Problem : A 100 ml, 0.5 M calcium nitrate solution is mixed with 200 ml of 1.25M calcium nitrate solution. Calculate the concentration of the final solution. Assume volumes to be additive.

Solution : We first calculate milli-moles for each solution. Then, add them up to get the total moles present in the solution. Finally, we divide that total moles by the total volume.

Milli-moles in the first solution
=
M
1
V
C
C
1
=
100
X
0.5
=
50
Milli-moles in the first solution
=
M
1
V
C
C
1
=
100
X
0.5
=
50

Milli-moles in the second solution
=
M
2
V
C
C
2
=
200
X
1.25
=
250
Milli-moles in the second solution
=
M
2
V
C
C
2
=
200
X
1.25
=
250

Milli-moles in the final solution
=
50
+
250
=
300
Milli-moles in the final solution
=
50
+
250
=
300

Total volume of the solution
=
100
+
200
=
300
m
l
Total volume of the solution
=
100
+
200
=
300
m
l

Hence, molarity of the final solution is :

M
=
Milli-moles of solute
V
C
C
=
300
300
=
1
M
M
=
Milli-moles of solute
V
C
C
=
300
300
=
1
M

Alternative method

Molarity of the first solution when diluted to final volume of 300 ml is obtained as :

M
1
V
C
C
1
=
M
2
V
C
C
2
M
1
V
C
C
1
=
M
2
V
C
C
2

⇒
0.5
X
100
=
M
2
X
300
⇒
0.5
X
100
=
M
2
X
300

⇒
M
2
=
50
300
=
1
6
⇒
M
2
=
50
300
=
1
6

Molarity of the second solution when diluted to final volume of 300 ml is obtained as :

M
1
V
C
C
1
=
M
2
V
C
C
2
M
1
V
C
C
1
=
M
2
V
C
C
2

⇒
1.25
X
200
=
M
2
X
300
⇒
1.25
X
200
=
M
2
X
300

⇒
M
2
=
250
300
=
5
6
⇒
M
2
=
250
300
=
5
6

Hence, molarity of the final combined solution is :

M
=
1
6
+
5
6
=
1
M
=
1
6
+
5
6
=
1