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Acid-Base Buffer Question

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#1 Luria Bertani

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Posted 13 August 2014 - 04:33 AM

Hello.

I am trying to make up different concentrations of 1M Bis-Tris buffer (MW: 209.24) so I can perform a protein crystallisation experiment. The idea is I make up different 50mL (final volume) x 1M stock solutions of the buffer at pH 5.5, 6, 6.5 and pH 7.

The pKa of Bis-Tris is 6.46, and the pKb is 2.88x10-8, and I prefer to do exact solutions rather than use approximations.

Putting the pH meter into the Bis-tris and then titrating in 12M HCl to adjust the pH but I find that this (a) Takes a long time, ( heats up the solution which then changes the pH a lot © can overshoot and then this causes a lot of waste.

So I am trying to calculate what volume of HCl I need to add to get the various pH's.

When I dissolve 10.462g of Bis-Tris and measure the pH I get a very basic solution. I want to know how much volume of 12M HCl I need to add to get my desired pH (5.5, 6, 6.5 and pH 7) at the end.

I have been able to get a partial solution:

Bis-Tris + H2O <=> Bis-Tris-H + OH

k= (Bis-Tris-H)(OH-) / (Bis-Tris)

When the bis-tris powder is added to water, an amount x is dissociated, giving rise to x moles of Bis-Tris-H and x moles of OH-

kb = (Bis-Tris-H)(OH-) / (Bis-Tris)

kb = (x)(x) / (1M - x)

kb = x2 / (1-x)

Solving a quadratic, x = 0.0017 (this corresponds to (OH-) concentration)

Taking -log (OH-) = pOH of 3.76, or a pH of 10.24 (basic).

This explains why my solution is basic when I dissolve the Bis-Tris buffer powder in water, but how do I work out how much 12M HCl acid to add to get to say pH 5.5??

Any help will be appreciated,

LB

#2 mdfenko

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Posted 14 August 2014 - 06:52 AM

the easiest way to do this would be to use equimolar bis-tris and bis-tris hydrochloride and adjust the mixture with them until you reach the desired pH (the volume will be different but the concentration of the buffering agent will remain constant)

or you can use the henderson-hasselbach equation to determine how much acid to add:

Henderson-Hasselbach Equation

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#3 Luria Bertani

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Posted 14 August 2014 - 08:18 AM

Thanks for the reply. I think I solved the problem today but it didn't quite make sense at the lab bench. I don't have the acid form of the chemical in the lab, only 12M HCl to titrate it with.

Here is my working

--------------------------------

Titration of a weak base with a strong acid

Bis-Tris is a weak base with a pKa of approx. 6.5. This means its pKb is 7.5 (because pKa + pKb must equal 14) or the Kb is 3.16x10^-8

When I dissolve it in water to make a 1M Bis-Tris solution:

BT + H2O <=> BTH  + OH -

Kb = (BTH) (OH-) / (BT)

I can rearrange this so that

(OH-) = kb ((BT) / (BTH)

I want say pH 5.5, this corresponds to a pOH of 8.5, or a hydroxide concentration of 3.16 x 10^-9

Adding strong HCl to my 1M solution will consume the base BT (by x amount) and produce x amount of BTH

(OH-) = kb(BT - x) / (x)

substituting the values and doing this algebraically, this comes to

(3.16x10^-9) = (3.16x10^-8)(1-x) / (x)

(3.47x10^-8)x = 3.16 x10^-8

therefore x = 0.91

0.91 moles of HCl x (1000mL / 12 moles) = 75mL HCl

The problem here is that I went back to the lab bench and checked my solution - something is not right because I have a 50mL tube and I used a pH meter to check the pH (it is 5.5) and pH paper to double check (it is pH 5.5) and it seems unreasonable to fit 75mL x 12M HCL in a 50mL tube.I was able to get the pH of the 1M solution down to pH 5.5 and did not overflow the tube, so I'm still puzzled.

#4 Luria Bertani

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Posted 14 August 2014 - 09:42 AM

Ok, I solved it and I checked it in the lab experimentally and it makes sense at the lab bench!

Solution:

We want pH 5.5 of Bis-Tris. We only have the base form and HCl around.

---

The pH 5.5 corresponds to a pOH of 8.5, this is an OH- concentration of 3.16x10^-9

The first step is to calculate the number of moles of base I have in my 50 mL tube

If I have 50mL (final volume) solution of 1M Bis-tris buffer, then I have 0.05 moles of base.

When I add HCl acid, I reduce the moles of base by x, and create x amount of protonated base BTH

BT + H2O <=> BTH  + OH -

Kb = (BTH)(OH-) / (BT)

rearranging to give OH-

(OH-) = kb ((BT) / (BTH) ; A pH of 5.5 is 3.16x10^-9

3.16x10^-9 = (2.88x10^-8)(0.05 moles of base - x moles HCl) / (x moles BTH created)

a bit of algebra

(3.16x10^-9)x = (1.44x10^-9) - (2.88x10^-8)x = 1.44X10^-9

= (3.196x10^-8)x = 1.44x10^-9

x = (1.44x10^-9) / (3.196x10^-8)

x = 0.045 (I presume that this is the number of moles of HCl required to reduce the pH to 5.5)

Assuming 12M HCl,

0.045 moles x (1000mL / 12 moles) = 3.75 mL of 12M HCL (or 7.5mL of 6M HCl if working with fuming HCl is to be avoided)

------

To verify, I went to the lab, measured out 0.05L x 1M x 209.24 g/L of Bis-Tris (Sigma) = 10.462g and dissolved in 40mL of water. The pH I measured using the pH meter was about 10.20, which is basic and what I expect.

I then measured out 7.5mL x 6M HCl (I decided the 12M HCl was too fuming and dangerous) and added it, and topped it up to the final volume of 50mL with water.

I gave it a good shake, and put the pH meter in.

The reading was pH 5.5, BANG ON.

Solved. :=)

Edited by Luria Bertani, 14 August 2014 - 09:44 AM.

#5 bob1

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Posted 14 August 2014 - 01:15 PM

Note that tris is slowly protonated, so it would pay to check the pH today, it may well be that you have overshot.

#6 mdfenko

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Posted 03 September 2014 - 04:06 AM

Note that tris is slowly protonated, so it would pay to check the pH today, it may well be that you have overshot.

luria bertani is using bis-tris, not tris (i'm not sure if bis-tris protonates at a different rate than tris, so, your statement may still be important to note).

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#7 bob1

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Posted 03 September 2014 - 12:49 PM

Note that tris is slowly protonated, so it would pay to check the pH today, it may well be that you have overshot.

luria bertani is using bis-tris, not tris (i'm not sure if bis-tris protonates at a different rate than tris, so, your statement may still be important to note).

Good point - I missed that in my reading of the posts.