The molar solubility of silver sulfide in a 0.162 M ammonium sulfide solution is _______ M.

Expert Solution

Ksp of Ag2S = 8*10^-51

(NH4)2S(aq) ----------------> 2NH4^+ (aq) + S^2- (aq)

0.162M 0.162M

Ag2S(s) -----------> 2Ag^+ (aq) + S^2- (aq)

2s s+0.162

Ksp = [Ag^+]^2[S^2-]

8*10^-51 = (2s)^2(s + 0.162)

8*10^-51 = (2s)^2( 0.162) [ s + 0.162 = 0.162 , s<<<<<<0.162]

s = 1.11*10^-25

The solubility of Ag2S in 0.162M m(NH4)2S = 1.11*10^-25

I have taken Ksp of Ag2S = 8*10^-51 , please any doubt comment on the BOX

queen_honey_blossom answered 2 years ago

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