This is a semi- solved problem from the 2014 AP Chemistry Free Response section: Q2.  The idea is to give the student (you) some clues to solve the problem without completely solving it for you.

(a) learly, the acid is whatever loses a proton (or H+ or H3O+ in disguise), i.e., CH3COOH and the conjugate base is what it becomes: CH3COO

(b) Hint: Set up an ICE equation with the equation on the left.

Remember that pH can give you the concentration of H3O+ to use in the ICE equation.

(i) Hint:
You have been given equal moles of propanoic acid and NaOH.
(Remember M*V = number of moles)
Does that mean pH =7 (neutrality)?

Not if this is a reaction between a weak acid and a strong base, in which case the salt will be basic.

(ii)  Hint: The concept is yet again a contrast between a strong acid and a weak acid.
Imagine a strong acid as generous, completely dissociating to give protons or H+.
A weak acid, on the hand, is miserly, letting go of less than 5% of its protons. 

It stands to reason that we will need a whole lot more of the propanoic acid than the HCl to attain the same pH.

Hint: This is a neutralization reaction. And neutralization reactions are always “single arrow” to be solved by stoichiometry.

The first part is your standard

Remember that in an earlier step you had found the Ka of the acid. Compare pKas and decide if you need a different indicator.

So, where shall I send the free guide to help you

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