# Titration + dilution calculation~HELP

A 25 ml aliquot of vinegar was diluted to 250mL. Titration of 50 mL aliquots of the diluted solution required an average of 34.88 ml of 0.096M NaOH.

(a) Express the acidity of the vinegar as percent (w/v) acetic acid.

Thanks~~~

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Consider the titration between 50 mL diluted vinegar and NaOH.

CH3COOH + NaOH → CH3COONa + H2O

Mole ratio CH3COOH : NaOH = 1 : 1

No. of moles of NaOH = MV = 0.096 x (34.88/1000) = 0.003348 mol

No. of moles of CH3COOH in 50 mL of diluted vinegar = 0.003348 mol

No. of moles of CH3COOH in 250 mL of diluted vinegar

= 0.003348 x (250/50)

= 0.01674 mol

No. of moles of CH3COOH in 25 mL of original vinegar = 0.01674 mol

Molar mass of CH3COOH = 12x2 + 1x4 + 16x2 = 60 g mol-1

Mass of CH3COOH in 25 mL of original vinegar

= mol x (molar mass)

= 0.01674 x 60

= 1.004 g

Acidity of the original vinegar

= (1.004/25) x 100%

= 4.016%

2008-06-15 11:09:44 補充：

The 「acidity of the original vinegar」should be the「concentration of the original vinegar」instead.

• H+ + NaOH --&gt; Na+ + H2O (1)

NO. OF MOLE OF NaOH: 34.88 /1000 X 0.096=0.00334848

BY (1) No.of mole of vinegar = no .of mole of NaOH =0.00334848

No. of mole = Volume X acidity

Acidity : 0.00334848/ (50/1000) =0.0669696

THE ORIGIN (NOT YEY DILUTE) VINEGAE ACIDIRY IS&quot;

0.0669696/ (25/250) = 0.669696

=0.67 M