The correct number of significant figures are,
Final volume of burette is 38.86ml
Volume of NaOH is 38.36ml
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) = ?
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
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
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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