Asked by Kelly
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]
Answers
Answered by
DrBob222
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. .
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. .
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