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Unformatted text preview: Solve for [H2S] [H2S]=14.9 x 10 7 [S2] Substitute the equation above and the Kb1 equation solved for HS into the MB equation, then we get ½ [Ag+] = [S 2] + 8.3 x 10 7 [S 2] + 14.9 x 10 7 [S 2] [S 2] = 2.3 x 109 [Ag+] From the Ksp, we get 2.3 x 109 [Ag+] 3 = 6x 1050 So, [Ag+] = 3 x 1014 and then solubility = ½ [Ag+] = 1.5 x 1014 Now, check the assumptions that you used for the 3% rule....
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This note was uploaded on 07/17/2008 for the course CHEM 211 taught by Professor Allen during the Summer '06 term at Ohio State.
 Summer '06
 ALLEN
 Analytical Chemistry

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