Exercise 2 | Chemical Calculations for Solutions

What is the molarity of a 500-mL solution if 30.0 mL of 6.0 M HCl is added to 470 mL of 2.0 M HCl?

Solution 1: 30.0 mL of 6.0M HCl

Solution 2: 470 mL of 2.0M HCl

Solution 3: final 500-mL solution


The number of moles in solution does not change:

n1 + n2 = n3

M1 V1 + M2 V2 = M3 V3

⇒ M3 = M1 V1 + M2 V2V3

M3 = 6.0 × 30.0 + 2.0 × 470500 = 2.24 mol.L-1