Question

Consider the titration of a 20.0 mL sample of 0.105 M HC₂H₃O₂ with 0.125 M NaOH.

Determine each quantity:

a. the initial pH

b. the volume of added base required ot reach equivalence point

c. the pH at 5.0 mL of added base

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

e. the pH at the equivalence point

f. the pH after adding 5.0 mL of base beyond the equivalence point

Please add explanations!

Answer 1

weakacid vs strongbase

a. initial pH

pH = 1/2(pka-logC)

    = 1/2(4.74-log0.105)

    = 2.86

b. at equivalence point , no of mole of aceticacid = no of mole of NaOH

no of mole of aceticacid = M*V = 20*0.105 = 2.1 mmole

no of mole of NaOH = 2.1 mmole

the volume of added base required ot reach equivalence point = n/M = 2.1/0.125

                                = 16.8 ml

c. the pH at 5.0 mL of added base , the solution is buffer

no of mole of aceticacid = M*V = 20*0.105 = 2.1 mmole

no of mole of NaOH = 5*0.125 = 0.625 mmole

pH = pka + log(salt(or)base/acid)

    = 4.74+log(0.625/2.1)

   = 4.21

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

   pH = pka

    pH = 4.74

e. the pH at the equivalence point , the solution turns salt. it is basic in nature

pH of salt solution = 7 + 1/2(pka+logC)

C = concentration of salt = n/v = 2.1/36.8 = 0.057 M

pH = 7+1/2(4.74+log0.057) = 8.75

f. the pH after adding 5.0 mL of base beyond the equivalence point

    concentration of excess base = ((21.8*0.125)-2.1)/41.8 = 0.015 M

pH = 14-(-log0.015) = 12.17