Question

Your group conducts a series of titrations of 25.00 mL of sodium hydroxide with a standardized solution of 0.1357 M HCl solution.

	Run1	Run2	Run3
Final volume HCl	22.58 mL	23.21 mL	22.32 mL
Iniial volume HCl	3.50 mL	3.25 mL	3.00 mL

From the data you collected above, what volume of NaOH would be used to titrate and completely neutralize 10.00 mL of 0.1350 M HBr?

Answer 1

For Run1,

Volume of HCl used = 22.58 - 3.50 = 19.08 mL

Molarity of NaOH = ( 19.08 x 0.1357 ) / ( 25 ) = 0.1036 M

For Run2,

Volume of HCl used = 23.21 - 3.25 = 19.96 mL

Molarity of NaOH = ( 19.96 x 0.1357 ) / ( 25 ) = 0.1083 M

For Run3,

Volume of HCl used = 22.32 - 3.00 = 19.32 mL

Molarity of NaOH = ( 19.32 x 0.1357 ) / ( 25 ) = 0.1049 M

Average Molarity of NaOH = ( 0.1036 + 0.1083 + 0.1049 ) / 3

Average Molarity of NaOH = 0.1056 M

Milimoles of HBr to be neutralized = 10 x 0.135 = 1.35

M₁V₁ = M₂V₂

0.1056 x V = 1.35

V = 12.784 mL