A solution is prepared by dissolving 10.8 g ammonium sulfate in enough water to make 100.0 mL of stock solution. A 10.00 mL sample of this stock solution is added to 50.00 mL of water. Calculate the concentration of ammonium ions and of sulfate ions in the final solution.

NH

+

4

:

0.272

M

SO

2

−

4

:

0.136

M

Explanation:

We can solve this problem using some molarity calculations:

molarity

=

mol solute

L soln

We should convert the given mass of

(NH

4

)

2

SO

4

to moles using its molar mass (calculated to be

132.14

g/mol

):

10.8

g (NH

4

)

2

SO

4

⎛

⎝

1

l

mol (NH

4

)

2

SO

4

132.14

g (NH

4

)

2

SO

4

⎞

⎠

=

0.0817

mol (NH

4

)

2

SO

4

This is the quantity present in

100

mL soln

, so let's calculate the molarity of the solution (converting volume to liters):

molarity

=

0.0817

l

mol (NH

4

)

2

SO

4

0.100

l

L soln

=

0.817

M

10

mL

of this solution is added to

50

mL H

2

O

, which makes a

60

-

mL

total solution.

We can now use the dilution equation

M

1

V

1

=

M

2

V

2

to find the molality of the new,

60

-

mL

solution:

(

0.0817

M

)

(

10

l

mL

)

=

(

M

2

)

(

60

l

mL

)

M

2

=

(

0.817

M

)

(

10

mL

)

60

mL

=

0.136

M

This means that there are

0.136

moles of

(NH

4

)

2

SO

4

per liter of solution.

Let's recognize that

1

mol (NH

4

)

2

SO

4

contains

2

mol NH

+

4

1

mol SO

2

−

4

The concentrations of each ion is thus

(

2

)

(

0.136

M

)

=

0.272

M

NH

+

4

(

1

)

(

0.136

M

)

=

0.136

M

SO

2

−

4

Explanation:

it takes on an electronical energy