# Moles of NaOH to Create Buffer Solution pH=6 in 0.42M Ethanoic Acid

• Chemistry
• stunner5000pt
In summary, the first and second questions involve calculating the number of moles of ethanoic acid and NaOH needed to create a buffer solution with a pH of 6. The Henderson-Hasselbach equation is used to determine the amount of NaOH needed, taking into account the concentration of the base and the acid. However, it is important to remember that a buffer solution consists of an acid and its conjugate base, in this case, acetic acid and ethanoate ions. Therefore, both substances must be present in the solution for the calculations to be accurate.

#### 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

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

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

Borek said:
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?

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