puffylover

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a student dissolves 20 mg of a solid NaOH in 30mL of distilled water. 50mL of an unknown concentration of acetic acid (Ka=1.5E-5) was required to completely neutralize the base present. Which of the following answer choices correctly solves for the concentration of acetic acid added?

my question is: they solved this problem without taking into consideration the Ka. Why is this?

Their answer: (20/40)/(50)
 
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RogueUnicorn

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there's a q&a forum that i know you know how to use
 

puffylover

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i don't know if your response was meant to come off as rude, but it kinda was.

i made a mistake.... sorry..
 

RogueUnicorn

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i'll make sure to end all my posts with cute smileys just for you then. :)
 

RogueUnicorn

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to actually answer your question, the key is to consider the solution to which you add the weak acid. what is the reaction, and how complete is it?


edit: :)
 

puffylover

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hmm.

so the solution is NaOH -> OH- + Na+

and your titrating with CH3COOH -> CH3COO- + H+

the reaction you're adding the weak acid to is complete cuz it's a strong acid, no?

are you saying that the weak acid fully dissociates in the beginning of the titration?

i just don't understand why .01 moles of CH3COOH will necessarily equal .01 moles of H+. Doesn't ICE tell you that CH3COOH=.01 moles - x and H+=x?
 
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loveoforganic

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You're reacting a strong base with a weak acid. The strong base will dissociate 100%. The weak acid will not. However, the few protons that do dissociate from the weak acid will be immediately consumed by the strong base, forming water. You now have no protons in solution, so the weak acid dissociates some more, releasing more protons. Once again, protons immediately consumed. Cycle repeats until you're left with just salt and water.
 

Jlaw

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a student dissolves 20 mg of a solid NaOH in 30mL of distilled water. 50mL of an unknown concentration of acetic acid (Ka=1.5E-5) was required to completely neutralize the base present. Which of the following answer choices correctly solves for the concentration of acetic acid added?

my question is: they solved this problem without taking into consideration the Ka. Why is this?

Their answer: (20/40)/(50)
can't you just calculate molarity of NaOH and then use (Molarity base)(Volume base)=(Molarity acid)(Volume Acid) to find the molarity at equivelance?
 

BerkReviewTeach

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can't you just calculate molarity of NaOH and then use (Molarity base)(Volume base)=(Molarity acid)(Volume Acid) to find the molarity at equivelance?
Yes, but why waste time with an unnecessary extra step?

The relationship you are basing that on is moles acid = moles base added at equivalence.

It's easier to get the moles of base from 20 mg/40 g/mole than it is to calculate the molarity of NaOH (which requires that same calculation of moles, but now divided by volume) just to mulitply it by 30 mL to get moles again.

What they did was:
Moles base = 20 mg/40 g/mole = 0.5 mmole NaOH
Moles acid = Macid x 50 mL

Setting them equal gives:
0.5 mmole NaOH = Macid x 50 mL
so Macid = 0.5 mmole/50 mL = 0.5 mole/50 L = 0.01 M

best explanation ever! thank u!
I concur! LoveofOrg has some of the best explanations around this place: insightful, concise, conceptual, and always polite.