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I got the above problem correct by setting up the net ionic equation and finding the limiting reactant. I found that there were 0.01076 moles of NaOH left, and then calculated it's molarity which led to the PH. I'm confused by when you have a strong base and weak acid, when calculating the pH when all of the acid is used up, why can you disregard the acid's conjugate base in the calculation of the pH. Doesn't the reaction still create a certain amount of the acids conjugate base?

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    $\begingroup$ You have 100% butanoate at this point in the titration, e.g. a mixture of weak base and strong base. The strong base lowers the proton concentration by a lot so that there are insufficient protons available to react with the weak base. Or, said differently, the pH is so much higher than the pKa that you can ignore the acid/base behavior of the weak base. $\endgroup$ – Karsten Theis May 3 at 1:34
  • $\begingroup$ To state @KarstenTheis's comment differently, yes there will be some butanoic acid in the solution. However because of the very high pH the concentration of butanoic acid is vanishingly small compared to the butanoate anion and hence insignificant in the problem. $\endgroup$ – MaxW May 3 at 5:27

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