Total volume is 7.50+5.00+7.50 =20.00 (+ 1 drop but I guess they don't count that).
You have 7.50 mL of 0.200 M KI so the concn in the mixed solution is (0.00750 L x 0.200 = ?? moles and that divided by 0.0020 L) = ?? M KI. All of the others are done the same way. If you are comfortable with millimoles, you can make it much simpler by 7.50 x 0.200 = ?? millimoles and that divided by 20 mL = xx M KI. Even simpler is to treat it as a dilution problem like so.
0.200 M x (7.50 mL/20.0 mL) = ?? M KI.
okay so im doing pre lab questions on reaction rates
I have a chart labeled effects of varying concentration
in reaction #1 flask A contains 7.50mL of .200 M KI , none of .20M KCI, 5.00 mL of .0050 M S2O3^2-, 1 drop of starch. Flask B contains 7.5mL of .100M S2O8^2-
It says for each reaction (i just gave the 1st one) determine the concentration of KI, (NH4)2S2O8 and Na2S2O3 that will be present when solutions A and B are mixed together. the total volume for each reaction is 20.0 mL. You must show one sample calculation for each component.
I'm not sure how to do this, if someone can give me the calculations for the first one, i can figure out the rest. ThankS!
2 answers
To correct DrBob222 it would be .020 L instead of 0.0020 L
1 answer