# Homework Help: Buffer calculations - Help

1. Feb 24, 2010

### ttor1236

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

Prepare 500ml acetic buffer with a concentration of 1 M and a pH o 6.0. (pKa for acetic acis is 4.76; molecular weight for actetic acid is 60g/mol and for sodium acetate is 136g/mol)

2. Relevant equations

3. The attempt at a solution

2. Feb 24, 2010

### JJMB

Use the Henderson-Hasselbalch to first find the ratio of acetate / acetic acid molarity. This is also the ratio of the number of moles of each.

The sum of the two molarities should equal 1.0 M.

You must find a way to rearrange one equation and plug it into the other. Post back if you need more help.

3. Feb 24, 2010

### ttor1236

Must I work out the [HA] and [A] and then work oout the molarity using n=m/MW???

4. Feb 24, 2010

### JJMB

Using Henderson-Hasselbalch, we find that [acetate] / [acetic acid] = 17.37

This ratio also applies to the number of moles of each, so moles acetate / moles acetic acid = 17.37

Rearranging this equation, we find that: moles acetate = moles acetic acid * 17.37

Since the concentration of the buffer is 1M, we know that: moles acetate + moles acetic acid / 0.5L = 1.0M

The next step is to take the rearranged equation and plug it into the equation directly above.

17.37 * moles acetic acid + moles acetic acid / 0.5L = 1.0M

Simplified: 18.37 * moles acetic acid / 0.5L = 1.0M

Solving for the moles of acetic acid gives: 0.0272 moles acetic acid that need to be used in the buffer.

Next solve for the number of moles of acetate: 0.0272 * 17.37 = 0.473 moles of acetate.

Take the number of moles of acetate and multiply it times the molecular weight to get the amount of grams you need to add to the buffer.

0.0272 moles of acetic acid would come out to 27.2 ml of 1M acetic acid.

You add the acetate and acetic acid to a clean 500 ml volumetric flask and dilute to volume with distilled h20.

I hope that helped.

5. Feb 25, 2010

### ttor1236

Thanks a mill.