[Mania28]

Hi,

I want to produce a Suspension with a Latex-colloid-concentration of 0,01 g/l. This Suspension should have a pH = 2 and an ionic strength of 0,01 mol/l. The pH should be adjusted with 0,01 or 0,1 M HCl and the ionic strength with NaCl solution. I want to produce a suspension of 4 ml.

At pH values between 4 and 8, I took 0,4 ml of the 0,1 g/l Latex-colloid stock solution and filled it up to 4 ml with 0,011 mol/l NaCl-Solution. So is the correct ionic strength in 4 ml Suspension adjusted. To adjust the right pH, I titrated 0,01 M HCl to this Suspension; the added amounts were so low (between 3 and 30 µl), that the change in concentration/ionic strength is irrelevant.

At a pH of 2 however, the added amounts of 0,01 M HCl are so high, that the ionic strength of the Suspension and the concentration of colloids changes dramatically. if I would take 0,1 M HCl, the added volume to the suspension would be lower, but the change in ionic strength would be the same.

I´d like to know, how it is possible to adjust pH = 2 and ionic strength of 0,01 mol/l in 4 ml Suspension with 0,01 g/l Latex-colloids. Is there a way to calculate the amounts correclty??

Please help me!!!!

[Borek]

0.01M HCl solution has (almost) right pH and right ionic strength, no need for any additions to keep pH or ionic strength at given levels.

Start with 3.6mL of HCl solution of such concentration that once you add 0.4 mL of Latex you will end with 0.01M of HCl.

[Mania28]

0.01M HCl solution has (almost) right pH and right ionic strength, no need for any additions to keep pH or ionic strength at given levels.

Start with 3.6mL of HCl solution of such concentration that once you add 0.4 mL of Latex you will end with 0.01M of HCl.

How do you know that 0,01 M HCl has the right pH? You are right, I measured it and it has pH = 2, but how can one calculate it?

Could you tell me how i can calculate the right HCl concentration of the 3,6 ml so that I end mit 0,01 M HCl?


 

[Borek]

How do you know that 0,01 M HCl has the right pH? You are right, I measured it and it has pH = 2, but how can one calculate it?

What is pH definition?

Could you tell me how i can calculate the right HCl concentration of the 3,6 ml so that I end mit 0,01 M HCl?

http://www.chembuddy.com/?left=concentration&right=dilution-mixing

Note: you may start with 0.1M solution, take required volume, add some water to dilute the acid, add Latex suspension and fill with water up to 4 mL. You don't have to prepare HCl solution of exactly required concentration.

[Mania28]

ok borek, thanks a lot for the *delete me*  but now i´ve got another (hopefully last) question:

If I want to adjust pH = 3 and an ionic strength of 0,01 mol/l (still with NaCl), in the mentioned 4 ml Suspension with 0,01 g/l Latex-colloids. This would be my approach:

pH = 3 conforms with 0,001 mol/l HCl

0,001 mol/l * 4 ml = x ml * 3,6 ml ; x = 0,0011 mol/l

This means, I would dilute 0,0011 mol/l HCl in 3,6 ml Water to get an pH = 3 in 4 ml Suspension.
But the further task is to adjust the whole suspension to an ionic strength of 0,01 M.
Since there would be an ionic strength of 0,0011 mol/l in the suspension because of the HCl, I have to calculate: 0,01 mol/l - 0,0011 mol/l = 0,0089 mol/l NaCl has to be added to the 3,6 ml.

If I have an 0,1 mol/l NaCl stock solution, the calculation would be:

0,0089 mol/l * 3,6 ml = x ml * 0,1 mol/l

x = 0,320 ml

This means, I would add 320 µl of 0,1 M NaCl stock solution to the 3,6 ml, and add 0,4 ml of 0,1 g/l Latex-colloids to get a 4 ml Suspension with an ionic strength of 0,01 mol/l and a pH = 3.

I would be very grateful for your *delete me*!!

[Borek]

pH = 3 conforms with 0,001 mol/l HCl

OK

0,001 mol/l * 4 ml = x ml * 3,6 ml ; x = 0,0011 mol/l

This means, I would dilute 0,0011 mol/l HCl in 3,6 ml Water to get an pH = 3 in 4 ml Suspension.

That's OK, but that's not an easy way. It will be easier to calculate how much HCl you need and use somewhat more concentrated solution to measure this amount of acid. Note that you don't have 0.0011M solution, so you have to either use 0.01M or 0.1M. You need 4mL*0.001M=4µmol of HCl. Convert it to volume of 0.01M HCl solution that contains this amount of acid. Final step will be filling up to 4 mL.

But the further task is to adjust the whole suspension to an ionic strength of 0,01 M.
Since there would be an ionic strength of 0,0011 mol/l in the suspension because of the HCl

You are on the right track, but you forgot something. In the final solution HCl will be no longer 0.0011M, it'll be 0.001M, so the ionic strength will be 0.001 as well.

This means, I would add 320 µl of 0,1 M NaCl stock solution to the 3,6 ml, and add 0,4 ml of 0,1 g/l Latex-colloids to get a 4 ml Suspension with an ionic strength of 0,01 mol/l and a pH = 3.

I would be very grateful for your *delete me*!!

Two mistakes. First was signalled above, second one - 3.6+0.32+0.4mL > 4mL

The real question is, how precisely you need to know pH. If you need it to be just 2, everything is OK, if you need 2.0 or 2.00 then you better check it with pH meter. And in the case of solutions with pH in the range 5-9 (or better 4-10) use buffers to stabilise pH, as in this range minute amounts of acids/bases can change pH dramatically.

[Mania28]

Ok borek, you helped me a lot!!!

Thanks from germany!

[Mania28]

Hi,

now I´ve got a new question concerning the preparation of a solution. I want to produce a 10 ml solution with a Latex-colloid concentration with 0,05 g/l Latex-colloids with an ionic strength of 0,1 mol/l (with NaCl) and pH =2 with HCl.

This would be my approach:

- pipet 5 ml of 0,1 g/l Latex-colloids in flask (to get 0,05 g/l colloids in 10 ml)

- in the remaining 5 ml the ionic strength should be 0,2 mol/l to get in 10 ml 0,1 mol/l
- in the remaing 5 ml the HCl concentration should be 0,02 mol/l to get 0,01 mol/l (pH =2) in 10 ml

my idea woul be to split up the remaining 5 ml in 2,5 ml "ionic strength solution" and 2,5 ml "pH-solution:

HCl: 0,01 mol/l * 10 ml = x * 2,5 ml; x= 0,04 mol/l. => Add 2,5 ml 0,04 M HCl to get pH = 2 in 10 ml. This also adds an ionic strength of 0,01 mol/l to the whole solution of 10 ml.

NaCl: 0,09 mol/l * 10 ml = x * 2,5 ml. => 0,36 mol/l NaCl in 2,5 ml to get an Ionic strength of 0,1 mol in 10 ml suspension.

The Problem is, that this way seems pretty complicated and I wanted to know if there is some easier way.