We did a acid-base titration lab where we added NaOH to a vinegar mixture with phenolphthalein. Not even sure I got the reactant part right xD! thanks in advance!
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Balanced equation:
CH3COOH + NaOH = H2O + NaCH3COO
NaCH3COO is sodium acetate
CH3COOH + NaOH = H2O + NaCH3COO
NaCH3COO is sodium acetate
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Further to your subsequent query,
You have the number of moles of NaOH.
No. of moles of NaOH = No. of moles of Acetic Acid = x
Acetic Acid has a Mr of (12+3+12+16+16+1 = 60gmol-1)
Mass of acetic acid = x*60 = 60x
So percentage composition = 60x/5.12g x 100% = Ans
You have the number of moles of NaOH.
No. of moles of NaOH = No. of moles of Acetic Acid = x
Acetic Acid has a Mr of (12+3+12+16+16+1 = 60gmol-1)
Mass of acetic acid = x*60 = 60x
So percentage composition = 60x/5.12g x 100% = Ans
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OH- + HC2H3O2 --> H2O + C2H3O2-
Phenolphthalein is used as an indicator and its presence in the titration is said to be negligible when it comes to the reaction.
Phenolphthalein is used as an indicator and its presence in the titration is said to be negligible when it comes to the reaction.
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The net ionic equation is [H+] + [OH-] ---> H2O, [Na+] and [CH3COO-] are spectator ions.