In black-and-white film developing, excess AgBr is removed from the film negativ

Question

In black-and-white film developing, excess AgBr is removed from the film negative by hypo an aqueous solution of sodium thiosulfate (Na2S2O3), through formation of the complex ion Ag(S2O3)2^3-. Calculate the solubility of AgBr in (b) 1.0M hypo. Kf of Ag(S2O3)2^3- is 4,7 x 10^13 and Ksp AgBr is 5 x 10^-13.

I know the answer in (b) is 0.45 but I have a question about the step when we set up ICE table. Why we didn't assume that (1.0 - 2x) is neggligble as we do in most other euilibruim questions?

*The answe can be found on Chegg just by copying the question and pasting it into the search box.

Thank you in advance :)

Explanation / Answer

The value of equilibrium constant for the overall equation (K = Kf*Ksp) is 23.5, which is a considerable number. That is the formation of products is favorable and reactants consume to measurable extent. If K value is very less, we can assume that s is negligible. But that is not the case here, also on approximation of 1.0-2s = 1.0 gives s as 4.9. Then 1.0-2s will be '-8.8' but not equal to 1.0 as approximated. Thus, based on K value and back calculations approximations should be done.

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