Question

How many liters of water do we need to add to 5.00 moles to get a 0.648 M solution?

Answer 1

Answer: This is a typical acid/base equilibrium problem, that involves the use of logarithms.

Explanation:We assume that both nitric acid and hydrochloric acid dissociate to give stoichiometric

H

3

O

+

.

Moles of nitric acid:

26.0

×

10

−

3

⋅

L

×

8.00

⋅

m

o

l

⋅

L

−

1

=

0.208

⋅

m

o

l

H

N

O

3

(

a

q

)

.

And, moles of hydrochloric acid:

88.0

×

10

−

3

⋅

L

×

5.00

⋅

m

o

l

⋅

L

−

1

=

0.440

⋅

m

o

l

H

C

l

(

a

q

)

.

This molar quantity is diluted to

1.00

L

. Concentration in moles/Litre =

(

0.208

+

0.440

)

⋅

m

o

l

1

L

=

0.648

⋅

m

o

l

⋅

L

−

1

.

Now we know that water undergoes autoprotolysis:

H

2

O

(

l

)

⇌

H

+

+

O

H

−

. This is another equilibrium reaction, and the ion product

[

H

+

]

[

O

H

−

]

=

K

w

. This constant,

K

w

=

10

−

14

at

298

K

.

So

[

H

+

]

=

0.648

⋅

m

o

l

⋅

L

−

1

;

[

O

H

−

]

=

K

w

[

H

+

]

=

10

−

14

0.648

=

?

?

p

H

=

−

log

10

[

H

+

]

=

−

log

10

(

0.648

)

=

?

?

Alternatively, we know further that

p

H

+

p

O

H

=

14

. Once you have

p

H

,

p

O

H

is easy to find. Take the antilogarithm of this to get

[

O

H

−

]

.

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