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Weak Acid Strong Base pH
Timeless Thinker:
25 mL of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH after 5 mL of NaOH have been added? Ka for acetic acid = 1.8 × 10–5.
So I created an ICE table in moles of each substance:
CH3COOH + OH- :rarrow: CH3COO-
I: .0025 .0005
C: -.0005 -.0005 +.0005
E: .0020 0 .0005
Then I converted these back to M using n/V
E: .0667 0 .0167
After this, I used Henderson Hasselbalch equation of pH= pKa+ log[A-]/[HA]
4.745+ -.602= 4.142
pH= 4.142
Should I be getting a negative number for my log[A-]/[HA]?
Borek:
--- Quote from: Timeless Thinker on April 05, 2020, 01:25:05 AM ---25 mL of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH after 5 mL of NaOH have been added? Ka for acetic acid = 1.8 × 10–5.
So I created an ICE table in moles of each substance:
CH3COOH + OH- :rarrow: CH3COO-
I: .0025 .0005
C: -.0005 -.0005 +.0005
E: .0020 0 .0005
--- End quote ---
That's not entirely correct. The problem here is how one defines Initial, Change and Equilibrium. You got a correct stoichiometry - and that's a right first step - but not E (equilibrium) concentrations. Equilibrium after the neutralization can slightly shift (and you could use ICE table to find that shift). Usually the change is negligible, compare https://www.chembuddy.com/?left=buffers&right=with-ICE-table
--- Quote ---Then I converted these back to M using n/V
--- End quote ---
Correct, but not necessary - you divided nominator and denominator by the same number, it cancels out.
--- Quote ---After this, I used Henderson Hasselbalch equation of pH= pKa+ log[A-]/[HA]
4.745+ -.602= 4.142
pH= 4.142
--- End quote ---
Looks OK.
--- Quote ---Should I be getting a negative number for my log[A-]/[HA]?
--- End quote ---
Why not? Sometimes it will be negative, sometimes positive.
Timeless Thinker:
But it's being titrated with a strong base, why would the pH go down?
AWK:
Goes down?
pH of 0.1 M acetic acid is close to 2.9.
Timeless Thinker:
But when I take the log of [A-]/[HA], is -.602, is that correct?
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