# Acid/Base & dilution

stunner5000pt
Homework Statement:
You need to make an acetate buffer with a bottle of 0.42 M of ethanoic acid and a bottle of 0.15 M of NaOH. The pKa of ethanoic acid is 4.76
a. how many moles of ethanoic acid are there in 100 mL of the 0.42 M solution
b. how many moles of NaOH must be added to the 100 mL 0.42 M ethanoic acid solution to create a buffer solution of ph = 6
c. what volume of NaOH should be added in mL to reach this pH
Relevant Equations:
Henderson Hasselbach equation
the first and second seem easy...

a. how many moles of ethanoic acid are there in 100 mL of the 0.42 M solution
n = CV = (0.42 M)(0.1 L) = 0.042 mol

b. how many moles of NaOH must be added to the 100 mL 0.42 M ethanoic acid solution to create a buffer solution of ph = 6
this is where the Henderson Hasselbach comes in

$$pH = pK_{a} +\log \frac{[base]}{[acid]}$$
$$6 = 4.76 + \log \frac{[base]}{0.42}$$
$$1.24 = \log \frac{[base]}{0.42}$$
$$10^{1.24} \times 0.42 = [base]$$
$$[base] = 3.32 M$$
n(moles of NaOH) = 3.32M x 0.1 = 0.332 mol

c. what volume of NaOH should be added in mL to reach this pH

i know that I cannot use C1V2 = C2V2 because the amount of solution that we add would dilute the final concentration
How does one keep the concentration the same as the target 3.32 in question (b)?

Your guidance & help is always appreciated!

Thank you

Mentor
Nope. NaOH is not the [base] in the HH equation.

What happens when you add NaOH to the solution of the acetic acid?

stunner5000pt
Nope. NaOH is not the [base] in the HH equation.

What happens when you add NaOH to the solution of the acetic acid?
ahh... so the [base] is composed of ethanoate ions & NaOH?
But how do we separate these two from the calculation?

Mentor
Nope again.

Can you have both acetic acid and NaOH in the solution?

Looks like you are missing an important part of the buffer definition - it is an acid and its _conjugate_ base.

tech99