# Titration Calculation question.

## Main Question or Discussion Point

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 $$NaOH$$?