Water/Wastewater Laboratory Calculations - Basic Mathematics for Water and Wastewater Operators - page 256

Solution:

NV = NV

11 22

01 . N × 1 000 mL

000 mL

= 1 2 . N × V

2

01 . N × 1

= V

2

1 . 2 N

83.3 mL = V

2

■ Example 16.39
Problem: A 0.1- N solution is needed to conduct an analysis, but a 2.1- N solution is
on hand. How many milliliters of the 2.1- N solution are required to make 1 L of a
0.1- N solution?

Solution:

NV = NV

11 22

01 . N × 1 000 mL

000 mL

= 2 1 . N × V

2

01 . N × 1

= V

2

2 . 1 N

47.6 mL = V

2

■ Example 16.40
Problem: If 475 mL of 5- N NaOH are diluted to 1 L, what is the new normality of
the solution?

Solution:

NV = NV

11 22

5 5 N × 0mL = N × 1 000 mL

2

5 5
1

N 0mL

000 mL
2.375

×

= N

2

= N

2

■ Example 16.41
Problem: How many milliliters of water should be added to 10 mL of 15- N H 2 SO 4
to make 0.4- N H 2 SO 4 ?