Question

what is the molar concentration of [I-] from the mixing of .02M Na2S2O3, 0.3M KI, and 0.1H2O2

Answer 1

you need to be given volumes for the solutions.

Answer 2

I have the volumes:
Na2S2O3- .001L
KI- .001L
H202- .003L

Answer 3

find the moles of \(I^-\), then find the new molarity taking into account the total volume (of the other solutions).

Answer 4

How can I find the moles of I- with this information?

Answer 5

use the molarity formula \(M=\dfrac{n_{solute}}{L_{solution}}\)

Answer 6

Would I first use this formula to calculate the moles of the ions I3- and S2O3^2- and then add those together to get the resulting moles for I-. Sorry if this is confusing you.

Answer 7

there is no need to use \(S_2O_3^{2-}\), the question is asking about \(I^-\), no?

you find the moles of \(I^-\), then use the molarity equation with the total volume.

Answer 8

Wait... I think that I get it. Once I find the moles for I3- I can use it to find the \[I ^{-}\] because of the equation \[3 I ^{-}(aq) + H2O2(aq) + 2 H3O ^{+}(aq)--- I3^{-} (aq) + 4 H2O(l)\]

Answer 9

and then find the molarity

Answer 10

i'm afraid you're overcomplicating the question.

\(M_{I^-}=\dfrac{n_{I^-}}{L_{solution}}=0.3 M=\dfrac{n_{I^-}}{0.001L}\rightarrow n_{I^-}=0.0003~moles\)

new molarity \(M_{I^-}=\dfrac{n_{I^-}}{L_{solution}}\rightarrow M_{I^-}\dfrac{0.0003~moles}{0.001~L+~0.001~L+0.003~L}\)

Answer 11

there was supposed to be an equal sign in the last part \(M_{I^-}=...\)

Answer 12

I know, I'm trying to fill out this table for a rate law and activation energy lab. I understood everything until I got to this part, but thank you for the help