Answer to Question #71046 in General Chemistry for Allison Mulligan

Question

What is the final molarity of HF, when 5.67 mL of a 2.58 M stock solution is diluted to 125 mL.

Answer 1

If C1=2.58 M, V1=5.67 ml, V2=125 ml then
n (HF) = C1*V1=2.58 M * 5.67 ml =14.6 mmole
C2=n(HF)/V2 = 14.6 mmole / 125 ml = 0.117 M