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30mL 3M acetic acid (pka 4.7) is added to 50mL 2M NaOH. what is the resulting pH

a) <2

b) between 2 and 7

c) 7

d) >7

ans =d

can someone explain how to actually solve for pH? the solution describes doing this as a limiting reagent problem in which you solve for the moles of acetic acid and sodium hydroxide.

I'm just a little lost because doesn't acetic acid form its own equilibrium? So doesn't acetic acid form an equilibrium, which means that the actual acid in solution is not that which you calculate from simply multiplying the initial concentration by the initial volume?

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closed as unclear what you're asking by Mithoron, A.K., Todd Minehardt, ashu, Jon Custer Apr 4 at 19:43

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    $\begingroup$ HINT - How many millimoles of acid and base are there? $\endgroup$ – MaxW Apr 2 at 20:43
  • $\begingroup$ It does but it only raises pH further - you'd need to treat the solution as buffer if there would be less NaOH added then the amount of acid . $\endgroup$ – Mithoron Apr 2 at 21:18
  • $\begingroup$ It's not really it's own equilibrium in the sense that all of the acid-base equilibria are coupled via the exchange of hydronium ions. $\endgroup$ – Zhe Apr 3 at 15:15
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As MaxW intimated, you are adding more millimoles of a strong base to fewer millimoles of a weak acid. Intuition, if nothing else, tells one that you will have an excess of strong base remaining and therefore the final solution will be basic; that is >7.

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