Titration Calculation question.

  • #1

Main Question or Discussion Point

Here's the question:
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95.0 mL of 2.1 M [tex]H_2SO_4 _(aq)[/tex] was needed to neutralize 52.0 mL of NaOH [tex]_(aq)[/tex] . What was the Molarity concentration of the sodium hydroxide solution?

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Here's what I did:

Reaction: [tex]H_2SO_4 + 2 NaOH -> Na_2SO_4 + 2H_2O[/tex]


STEP 1: mol [tex]H_2SO_4[/tex] = (M [tex]H_2SO_4[/tex]) ( V [tex]H_2SO_4[/tex] ) = ( 2.1 M ) ( 0.095 L ) = 0.1995 mol

STEP 2: mol [tex]NaOH[/tex] = 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 [tex]H_2SO_4[/tex] by 2 to get the moles of [tex]NaOH[/tex]?
 

Answers and Replies

  • #2
chem_tr
Science Advisor
Gold Member
609
2
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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