[chaos]

Hello everyone.

The experiment that I did involved 6 different reaction mixtures. They all contained water (mL), 0.05 M HC2H302- NaC2H3O2 buffer (mL), 0.05 M KI (mL), 0.1% Starch (mL), 0.05M Na2S2O3, and 0.8M H2O2 (mL) which was added at the end so that a reaction would eventually occur when I stirred the mixture.

Each mixture had a different mL of each substance to test how each effected the rate of the reaction. In one of the reactions we added 0.3M HC2H3O2 (mL) to see how it effected the reaction.

Now my question deals with filling in the data in the report sheets under the section 1. Calculating the Reaction Orders.

Part C asks to calculate the order of the reaction with respect to H+. It hints that Ka for acetic acid = 1.8 x 10^-5 M, calculate the H+ concentration from Equation(5), using the known concentrations of acetic acid and sodium acetate contained in reaction mixtures 1 and 4. (In mixture 4 be sure that you account for the acetic acid contained in both the buffer and the added 0.3M acetic acid).

Equation (5) is Ka = ([H+][C2H3O2-])/ [HC2H3O2]

Ok, so the table asks for Acetic Acid Concentration (mol/L), Sodium Acetate Concentration (mol/L) and the calculated H+ Concentration (mol/L).

For the Acetic Acid concentration would it just be the given 0.05M or would that have to be calculated through multiplying it out with the L used in the experiment divided by the total L used.

Same thing goes for the Sodium Acetate Concentration.

And lastly, what exactly does this mean (In mixture 4 be sure that you account for the acetic acid contained in both the buffer and the added 0.3M acetic acid) mean?

Here is what I tried doing when solving for H+. The given Acetic Acid Concentration and Sodium Acetate Concentration was 0.05M since they were both in the buffer and we only used that in the 1st reaction. The equation would be 1.8 x 10^-5 M = ([H+][0.05M])/[0.05M]

Now the 4th reaction I set up the equation like this 1.8 x 10^-5 M = ([H+][0.05M])/[.3M] because in the 4th reaction we added 0.3M HC2H3O2.

Any assistance would be great. Thank you very much.