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I was looking over question 7 in the second to last practice final and I don't understand how it was known to use CH3COO- + H20 = CH3COOH + OH-? Knowing what chemical equation to use is something I still don't understand, if anyone has some pointers on knowing what equation to use I'd really appreciate it!
So, the question is asking you to identify which indicator you should use in a titration (which we didn't go over in lecture), but basically you're looking for the pH at the stoichiometric point, when moles of acid = moles of base. Since we are told that we're titrating CH3COOH with NaOH (weak acid and strong base), we know that the pH will be slightly basic at the stoichiometric point. This is because only H2O, Na+, and CH3COO- are present at the stoichiometric point (since all the H3O+ and OH- has neutralized/cancelled itself out). We know that Na+ has no effect on pH, but CH3COO- does, since it is the conjugate base of a weak acid (conjugate seesaw). Therefore, CH3COO- reacts with H2O as shown in the reaction you typed, producing some OH-, making the solution somewhat basic at the stoichiometric point. Hope this helps!
Thank you so much! This helped clarify the problem
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