Calculations for preparing 100.0mL of a 1.25 x 10-4 M aqueous solution of Fe(NO3)3(aq) by diluting a stock solution of 0.0025M Fe(NO3)3 (aq). You will have a 100.0 mL volumetric flask to dilute in.

Question

net ionic equation:

Fe^3+(aq)+SCN-(aq)=FeSCN^2+(aq)

Answer 1

V₁M₁ = V₂M₂

V₁ = initial volume = x ml

M₁ = initial molarity = 0.0025 M

V₂ = final volume = 100.0 ml

M₂ = final molarity = 1.25x10-4 M

Solving for V₁ we have...

V₁ = V₂M₂ / M₁

V₁ = (100.0 ml)(1.25x10-4 M) / 0.0025 M

V₁ = 5.00 mls

Take 5.00 mls of 0.0025 M Fe(NO₃)₃ and dilute to a final volume of 100.0 mls