1.0 dm3 of distilled water was used to wash 2.0g of a precipitate of AgCl. If the solubility product of AgCl is 2.0 x 10\({^-10}\)moldm-6, what quantity of silver was lost in the process?
- 2.029 * 10-3mol dm-3
- 1.414 * 10-3mol dm-3
- 2.029 * 10-5mol dm-3
-
1.414 * 10-5mol dm-3
The correct answer is: D
Explanation
In order to calculate this, it suffices to calculate the solubility of Ag, so as to know the amount lost.
\(K_{sp} = [Ag^{+}][Cl^{-}]\)
Let the solubilty of Ag and Cl = d
\(2 \times 10^{-10} = d \times d= d^{2}\)
\(d = \sqrt{2\times10^{-10}}\)
= \(1.414 \times 10^{-5}moldm^{-3}\)
There is an explanation video available .