Question

# Consider the titration of a 21.0 −mL sample of 0.100 M HC2H3O2 with 0.125 M NaOH....

Consider the titration of a 21.0 −mL sample of 0.100 M HC2H3O2 with 0.125 M NaOH. Determine each of the following.

A. the initial pH = 2.87

B. the volume of added base required to reach the equivalence point = 16.8 mL

C. the pH at 6.00 mL of added base = 4.49

D. the pH at one-half of the equivalence point = 4.74

F. the pH after adding 6.00 mL of base beyond the equivalence point.... I need help with this portion of the problem. I feel like messing up a simple step somewhere. Please help and show your work. Thanks!

Ka = 1.8*10^-5

pKa = 4.75

then

a)

initially

Ka = [H+][A-]/[HA]

[H+] = [A-] = x

[HA] = 0.1-x

then

1.8*10^-5 = x*x/(0.1-x)

x = 0.0013

p H= -loG(0.0013 = 2.8860

b)

for equivalence point

mol of acid = mol of base

mol = MV

MV1 = MV2

21*0.1 = 0.125*V

V = 21*.1/0.125 = 16.8

d)

pH 1/2 half equivalence point

pH = pKa + log(conjugate/acid)

conjugate = acid in half equivalence so

pH = pKa + log(1)

pH = pKa = 4.75

F)

after 6 ml beyond equivalence point, thjis will be pretty basic so

mmol of base = MV = 0.125*(33.6+6) = 4.95 mmol of base

mmol of acid = MV = 21*0.1 = 2.1 mmol

mmol of excess base = 4.95-2.1 = 2.85 mmol

then

total vol = V1+V2 = (21+33.6+6) =60.6 ml

then

[OH-] = mmol excess / total V = 2.85 /60.6 = 0.047029M

pOH = -loog(OH) = -log(0.047029 = 1.3276

pH = 14-pOH = 14-1.3276 = 12.6724

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