# Homework Help: H-H equation question

1. Oct 15, 2011

### vande060

1. The problem statement, all variables and given/known data

prepare 40mM TES buffer pH 7.4
prepare 100mls of it
mw TES = 229.25
pka = 7.5

2. Relevant equations

ph = pka +log(A-/HA)

3. The attempt at a solution

My problem is that I don't know what the MW of the acid or base is for this buffer. Is the given MW the acid, and I subract a proton weight for the base or something? I can do the calculation though:

7.4 = 7.5 +log(A-/HA)

-.1 = log(A-/HA)

A-/HA = .79

56% base
44% acid

.56* 40mM * MW of base *.1L = moles of base
.44*40mM *MW of acid *.1L = moles of acid

So how should I figure out the MW of acid and base?

2. Oct 15, 2011

### Staff: Mentor

http://en.wikipedia.org/wiki/TES_(buffer [Broken])

Given molar mass refers to the acid.

Last edited by a moderator: May 5, 2017
3. Oct 15, 2011

### vande060

Thank you,

Ok, thats kind of what I thought. So the acid MW is 229.25, and the base would be (229.25 - 1.008) is that correct?

Last edited by a moderator: May 5, 2017
4. Oct 15, 2011

### Staff: Mentor

You don't need molar mass of the conjugate base. Use enough acid to prepare 40 mM solution, neutralize it with any strong base.

5. Oct 15, 2011

### vande060

so I would go:

229.25g/mol * .04moles/liter *.1L = .917g TES added

then to calulate strong base needed to convert 56% of acid to base

.04moles/liter *.1L = .004 moles TES * .56 = .00224 moles of strong base needed

6. Oct 15, 2011

### Staff: Mentor

Check your math. If more than half of the acid is converted to base pH is higher than pKa.

But in general you are on the right track.

7. Oct 15, 2011

### vande060

whoops it should be 44% converted to base