Question

Calculate the moles of insoluble CH3COOAg in mixture #1 at equilibrium.

Mixture #1 Initial (Agt] 160mm Initial (CH3C00] 1140 m4 Final [Ag+] comm Final (CH3C00] goma TABLE 6.3 Solubility Analysis of CH3COC Record the concentrations in mol pure CH3COOAG. Mixture #1 Equilibrium (Ag+] 0.04m Equilibrium [CH3C00] 0.12 m Ksp 0.0051 Average K. ooo!

Answer 1

Calculate the moles of insoluble CH3COOAg in mixture #1 at equilibrium :

Ag+(aq) + CH3COO- (aq) <===> CH3COOAg (s)

Ksp = [Ag+] [CH3COO-] = 0.042 M * 0.12 M = 5.04 x 10-3

for mixture #1 :

             CH3COOAg (s) <===> Ag+(aq) + CH3COO- (aq)

initial            1                            0               80 mM = 0.08 M

eq.              1-x                           x             x + 0.08

Ksp = [Ag+] [CH3COO-] = x * x+0.08 = 5.04 x 10-3

x2 + 0.08*x - 5.04 x 10-3 = 0

on solving :   x ~ 0.042 M

thus ,

insoluble CH3COOAg , for 1 L solution ~ 0.02 moles