Chemistry kinetics question?

Question

A sample was prepared by mixing 18.0mL of 5.00 X 10^-3 M Crystal Violet (CV) with 2.00mL of 0.333 M NaOH. What are the resulting concentrations of CV and OH-?

Answer 1

The final volume of the mixture is 18.0 ml + 2.00 ml = 20.0 ml.

To find final concentration of each, use C1V1 = C2V2

For Crystal Violet: (5.00x10^-3 M)(18.0 ml) = (x M)(20.0 ml); x = 4.5x10^-3 M final concentration

For NaOH: (0.333 M)(2.00 ml) = (x M)(20.0 ml); x = 0.0333 M NaOH = 0.0333 M OH- final conc.

Answer 2

This is a classic kinetics experiment to determine reaction order. The first answer isn't correct as it doesn't account for reaction, but it does show that the concentration of [OH^-]>>[CV⁺] which helps later. The empirically derived rate law is...

Rate = k[OH^-][CV⁺], i.e., the exponents in the rate expression are both first order in the reactants.

This can be run in a pseudo first order overall by having [OH^-]>>[CV⁺] as shown.

so,

Rate = k'[CV⁺], whose integrated rate expression is ln[CV⁺]_t = ln[CV⁺]₀ - k't. The subscripts denote the concentration of crystal violet at time t and at time zero, and t is time of reaction. k' is the pseudo first order overall rate constant and needs to be determined empirically. k'~3.4x10^-3 M/s.