2.0dm3 of distilled water was used to wash 2.0g of precipitate of Agcl. If the solubility product of Agcl is 2.0×10^-5. what is the quantity of silver was lost in the process?

Answer 1

The solubility product constant (Ksp) of AgCl is 2.0×10^-5.

First, we need to calculate the concentration of silver ions (Ag+) in the saturated solution of AgCl using the solubility product constant:

AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)

Ksp = [Ag+][Cl-]
2.0×10^-5 = [Ag+]^2

[Ag+] = √(2.0×10^-5)
[Ag+] = 0.00447 mol/L

Now, we know that 2.0 dm3 (or 2000 mL) of distilled water was used to wash the precipitate. This means that 0.00447 mol of silver ions were dissolved in 2.0 dm3 of water.

Therefore, the total quantity of silver lost in the process is:

Total quantity of silver lost = 0.00447 mol/L x 2.0 dm3
Total quantity of silver lost = 0.00894 mol

Therefore, the quantity of silver that was lost in the process of washing the precipitate of AgCl is 0.00894 mol.