Question

Determining pH at equivalence point in a titration of a weak base and strong acid.

Calculate the pH at the equivalence point for the titration of 0.150 M methylamine (CH3NH2) with 0.150 M HCl. The Kb of methylamine is 5.0× 10–4.
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Answer #1
Concepts and reason

The concept used to solve this question is to calculate the at the equivalence point of the titration.

Fundamentals

is the measure of hydrogen ion concentration. Lower the, higher is the hydrogen ion concentration and lower is the hydroxide ion concentration.

The formula to calculate the is as follows:

pH = -log[1,01]

An equilibrium constant for the dissociation reaction is the equal to the product of the concentrations of the respective ions divided by the concentration of the undissociated molecule. It represents the extent of dissociation.

Consider a general reaction of acid, HA+H,0—>H,0* +A
.

The equilibrium constant for the dissociation of is given as follows:

(vh]
[-v][.oʻH]
*****

Consider a general reaction of base, B+H,0
BH* +OH
.

The equilibrium constant for the dissociation of is given as follows:

(В)
L-HO ][ Ня)
У

The formula relating and is as follows:

K. = KK
1.0x10-14 = KK,

An equivalence point in a titration is the point at which the moles of acid equals to the moles of base added.

The reaction between methyl amine and hydrochloric acid is as follows:

CH,NH, +HCI— *CH,NH; +CH

At equilibrium, moles of equal to the moles of CH,NH,
.

At equivalence point, total volume is doubled and the concentration is halved.

So, initial concentration of [CH,NH;]=0.075 M

Construct an ICE table for the dissociation reaction of CH, NH
as follows:

CH,NH, ĐH,0
Initial: 0.075M
Change:
-X
Equilibrium: (0.075-x)
=CH,NH, TH,Oh
0
+ x + x

K_ICH,NH,] H100
[CH,NH;]
K. [CH,NH,] H,0]
|CH, NH]
1.0x10-1* (x)(x)
5.0x104 (0.075-x)
(*
)
2.0x10-11_
(0.075 - x)

Solve for x.

x=[H30*]
=1.22x10

The at the equivalence point is calculated as follows:

pH = -log(1.22x10 M)
= 5.91

Ans:

The at the equivalence point is .

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