Titration Calculation question.

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To determine the molarity of the sodium hydroxide (NaOH) solution, the neutralization reaction between sulfuric acid (H2SO4) and NaOH is considered, where one mole of H2SO4 reacts with two moles of NaOH. The calculation begins by determining the moles of H2SO4 used, which is calculated as 0.1995 moles from the given volume and concentration. Since the reaction requires two moles of NaOH for every mole of H2SO4, the total moles of NaOH are calculated to be 0.399 moles. Finally, the molarity of NaOH is found by dividing the moles of NaOH by the volume of the NaOH solution, resulting in a concentration of 7.67 M. The necessity of multiplying the moles of H2SO4 by two is confirmed as essential due to the stoichiometry of the reaction.
philistinesin
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Here's the question:
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95.0 mL of 2.1 M H_2SO_4 _(aq) was needed to neutralize 52.0 mL of NaOH _(aq) . What was the Molarity concentration of the sodium hydroxide solution?

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Here's what I did:

Reaction: H_2SO_4 + 2 NaOH -> Na_2SO_4 + 2H_2O


STEP 1: mol H_2SO_4 = (M H_2SO_4) ( V H_2SO_4 ) = ( 2.1 M ) ( 0.095 L ) = 0.1995 mol

STEP 2: mol NaOH = 2 * 0.1995 mol = 0.399 mol

STEP 3: Molarity = mol / Liters = ( 0.399 mol ) / ( 0.052 L ) = 7.67 M

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Is that answer right? Should I have done STEP 2 ( I guess that's what's confusing me the most right now), was it necessary to multiply the moles of H_2SO_4 by 2 to get the moles of NaOH?
 
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This seems correct. Step 2 is very necessary, since the reaction needs two moles of hydroxide for one mole of diacid.
 
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