solution? Then there are various methods to prepare it, but then it depends on the type of adsorbate you require. i.e If you need to prepare for Iron you have separate procedure and similarly for other metals of salts.