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ALab Data
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Standardized NaOH (M) 0.4171
Initial volume of buret (mL) 0.50
Volume of vinegar (mL) 10.00
Observations
solution increased.
Final volume of buret (mL)
Volume of NaOH (mL)
Molarity of acetic acid (M)

Respuesta :

The correct number of significant figures are,

Final volume of burette is 38.86ml

Volume of NaOH is 38.36ml

Given data:

Initial volume of burette V(I)= 0.50ml

Molarity of NaOH is M1 = 0.4171M

Volume of Vinegar V2 = 10.00ml

Molarity of acetic acid M2= 1.6M

Final volume of burette V(f)= ?

volume of NaOH V1 = V(f) - V(I) = ?

Determining the moles of acetic acid (CH3COOH)

Molarity = moles of solute / volume of solution

Molarity of acetic acid = moles of CH3C00H / Volume of vinegar

1.6 = moles of CH3C00H / 10.00

moles of CH3C00H = 1.6 * 10.00

Moles of CH3COOH is 16.00mols

Determining the moles of Sodium Hydroxide (NaOH)

In a volumetric titration the amount of Standard solution (titrant) used in the titration is stoichiometrically equivalent to amount of unknown solution (analyte) used for the titration at the endpoint.

Hence,

Moles of NaOH = Moles of CH3COOH

Moles of NaOH is 16.00mols

Determining the volume of Sodium Hydroxide (NaOH)

Molarity of NaOH = moles of NaOH / Volume of NaOH

0.4171 = 16.00 / V(f)-V(I)

0.4171 = 16.00 / V(f)- 0.50

V(f)  = ( 16 / 0.4171 ) + 0.50

V(f) = 38.86 ml

Final volume of NaOH V(f) is 38.86ml

Volume of NaOH V1 = 38.86 - 0.50

Volume of NaOH is 38.36ml

DISCLAIMER: Question is incomplete

Initial volume of burette = 0.50ml

Molarity of NaOH  = 0.4171M

Volume of Vinegar = 10.00ml

Molarity of acetic acid = 1.6M

Final volume of burette = ?

volume of NaOH  = ?

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