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

  • Thread starter Thread starter stunner5000pt
  • Start date Start date
  • Tags Tags
    Dilution
AI Thread Summary
To create a buffer solution with a pH of 6 using 0.42 M ethanoic acid, there are 0.042 moles of ethanoic acid in 100 mL. The required concentration of the base, derived from the Henderson-Hasselbalch equation, is 3.32 M, which translates to 0.332 moles of NaOH needed. However, NaOH is not the base in the equation; the base is actually the ethanoate ions formed when NaOH reacts with ethanoic acid. The discussion emphasizes the importance of recognizing that a buffer consists of an acid and its conjugate base. Understanding the relationship between the components is crucial for accurate calculations in buffer preparation.
stunner5000pt
Messages
1,443
Reaction score
4
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
 
Physics news on Phys.org
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.
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
I don't get how to argue it. i can prove: evolution is the ability to adapt, whether it's progression or regression from some point of view, so if evolution is not constant then animal generations couldn`t stay alive for a big amount of time because when climate is changing this generations die. but they dont. so evolution is constant. but its not an argument, right? how to fing arguments when i only prove it.. analytically, i guess it called that (this is indirectly related to biology, im...
Back
Top