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Topic: Neutralizing  (Read 2196 times)

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Offline WhoCares357

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Neutralizing
« on: October 17, 2010, 04:23:44 PM »
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How many grams of dry NaOH must be added to 200 mL of 0.4 M H2SO4 to neutralize it completely? Molar mass of NaOH is 40 g/mol.

I'm going through a practice test (my professor doesn't post answers for it) and I wanted to see if I did this correctly:

I set up the equation NaOH + H2SO4 <-> HSO4- + H2O (l)
Then I set up an sRf table:

NaOH + H2SO4 <-> HSO4-
x            .08 mol      0
 -x         .08 - x       +x
 0           0              x

From this I figured the moles of NaOH = .08. Is this correct or did I totally miss the point of the question?
« Last Edit: October 17, 2010, 04:33:46 PM by WhoCares357 »

Offline JustinCh3m

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Re: Neutralizing
« Reply #1 on: October 18, 2010, 07:12:56 PM »
you need to recognize that there are 2 moles of H+ for every 1 mole of H2SO4 (diprotic acid).  You need to take this into consideration when deciding how much NaOH is needed for neutralization to occur...

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Offline Borek

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Re: Neutralizing
« Reply #2 on: October 19, 2010, 03:10:18 AM »
Is this correct or did I totally miss the point of the question?

This is simple stoichiometry, no need for ICE table - that is used when solving equilibrium problems. Start with properly balanced reaction equation.
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