Question

Consider the titration of a 26.0-mL sample of 0.180 M CH3NH2
with 0.145 M HBr. (The value of *K*b for CH3NH2 is
4.4×10−4.)

**Part A**

Determine the initial pH.

**Part B**

Determine the volume of added acid required to reach the equivalence point. answer to 3 sig figs

**Part C**

Determine the pH at 6.0 mL of added acid

**Part D**

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

**Part E**

Determine the pH at the equivalence point.

**Part F**

Determine the pH after adding 5.0 mL of acid beyond the equivalence point.

Answer #1

I...........0.180.............0.....0

C...........-x...............+x......+x

E........0.180-x...........x.........x

Kb = x^2 / (0.180-x)

4.4*10^-4 = x^2 / (0.180-x)

x = ~7.92*10^(-3)

[OH-] = ~7.92*10^(-3)

pOH = -log(~7.92*10^(-3))= 0.101

pH = 14 - pOH

pH = 13.8

b) The equivalence point is reached when moles of acid equals moles
of base.

M1V1 = M2V2

(0.180)(26) = (0.145)(V2)

32.276 mL = V2

c) When you add 4mL of acid, the volume changes.

Total volume = 26mL + 6mL

Initial mol CH3NH2

0.180 mol CH3NH2 / L * 0.0256L = 0.00468 mol CH3NH2

Added mol HBr

0.145 mol HBr / L * 0.004L = .00058 mol HBr

Determine the Ka value

Kw = Ka * Kb

Ka = (1.0*10^-14) / (4.4 *10^-4) = ~ 2.27*10^-11

pH = pKa + log ([base]/[acid])

pH = -log(2.27*10^-11) + log [(1.466*10^-4)/(2.0690*10^-5)]

pH = 10.64 + 0.8503

pH = ~11.49

d) Half of the equivalence point.

The mol ratio of base : acid will be equal.

Therefore it's 1:1. pH = pKa

pH = pKa + log ([base]/[acid])

pH = 10.64 + log (1/1)

pH = 10.64

Consider the titration of a 26.0-mL sample of 0.170 M CH3NH2
with 0.145 M HBr. (The value of Kb for CH3NH2 is 4.4×10−4.)
Initial pH- 11.94
Volume of added acid required to reach the equivalence point:
30.5 mL
Determine the pH after adding 5.0 mL of acid beyond the
equivalence point.

Consider the titration of a 27.0 −mL sample of 0.180 M CH3NH2
with 0.145 M HBr. Determine the pH after adding 6.0 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
point

Consider the titration of a 28.0?mL sample of 0.180M CH3NH2 with
0.150M HBr. 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 6.0mL of added acid
D.) the pH at one-half of the equivalence point
E.) the pH at the equivalence point
F.) the pH after adding 6.0mL of acid beyond the equivalence
point
Some help & direction would be great very confused! Thank...

Consider the titration of a 27.0 −mL sample of 0.175 MCH3NH2
with 0.150 M HBr. Express answers using two decimal places.
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 one-half of the equivalence point
E) the pH at the equivalence point
F) the pH after adding 4.0 mL of acid beyond the equivalence
point

Consider the titration of a 81.9 mL sample of 0.172 M CH3NH2 ,
methylamine, with 0.2 M HBr. The Kb of methylamine is 4.4x10−4.
Determine the volume (mL) of added acid required to reach the
equivalence point

Calculate the pH at the equivalence point for the titration of
0.180 M methylamine (CH3NH2) with 0.180 M HCl. The Kb of
methylamine is 5.0× 10–4.
pH=?

Consider the titration of a 27.8 −mL sample of 0.125 M RbOH with
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 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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