# Moles of a conjugate base in a buffer solution

1. Jul 8, 2015

### alr1014

1. The problem statement, all variables and given/known data
You need to produce a buffer solution that has a pH of 5.70. You already have a solution that contains 0.0200 moles of acetic acid. Using the Henderson-Hasselbalch equation calculate the moles of sodium acetate needed to create a buffer with the desired pH? The Ka of acetic acid is 1.8 × 10^-5.

2. Relevant equations
pKa = -log (Ka)
pH= pKa + log (base / acid)
3. The attempt at a solution
pKa= -log (1.8 × 10^-5) = 4.74
5.70 = 4.74 + log (base/acid)
Unfortunately, I have been staring at this problem for about an hour and am unsure how to find the concentration of the acid to discover the consent ration of the base. I am not sure how to proceed since the problem does not give a volume to find the molarity.
Any help with this would be greatly appreciated!!

2. Jul 9, 2015

### Staff: Mentor

Hint: you don't need the acid concentration.

$$C_A = \frac {n_A} {V}$$
$$C_B = \frac {n_B} {V}$$

$$\frac {C_A}{C_B} = ...$$

3. Jul 9, 2015

### alr1014

I appreciate your responce, any responce really. Although, your equation set seems to need a volume as well, and the problem I am to work out does not include one. That is a big area I am being messed up. As well as the problem states to use that particular equation to find the moles of base needed.

4. Jul 9, 2015

### Staff: Mentor

Plug concentrations into ratio and see what happens. V/V = ?

5. Jul 9, 2015

### alr1014

Since the problem called for using the henderson-hasselbalch equation, and that entails using the concentration of the acid, I used the volume used in the lab for the titration (25 mL). Not sure if that is right... but I got an answer. I will check with a few of my class mates before lab in the morning. Thank you for your help.

6. Jul 9, 2015

### Staff: Mentor

You don't need any specific volume.

V/V = 1 - volume cancels out. Ratio of concentrations (if these are concentrations in the same solution) is identical to the ratio of numbers of moles.

7. Jul 9, 2015

### alr1014

That very simple answer was probably the best answer you could have given me! Tha k you so very much for having the patience for dealing with slightly dense people like me!

8. Jul 9, 2015

### Staff: Mentor

I hoped you will see it by yourself. You will learn much more when you put an effort and find such things by yourself, than when they are presented on a silver plate.