211 Ag2S problem key

211 Ag2S problem key - Solve for [H2S] [H2S]=14.9 x 10 7...

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Practice Midterm I, Problem #7: one way to do this problem is… AgsS=2Ag+ + S2- S2- + H2O= HS- HS- + H2O= H2S + OH- H2O= H+ + OH- Solve/Write out all Ks: Ksp= 6 x 10 -50 Kw=10 -14 Kb1=8.3 Kb2=1.8 x 10 -7 MB: ½ [Ag+] = [S2-] + [HS-] + [H2S] CB: [Ag+] + [H+] = 2 [S2-] + [HS-] + [OH-] Assumptions: OH- ~ H+ since Ksp is very small! Ag+ << H+ & 2 S2- + HS- <<OH- So OH-~1 x 10 -7 M Substitute into Kb’s Combine Kb’s by substitution
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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 10-9 [Ag+] From the Ksp, we get 2.3 x 10-9 [Ag+] 3 = 6x 10-50 So, [Ag+] = 3 x 10-14 and then solubility = ½ [Ag+] = 1.5 x 10-14 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.

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