Find number of moles of HCl to change pH of Histidine buffer.

[Sprinkles]

There is 400 mL of 0.2 M histidine buffer- pH 11. Find the moles of HCl that are needed to change the pH to 3.2.

 

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First, this is how you would go from acid to base (acid-->base):

 

1. There is 400 mL of 0.2 M histidine buffer- pH 3.2 Find the moles of OH that are needed to change the pH to 11.

 

H+2 <--> H+1 <-->H0 <-->H-1

 

1) Find a and b using H-H, and pKa values (2.1-pKa of COOH and 3.2 in one equation, and 9.6-pKa of NH3+ and 11 in the second equation).

ADD PLUS ONE. (H+1<---->H0)

a+b+1=total1) [This total is the total # of equivalents.]

 

2) Multiply .4 L by 0.2 M.=total2)

 

3) Multiple total1) and total2) together to get the final answer.

 

 

Now my question:

 

What if we had to go backwards- from base to acid?

 

There is 400 mL of 0.2 M histidine buffer- pH 11. Find the moles of HCl that are needed to change the pH to 3.2.

 

 

Would I subtract -1 now? (Please see 1).) I am just confused on how to go backwards (base to acid). How should the number change? When going from a-->b you need moles of OH- to increase the pH, but from b-->a you would need moles of HCl to decrease the pH, so the number of moles should be different, I assume. However, when calculating it, I get the same number of moles for both H+ and OH-.

 

 

Thank you!

Edited December 4, 2011 by Sprinkles

[Fuzzwood]

You are starting in a basic environment, meaning your histidine is fully deprotonated. What charge is it going to have?

[Sprinkles]

You are starting in a basic environment, meaning your histidine is fully deprotonated. What charge is it going to have?

 

Thank you for the reply!

 

 

It will have a charge of 0, but how would this affect the total number of moles of H+ needed?

 

For example, we would still need to do two H-H equations (1st H-H: use pKa of COOH, 2.1 and pH, 3.2 / 2nd H-H: use pKa of NH3+, 9.6 and pH 11). Find "a" by using the 1st H-H, and find "b" using the 2nd H-H.

 

The total number of moles of Histidine= 0.4 L x 0.2 M= 0.08 moles of Histidine

[*Assuming that the addition of acid doesn't interfere w/volume of solution.*]

 

 

To find the total number of moles of H+ = ("a")(0.08) + ("b")(0.08) (+ or -) _______

 

I'm getting stuck with the last blank. Would I subtract or add 0.08 now?

 

 

Edit: To find the moles of OH- needed to change pH from 3.2 to 11, you would add another 0.08 moles of Histdine (in that last blank).

So, to find the moles of H+ needed to change pH from 11 to 3.2, does this mean that you need to subtract 0.08 moles of Histidine?

 

I am getting confused, because I get the same answer for both the number of moles of OH- needed to go up the curve, and for the number of moles

of H+ needed to go down the curve. My reasoning is because it's all along the same curve, and it's an equilibrium. Therefore, shouldn't the necessary

moles of H+ and OH- be the same?

Edited December 4, 2011 by Sprinkles