Answer to Question #270324 in Organic Chemistry for Alisha

Question

A hypothetical acid HA has a dissociation constant 1.8 𝑥 10ିଷ.

a. Write a balanced dissociation reaction of the acid

b. Write the equilibrium constant expression for the acid

c. If the initial concentration of the acid is 0.5 M, determine its hydrogen ion

concentration.

d. Calculate the ionization constant

e. Comment on the strength of the acid.

Answer 1

a) HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq).

b)We assess the equilibrium:

H

(

O

=

)

C

O

H

(

l

)

+

H

2

O

(

l

)

⇌

H

(

O

=

)

C

O

−

+

H

3

O

+

And we write the equilibrium equation in the usual way:

K

a

=

Concentration of products

Concentration of reactants

K

a

=

[

H

(

O

=

)

C

O

−

]

[

H

3

O

+

]

[

H

(

O

=

)

C

O

H

(

l

)

]

.

This site reports that K

a

 for formic acid =

1.70

×

10

−

5

c)I'm assuming that you're working with a monoprotic weak acid here so that the ionization equilibrium can be written like this

HA

(

a

q

)

+

H

2

O

(

l

)

⇌

A

−

(

a

q

)

+

H

3

O

+

(

a

q

)

Now, you know that the solution has

pH

=

5

As you know, the pH

 of the solution is defined as

pH

=

−

log

(

[

H

3

O

+

]

)

−−−−−−−−−−−−−−−−−−−−

You can rearrange this equation to find the equilibrium concentration of hydronium cations.

log

(

[

H

3

O

+

]

)

=

−

pH

This is equivalent to

10

log

(

[

H

3

O

+

]

)

=

10

−

pH

which gets you

[

H

3

O

+

]

=

10

−

pH

In your case, you will have

[

H

3

O

+

]

=

10

−

5.0

=

1.0

⋅

10

−

5

 M

Now, notice that every mole of HA

 that ionizes produces 1

 mole of A

−

, the conjugate base of the acid, and 1

 mole of hydronium cations.

This means that, at equilibrium, the solution has

[

A

−

]

=

[

H

3

O

+

]

→

 produced in a 1

:

1

 mole ratio

In your case, you have

[

A

−

]

=

1.0

⋅

10

−

5

 M

The initial concentration of the acid will decrease because some of the molecules ionize to produce A

−

 and H

3

O

+

.

So, in order for the ionization to produce [

H

3

O

+

]

, the initial concentration of the acid must decrease by [

H

3

O

+

]

.

This means that, at equilibrium, the concentration of the weak acid will be equal to

[

HA

]

=

[

HA

]

initial

−

[

H

3

O

+

]

In your case, you will have

[

HA

]

=

0.01 M

−

1.0

⋅

10

−

5

.

M

[

HA

]

=

0.00999 M

By definition, the acid dissociation constant, K

a

, will be equal to

K

a

=

[

A

−

]

⋅

[

H

3

O

+

]

[

HA

]

Plug in your values to find

K

a

=

1.0

⋅

10

−

5

M

⋅

1.0

⋅

10

−

5

.

M

0.00999

M

K

a

=

1.001

⋅

10

−

8

 M

Rounded to one significant figure and expresses without added units, the answer will be

K

a

=

1

⋅

10

−

8

d)HCN] is 0.5M

[H3

​O+

] is 0.0068M

[CN−

] is 0.0068M

We can assume here that, [H3O+

]=[CN−

]

The equilibrium constant would be as follows:

Ka

​=[HCN]

[H3

​O+

][CN−

]

​=0.5

0.00682

​ =9.2×10−5

e) it is a week acid