How many mL of .200M acetic acid are mixed with 13.2mL of.200M sodium acetqte to give a buffer with pH=4.2
pKa = 4.76 are thereabout. Look it up if you know it.
4.2 = 4.76 + log (base/acid)
-0.56 = log (B/A)
-0.56 = log (0.00264/A)
0.275 = 0.00264/A
So the acid (acetic acid) must be
0.00264/0.275 = 0.00960 moles (technically molar but the volume always cancels and you can ingore that for the moment.)
M x L = moles to find the L necessary.
Check my arithmetic. My calculator rounded here and there so you need to go through it again to make sure.
how did you calculate pKa because this is a practise question for my test and we won't be given the value. so how would you calculate it?
Use the Henderson-Hasselbalch equation.
I'm making a mistake in using that
equation:
n (CH3COO-)= .0132*.2 = .oo264
..does that mean that CH3COOH also has the same n?
ph= pKa +log(Conjugate Base/ Weak Acid)
4.2= pKa + log (.00264/ .00264)
since i have n the same it doesn't work. How do you calculate n?
Don't you have Ka for acetic acid = 1.75 x 10^-5 or something like that.
Then pKa = -log Ka = -log 1.75 x 10^-5 = -(-4.75696) = 4.75696 which I rounded to 4.76. And I get something like 48 mL of the acetic acid required but check my math.
Neither Ka nor Pka was given
On tests I have seen, they usually give Ka or pKa in the problem OR there will be an appended set of tables that will contain constants necessary to take the exam. I don't know of any test in which you are expected to know those constants.
okie thank you =)
To answer this question, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the concentration of its conjugate acid and conjugate base:
pH = pKa + log([A-]/[HA])
Where:
- pH is the desired pH of the buffer solution (4.2 in this case).
- pKa is the dissociation constant of acetic acid, which is approximately 4.75.
- [A-] is the concentration of the conjugate base (sodium acetate).
- [HA] is the concentration of the acid (acetic acid).
Let's solve this step by step:
Step 1: Find the concentration of sodium acetate ([A-]):
We know the volume of sodium acetate solution is 13.2 mL, and the concentration is 0.200 M. We can calculate the number of moles of sodium acetate:
moles of sodium acetate = concentration * volume
moles of NaC2H3O2 = 0.200 M * 13.2 mL / 1000 mL/L
Step 2: Find the concentration of acetic acid ([HA]):
To achieve pH 4.2, we can rearrange the Henderson-Hasselbalch equation to solve for [HA]:
[HA] = [A-] * 10^(pH - pKa)
Substituting the known values:
[HA] = [0.200 M * 13.2 mL / 1000 mL/L] * 10^(4.2 - 4.75)
Step 3: Calculate the required volume of acetic acid:
We know the concentration of acetic acid is 0.200 M. We can calculate the volume required using the following equation:
volume of acetic acid = (moles of acetic acid) / (concentration of acetic acid)
volume of acetic acid = [HA] / (0.200 M)
Now we have all the necessary information to calculate the volume of acetic acid. Plug in the values to get the final answer.