Question

Calculate the concentration of Ag+(aq) ions present at equilibrium when 10 grams of AgCl(s) is mixed with 300 mL of .05 M NaCl(aq).

AgCl(s) <---> Ag+(aq) + Cl-(aq) Kc=1.8E-10

Please show all work! Thanks!

Answer 1

here AgCl ionisation we hav to count in presence of common ion Cl- (from NaCl). the concentration of Cl- need to take from AgCl and NaCl also . total concnetration of Cl- = (x + y)

x = form AgCl

y = form NaCl .

but we know NaCl is the strong electrolyte . AgCl is the sparingly soluble salt

NaCl > > > > AgCl

so Cl- concentration = y = concentration from NaCl

AgCl <--------------------------> Ag+ + Cl-

0.05 M

Ksp = [Ag+][Cl-]

1.8 x 10^-10 = [Ag+] (0.05)

[Ag+] = 3.6 x 10^-9 M