Question

Would a precipitate of silver acetate form if 22.0 mL of 0.100 M AgNO3 were added to 45.0 mL of 0.0260 M NaC2H3O2? For AgC2H3O2, Ksp = 2.3 × 10-3. Q =

Answer 1

mole of Ag⁺ in 22.0 ml = 0.1 mol/L * 22*10^-3 L = 2.2*10^-3 mole

total volume of the solution = 45.0 + 22 = 67.0 ml = 67*10^-3 L

final [Ag⁺] = 2.2*10^-3 mole / 67*10^-3 L= 0.033 M

similarly, mole of C2H3O2- = 0.0260 mole/L * 45*10^-3 L = 1.17*10^-3 mole

[C2H3O2-] =1.17*10^-3 mole / 67*10^-3 L = 0.017 M

ionic product of AgC2H3O2 = [Ag⁺] [C2H3O2-] =  0.033* 0.017 = 0.561*10^-3 < Ksp value.

so, there will be no precipitation of silver acetate because value of ionic product is less than Ksp value.