Two hydroxides doesn't matter in this problem

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The discussion centers on a chemistry problem involving the neutralization of barium hydroxide (Ba(OH)2) with hydrochloric acid (HCl). The initial setup for calculating the required volume of HCl was based on the stoichiometry of the reaction, which indicates that two moles of HCl are needed to neutralize one mole of Ba(OH)2. The calculation performed was correct, leading to a volume of 40.0 mL of 0.150-molar HCl. However, the answer provided by the teacher was 20.0 mL, which was later confirmed to be a mistake. The key takeaway is that the presence of two hydroxide ions in Ba(OH)2 is significant for determining the stoichiometric ratio, and the correct approach validates the initial calculation.
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I'm having trouble understanding why the fact that there are two hydroxides doesn't matter in this problem.

What volume of 0.150-molar HCl is required to neutralize 25.0 milliliters of 0.120-molar Ba(OH)2?

I thought the setup would look like this simple enough:

(25.0 mL)(.120 M)(2) = (.150 M)(V)

and the volume came out to be 40.0 mL. I assumed that the neutralization reaction looked like this:

2HCl + Ba(OH)2 --> 2H2O + BaCl2

so for each mole of Ba(OH)2 there would be two moles of HCl. However, I was told the answer was 20.0 mL. Why is the presence of two hydroxides negligible, or is the answer I was given incorrect?
 
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First of all, work out the reaction, what is the stoichiometric ratio of this reaction.

Secondly, work out the number of moles in both substances
then it should be basic plugging and chugging.
 
kazimmerman said:
I thought the setup would look like this simple enough:
(25.0 mL)(.120 M)(2) = (.150 M)(V)
I see nothing wrong with this. For each mole of Ba(OH)2 you need
2 moles of HCl to neutralize it. V=40 mL.
 
Thanks for the replies. I had it confirmed with my teacher today and he simply made a mistake when grading the test. ;)
 
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