40.0 ml of an acetic acid of unknown concentration is titrated with 0.100 M NaOH. After 20.0 mL of the base solution has been added, the pH in the titration flask is 5.10. What was the concentration of the original acetic acid solution? [Ka(CH3COOH) = 1.8 ´ 10–5]
Use the Henderson-Hasselbalch equation.
5.10 = 4.74 + log(base/acid)
Calculate ratio base/acid. You know you have added 2 mmoles NaOH which will allow you to calculate acid at the pH = 5.10 point. Determine the amount mL required to neutralize that many mmoles acid, add to the 20 mL already used, then
mL x M = mL x M to arrive at the original M acid. Post your work if you get stuck. I get something like 0.072 but that is jut a quickie calculation. You need to be more precise than that. .
To find the concentration of the original acetic acid solution, we can use the concept of acid-base titration and the equilibrium constant expression for acetic acid.
First, let's understand the steps involved in this calculation:
1. Calculate the moles of NaOH added to the acetic acid solution.
2. Determine the initial moles of acetic acid present in the 40.0 ml solution.
3. Calculate the moles of acetic acid that reacted with NaOH.
4. Determine the remaining moles of acetic acid.
5. Calculate the concentration of the original acetic acid solution.
Now, let's go through each step:
1. Calculate the moles of NaOH added:
Moles of NaOH = volume of NaOH solution (in liters) × molarity of NaOH
Given that 20.0 mL of 0.100 M NaOH is added:
Moles of NaOH = 0.020 L × 0.100 mol/L = 0.002 mol NaOH
2. Determine the initial moles of acetic acid:
Since acetic acid is a monoprotic acid (donates one proton), the moles of acetic acid are equal to the moles of NaOH added.
Moles of acetic acid = 0.002 mol
3. Calculate the moles of acetic acid that reacted with NaOH:
Since acetic acid reacts with NaOH in a 1:1 ratio, the moles of acetic acid that reacted are also 0.002 mol.
4. Determine the remaining moles of acetic acid:
Moles of acetic acid remaining = initial moles of acetic acid - moles of acetic acid reacted
Moles of acetic acid remaining = 0.002 mol - 0.002 mol = 0 mol
5. Calculate the concentration of the original acetic acid solution:
Concentration of acetic acid = moles of acetic acid remaining (in mol) / volume of acetic acid solution (in liters)
Given that the volume of the acetic acid solution is 40.0 mL (which is 0.040 L):
Concentration of acetic acid = 0 mol / 0.040 L = 0 M
Therefore, the concentration of the original acetic acid solution is 0 M.