Determine the volume of water in mL needed to prepare 5.0mL (V2) of a 1.8 M (M2) HCl solution using 5.4 M (M1) HCl (M1).

Question

Less quantitative dilutions can also be done without a volumetric flask, The precision of concentration is dependent on how precise the volumes are transferred. For example a dilute solution of HCl can be made using a precise amount of 5.0 M HCl and distilled water. In this case the total volume of dilute HCl (V2) is the total volume of 5.0 M HCl plus the volume of water.

Determine the volume of water in mL needed to prepare 5.0mL (V2) of a 1.8 M (M2) HCl solution using 5.4 M (M1) HCl (M1). 

Answer 1

V1M1 = V2M2

(x ml)(5.4 M HCl) = (5 ml)(1.8 M HCl)

x = 1.7 mls of HCl

5.0 ml - 1.7 ml = 3.3 ml water needed