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13 Dec 2019

Consider the titration of a 25.0 −mL sample of 0.110 M HC2H3O2 with 0.130 M NaOH.

Ka(H2C2H3O2) = 1.8x10-5

A) Initial PH

B) the volume of added base required to reach the equivalence point

C) the pH at 4.00 mL of added base

D) the pH at one-half of the equivalence point

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Elin Hessel
Elin HesselLv2
17 Dec 2019

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Related questions

A titration of 25.0 mL of .11 M HC2H3O2 is performed with .22 M NaOH. Ka=1.8x10^-5.

Determine the initial pH.

Determine the amount of base required to reach the equivalence point of the titration.

Determine the pH after 10 mL of added base.

Consider the titration of a 25.0?mL sample of 0.110M HC2H3O2 with 0.125M NaOH. Determine each quantity:

a) the initial pH

b) the volume of added base required to reach the equivalence point

c) the pH at 6.00mL of added base

d) the pH at one-half of the equivalence point

e) the pH at the equivalence point

Q12. Considder the titration of a 25.0 mL sample of 0.100 M HC2H3O2 with 0.130 M NaOH

part A- the initial pH ( express answer in two decimal places)

part B- the volume of added base required to reach the equivalence piont

part C- the pH at 6.00mL of added base. (two decimal places)

part D- the pH at one-half of the equivalence piont (two decimal places)

part E-the pH at the equilence point (two decimal piont)

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