What is the final pH of the second dilution after adding 1 drop of 0.1 M NaOH?
There were 2 ml of buffer per well Calculate the final pH of the undiluted buffer after 0.25 ml of 0.1M NaOH is added. The Ka of CH3 COOH = 1.8 x 10 power -5. Use the CH3COOH bufffer concentration equation, an ICE table and a balanced dissociation equilibrium equation to help determine the pH.
Write the equation for the dissociation equilibrium of acetic acid.
Determine the volumes of 0.10M CH3COOH and 0.10M CH3COONa required to prepare 10 ml of the following pH buffers. (Note pka of CH3COOH = 4.7).
Please reply me thanks!
Your post is not clear enough for me to understand the question. The only part I could understand was the dissociation of acetic acid.
The dissociation of acetic acid is as follows:
CH3COOH ==> H^+ + CH3COO^-
In Activity2, there were 2 ml of bufffer per well. CALCULATE the final pH of the undiluted buffer after 0.25 ml of 0.1M NaOH is added. The Ka of CH3COOH = 1.8 x 10 power -5. Use the CH3COOH buffer concentration calculated in Data Tabe 2, a balanced CH3COOH + NaOH reaction equation, an ICe table and a balanced dissociation equilibrium equation to help determine the pH.
What is the final pH of the second dilution after adding 1 drop 0.1 M NaOH.
To determine the final pH of the second dilution after adding 1 drop of 0.1 M NaOH, we need more information. Specifically, we need to know the initial pH and concentration of the buffer solution before the NaOH is added.
Moving on to the next part, we can determine the final pH of the undiluted buffer after adding 0.25 ml of 0.1 M NaOH.
To calculate the final pH, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
Where:
pH = the final pH of the buffer solution
pKa = the dissociation constant of the acid (in this case, CH3COOH)
[A-] = the concentration of the conjugate base (CH3COO-) after the addition of NaOH
[HA] = the concentration of the acid (CH3COOH) after the addition of NaOH
To solve this problem, we'll need to set up an ICE table for the reaction:
CH3COOH + OH- --> CH3COO- + H2O
Initial:
CH3COOH: [CH3COOH]
OH-: [NaOH]
Change:
CH3COOH: -[NaOH]
OH-: -[NaOH]
CH3COO-: +[NaOH]
H2O: +[NaOH]
Equilibrium:
CH3COOH: [CH3COOH] - [NaOH]
OH-: [NaOH] - [NaOH]
CH3COO-: [NaOH]
H2O: [NaOH]
Now, we can substitute the equilibrium concentrations into the Henderson-Hasselbalch equation:
pH = pKa + log ([CH3COO-] / [CH3COOH - NaOH])
Since the volume of NaOH added is negligible to the total volume (10 mL), we can assume that the number of moles of NaOH added is equal to the number of moles of CH3COOH neutralized.
We can use this information to calculate the volume of CH3COOH and CH3COONa needed to prepare the 10 mL pH buffer solutions.
To calculate the volumes, we can use the equation:
pH = pKa + log ([CH3COO-] / [CH3COOH])
We can rearrange the equation to solve for the ratio of [CH3COO-] / [CH3COOH]:
[CH3COO-] / [CH3COOH] = 10^(pH - pKa)
Then, we can multiply the ratio by the total volume to determine the volume of CH3COOH and CH3COONa:
Volume of CH3COOH = [CH3COOH] x Total Volume
Volume of CH3COONa = [CH3COO-] x Total Volume
Let's plug in the values you provided, assuming we have a pH buffer of 5:
pH = 5
pKa = 4.7
Total Volume = 10 mL
[CH3COO-] / [CH3COOH] = 10^(5 - 4.7) = 10^0.3 = 2
Volume of CH3COOH = 2 mL
Volume of CH3COONa = 2 mL
Please note that you may need to adjust the volumes slightly to ensure the final volume adds up to 10 mL.
I hope this helps! Let me know if you have any other questions.
To answer your questions, let's break down the steps one by one:
1. What is the final pH of the second dilution after adding 1 drop of 0.1 M NaOH?
To calculate the final pH after adding NaOH, you need to consider the buffering capacity of the solution. However, the information provided does not include the initial pH or the concentration of the buffer. Therefore, it is not possible to determine the final pH without this information.
2. Write the equation for the dissociation equilibrium of acetic acid.
The dissociation equation for acetic acid (CH3COOH) can be written as follows:
CH3COOH ⇌ CH3COO- + H+
3. Determine the volumes of 0.10 M CH3COOH and 0.10 M CH3COONa required to prepare 10 ml of the following pH buffers. (Note: pKa of CH3COOH = 4.7).
To determine the volumes of CH3COOH and CH3COONa required to prepare a buffer of a specific pH, you need to consider the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
where [A-] is the concentration of the conjugate base (CH3COO-) and [HA] is the concentration of the acid (CH3COOH).
Let's assume you want to prepare a buffer at a pH of 5. We can set up the equation as follows:
5 = 4.7 + log([A-]/[HA])
Since the ratio [A-]/[HA] is equal to the ratio of their volumes (Va/Vb), we can rearrange the equation to solve for the volume of one component in terms of the other:
log(Va/Vb) = 5 - 4.7
Va/Vb = 10^(5 - 4.7)
Once you have the ratio of Va/Vb, you can choose any arbitrary values for Va and calculate Vb accordingly. For example, if you choose Va = 5 ml, then using the calculated ratio of Va/Vb, you can determine Vb:
Vb = Va / (Va/Vb) = 5 / (10^(5 - 4.7))
Repeat this calculation for different desired pH values to obtain the required volumes of CH3COOH and CH3COONa.
Please note that this explanation assumes ideal behavior of the acid-base system and neglects the activity coefficients, so the calculated volumes may not be exact. It is always best to perform actual laboratory calculations or consult literature tables for precise buffer preparation.
I hope this helps! Let me know if you have any further questions.