Question

Determine the $K_{sp}$ value for $AgCl$, when enough amount of $AgCl\left(s\right)$ is dissolved in water at $298K$ to produce a saturated solution of having a concentration of $A{g}^{+}$ ions is $1.3\times {10}^{-5}M$?

• A. $2.6\times {10}^{-10}$
• B. $1.3\times {10}^{-10}$
• C. $1.3\times {10}^{-5}$
• D. $1.8\times {10}^{-5}$
• E. $1.8\times {10}^{-10}$

Answer 1

Answer: (E)

$1.8\times {10}^{-10}$

1) Write the dissociation equation:

$AgCl\left(s\right)\rightleftharpoons A{g}^{+}\left(aq\right)+C{l}^{-}\left(aq\right)$

2)Writethe $K_{sp}$ expression :

$K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$

concentration of $\left[A{g}^{+}\right]$$=$$\left[C{l}^{-}\right]$ in the solution so

$K_{s}p$$=$$(1.3\ast {10}^{-}5{)}^2$$=$$1.7\ast {10}^{-}10$