In: Chemistry
What is the solubility of AgCl om 0.01 M AgNO3? What and how large is the error this solubility would cause in the results of gravimetric experiment? you may assume the total volume of the solution to be 250 mL. Please comment on the significance of this amount of chloride lost due to solubility.
Outline to solve the problem:
The solubility product of the sparingly soluble salt AgCl = Ksp = s*s = 1.8×10–10, i.e. s = 1.342*10-5 mol/L
In the presence of common ion (Ag+) containing strong electrolyte AgNO3 with 'c' M concentration,
Ksp = (s' + c)*s' = 1.8×10–10
Here, s' < s and s' <<< c, because of the common ion effect.
i.e. Ksp = c*s' = 1.8×10–10
Now, coming to the problem, c = 0.01 M
Therefore, 0.01*s' = 1.8×10–10
i.e. s' = 1.8×10–8 mol/L < s (= 1.342*10-5 mol/L)
Given that the volume of the solution = 250 mL = 250*10-3 L
For 1 L of solution = 1.8*10-8 mol of AgCl is soluble
For 250*10-3 L of solution = 250*10-3*1.8*10-8 mol= 4.5*10-9 mol of AgCl is soluble.
i.e. The solubility of AgCl om 0.01 M AgNO3 = 4.5*10-9 mol / 250 mL