Chemistry Help!!!!!

Question

5.00 ml of 5.05×10^-5 M CV⁺ is mixed with 4.00 ml of 0.0200 M NaOH and 1.00 mL of DI water. The absorbance of the solution was 0.645 at 20.00 seconds and 0.437 at 40.00 seconds.

What is the concentration of CV⁺ at 20.00 seconds?

 

What is the concentration of CV⁺ at 40.00 seconds? 

 

What is the rate of the reaction at the consecutive times of 20.00 and 40.00 seconds?

 

What is the initial concentration of CV⁺ in the mixture?

 

What is the initial concentration of OH^- in the mixture?

I'm stuck on all of it- no idea where to start! Thanks

Answer 1

Let us find the initial concentrations of CV⁺ and OH^-. Final volume = 5 ml + 4 ml + 1 ml = 10.00 mls.

CV⁺ : (5.00 ml)(5.05x10^-5 M) = (10.00 mls)(x M) and x = 2.53x10^-5 M initial [CV⁺]

OH- : (4.00 ml)(0.0200 M) = (10.00 mls)(x M) and x = 0.00800 M = 8x10^-3 M initial [OH-]

Now, it is not possible to find the concentration at 20 sec without knowing the molar extinction coefficient and the use Beer Lambert Law where Absorbance = ecl = extinction x concentration x path length. None of this information is given. Do you have this information? Or, if you had the absorbance of the solution at zero time, one could estimate the concentrations at 20 and 40 seconds, and hence obtain the rate at these 2 time points.

Absent any of this additional information, I really can't see how to estimate the concentrations at 20 and 40 seconds, not the rate of reaction at these times.

NOTE: Please follow up if you have additional information.