25.0 mL of 0.100 M acetic acid (Ka= 1.8 x 10^-5) is titrated with 0.100 M NaOH. Calculate the pH after the addtion of 27.00 mL of 0.100M NaOH.
my work
CH3COOH + H2O <-> H3O^+ + CH3COO^-
25mL x 0.100 mmol/ml = 2.5mmol CH3COOH
27mL x 0.100 mmol/ml = 2.7mmol NaOH
There are more base than acid.... so what do I do?
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• Chem - DrBob222, Tuesday, June 29, 2010 at 4:30pm
• So the base neutralizes all of the acetic acid and one is left with 2.5 mmoles of the salt (sodium acetate) + an excess of (2.7-2.5 = 0.2 mmoles NaOH. The volume is 25 + 27 mL = ??
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Hi, I am still confuse on how to work this problem out
2.7-2.5=0.2
So I have 0.2 NaOH
Can I convert it to CH3COOH?
CH3COOH + OH^- --> H2O + CH3COO^-
0.2 OH^- = 0.2 CH3COO^-
Then, 2.7 mmol would be how much CH3COO produced?
Total Volume: 52 mL
[CH3COO]= 2.7/52= ....M
[CH3COOH] = 0.2/52=....M
I would like to know if I am going in the right direction in this problem, or if I'm doing it completely wrong. Thanks.
You are on the right track! Let's continue with the problem.
Since you have 0.2 mmol of NaOH left after neutralizing all the acetic acid, you can assume that this 0.2 mmol will react with the water to produce hydroxide ions (OH-) and acetate ions (CH3COO-), according to the equation:
CH3COOH + OH- → H2O + CH3COO-
So, all the remaining NaOH will convert to CH3COO- ions.
Now, you need to calculate the concentrations of CH3COOH and CH3COO- after adding the NaOH. The total volume after adding the NaOH is 25 mL (the initial volume of acetic acid) + 27 mL (the volume of NaOH added) = 52 mL.
To calculate the concentration of CH3COOH, divide the remaining 0.2 mmol by the total volume in liters:
[CH3COOH] = 0.2 mmol / 52 mL * (1 L / 1000 mL)
To calculate the concentration of CH3COO-, divide the initial 2.7 mmol by the total volume in liters:
[CH3COO-] = 2.7 mmol / 52 mL * (1 L / 1000 mL)
Now that you have the concentrations of CH3COOH and CH3COO-, you can calculate the pH using the Henderson-Hasselbalch equation:
pH = pKa + log ([CH3COO-] / [CH3COOH])
The pKa of acetic acid is given as 1.8 x 10^-5.
Substitute the values into the equation and calculate the pH.
I hope this helps! Let me know if you have any further questions.