Unexposed silver halides are removed from photographic film by reaction with sod

Question

Unexposed silver halides are removed from photographic film by reaction with sodium thiosulfate (Na2S2O3, called hypo) to form the complex ion; Ag(S2O3)2^3-. What mass of Na2S2O3 is required to prepare 1.00 L of a solution that will dissolve 1.00 g of AgBr by the formation of Ag(S2O3)2^3-? Show work.

Kf for Ag(S2O3)2^3- = 4.7e13

Ksp for AgBr = 7.7e-13

(hint: merge Ksp and Kf into one overall equation showing the dissolution of AgBr by thiosulfate to form bisthiosufatoargentate (I) complex ion and Br-.)

Explanation / Answer

Unexposed silver halides are removed from photographic film by reaction with sodium thisoulfate (Na2S2O3, called hypo) to form the complex ion; Ag(S2O3)2(3-). What mass of Na2S2O3 is required to prepare 1.00 L of a solution that will dissolve 1.00 g of AgBr by the formation of Ag(S2O3)2(3-)? ~~~~~~Given Values~~~~~~ Kf Ag(S2O3)2[3-] = 4.7 * 10^13 Ksp AgBr = 3.3 * 10^-13 So I guess my real question would be how to set this problem up. I'm assuming that it will follow a template similar to: Kf = [Ag(S2O3)2(3-)][Br(-)] / (Ksp * [Na2S2O3]^2) I'm assuming that the reaction takes place so that: AgBr + 2Na2S2O3 ---> Ag(S2O3)2(3-) + 2Na + Br(-) With [AgBr]and Ag(+), which forms complex ion Ag(S2O3)2(3-), and Br(-)) = 1.00 g / 1 mol AgBr / 1.00 L I tried it this way and got an excessively large number. I feel like I'm missing a very easy concept here, maybe even something I've learned in years past (the lab has had other problems of a similar type that I've struggled with). The interesting thing is, I aced the test involving Ksp and Kf in lecture. Such is life, I guess! Any help would be greatly appreciated. Thanks.

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