Ag S(s)= 2Ag-(ag) + s fag) 2Ag fag) + 4CI (aq)=2A9C1; (aq) K (1) (2) ...... s (aq) + H* (aq)= HS-(ág) (3) ...... к, al HS-fag) +H* (aq)=H,S(aq) ..(4) ...... K2 Ag,S(s)+ 2H (aq) + 4C1 (aq) 2A9CI, (aq)+ H,S(aq) K =K K: K K2 K = 6.0x101 x(1.1×10°). -51 eq 9.5 x 10 x1.0×10-19 = 7.64 x10-15

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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Determine the molar solubility of Ag2S in a 0.10M NaCl solution buffered to a pH of 6.50 

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The following equilibrium constant for the reaction Ag2S(s)+ 4Cl-(aq) +2H+=2Ag2-(aq)+H2S(aq) was calculated to be 7.6x10-15 

Ag S(s)= 2Ag-(ag) + s fag)
2Ag fag) + 4CI (aq)=2A9C1; (aq) K
(1)
(2)
......
s (aq) + H* (aq)= HS-(ág)
(3)
......
к,
al
HS-fag) +H* (aq)=H,S(aq)
..(4)
......
K2
Ag,S(s)+ 2H (aq) + 4C1 (aq) 2A9CI, (aq)+ H,S(aq) K =K K:
K K2
Transcribed Image Text:Ag S(s)= 2Ag-(ag) + s fag) 2Ag fag) + 4CI (aq)=2A9C1; (aq) K (1) (2) ...... s (aq) + H* (aq)= HS-(ág) (3) ...... к, al HS-fag) +H* (aq)=H,S(aq) ..(4) ...... K2 Ag,S(s)+ 2H (aq) + 4C1 (aq) 2A9CI, (aq)+ H,S(aq) K =K K: K K2
K = 6.0x101 x(1.1×10°).
-51
eq
9.5 x 10 x1.0×10-19
= 7.64 x10-15
Transcribed Image Text:K = 6.0x101 x(1.1×10°). -51 eq 9.5 x 10 x1.0×10-19 = 7.64 x10-15
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