What is the molarity of each resulting solution when the following mixtures are prepared?

Question

(a) 587.7 mL H₂O is added to 20.0 mL 6.25 M HNO₃

(b) 31.7 mL 1.75 M HCl is added to 80.0 mL 0.336 M HCl

Answer 1

When you dilute a solution with water, as in part (a), you can simply use V1M1 = V2M2

V1 = initial volume = 20.0 ml

M1 = initial molarity = 6.25 M

V2 = final volume = 587.7 ml + 20.0 ml = 607.7 ml

M2 = final molarity

Solving for M2, we have

(20.0 ml)(6.25 M) = (607.7 ml)(x M)

x = 0.206 M

(b). 31.7 ml x 1 L / 1000 ml x 1.75 mol / L = 0.05548 moles HCl

80.0 ml x 1 L / 1000 ml x 0.336 mol / L = 0.02688 moles HCl

Total moles HCl = 0.05548 + 0.02688 = 0.08236 moles

Total volume = 31.7 ml + 80.0 ml = 111.7 mls = 0.1117 L

Final [HCl] = 0.08236 mols / 0.1117 L = 0.737 M HCl