Titration Calculation question.

[philistinesin]

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 $$NaOH$$?

 

[chem_tr]

This seems correct. Step 2 is very necessary, since the reaction needs two moles of hydroxide for one mole of diacid.

 

[ravenprp]

Your calculations are correct and your answer of 7.67 M is also correct. The reason for multiplying the moles of H_2SO_4 by 2 is because the balanced chemical equation shows that for every 1 mole of H_2SO_4, 2 moles of NaOH are needed for neutralization. This is why we multiply the moles of H_2SO_4 by 2 in order to find the moles of NaOH used in the reaction. Keep up the good work!