Question

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!

Answer #1

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

Consider the titration of a 21.0 −mL sample of 0.100 M HC2H3O2
with 0.120 M NaOH. Determine each of the following.
the volume of added base required to reach the equivalence
point
the pH at the equivalence point

Consider the titration of a 22.0-mL sample of 0.100 M HC2H3O2
with 0.125 M NaOH. (The value of Ka for HC2H3O2 is 1.8×10−5.)
Determine the pH at 4.0 mL of added base.Determine the pH at the
equivalence point.

Consider the titration of 24.0 mL of 0.100 M
HC2H3O2 with 0.120 M NaOH.
Determine the pH after adding 4.00 mL of base. I have found that
the volume of the base needed to reach the equivalence point is
20.0 mL.
Show your work and be as clear as possible. Thank you in
advance!

Consider the titration of a 23.4 −mL sample of 0.125 M RbOH with
0.100 M HCl. Determine each of the following.
A) the initial pH
B) the volume of added acid required to reach the equivalence
point
C) the pH at 5.3 mL of added acid
D) the pH at the equivalence point
E) the pH after adding 5.4 mL of acid beyond the equivalence
point

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0.110 M HCl. Determine each of the following. A) the initial pH
B)the volume of added acid required to reach the equivalence point
C) the pH at 6.0 mL of added acid D) the pH at the equivalence
point E) the pH after adding 4.0 mL of acid beyond the equivalence
point

Consider the titration of a 26.9 −mL sample of 0.115 M RbOH with
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of added acid? D. the pH at the equivalence point? E. the pH after
adding 4.1 mL of acid beyond the equivalence point?

Consider the titration of a 25.0 −mL sample of 0.180 M CH3NH2
with 0.155 M HBr. Determine each of the following.
a. the pH at one-half of the equivalence point
b. the pH at the equivalence point
c. the pH after adding 6.0 mL of acid beyond the equivalence
point
i already found that the initial pH is 11.95, the volume of acid
added to reach equivelance point is 29.0mL, and the pH og 6.0mL of
added acid is 11.23

Consider the titration of a 26.0 −mL sample of 0.175 M CH3NH2
with 0.150 M HBr. Determine each of the following.
1. the pH at 4.0 mL of added
acid
2.the pH at the equivalence point
3.the pH after adding 6.0 mL of acid beyond the equivalence
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What is the pH at the EQUIVALENCE POINT of the titration of
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Part 1
Calculate the pH after 35.0 mL of NaOH has been added.
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