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Question

What is equilibrium constant for the reaction:
3Ag2S(s)+Ba3(AsO4)2(s)+3H+2Ag3AsO4(s)+3HS+3Ba2+
Given Ka2 for H2S is 1.10×1015 and KSP for Ba3(AsO4)2,Ag2S and Ag3AsO4 are 7.70×1051,2.0×1049 and 1.0×1022 respectively.

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Solution

Ka2=[H+][S2][HS]=1.10×1015
KspBa3(AsO4)2=[Ba2+]3[AsO34]2=7.70×1051
KspAg2S=[Ag+]2[S^{2-}] = 2.0 \times 10^{-49}K_{sp}Ag_3AsO_4 = [Ag^+]^3[AsO_4^{3-}] = 1.0 \times
10^{-22}$

The equilibrium constant expression is K=[HS(aq)]3[Ba2+(aq)]3[H+(aq)]3
K=([H+][S2]Ka2)3×KspBa3(AsO4)2[AsO34][H+]3.

K=[S2]3×KspBa3(AsO4)2(Ka2)3×(KspAg3AsO4[Ag+]3)2.


K=(KspAg2S[Ag+]2)3×KspBa3(AsO4)2(Ka2)3×(KspAg3AsO4[Ag+]3)2.


K=(KspAg2S)3(KspBa3(AsO4)2)(kspAg3AsO4)2

K=(2.0×1049)3(7.70×105)(1.6×1022)2=4.63×10108=x×10108

x=4.63

100x=463.

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