Ionic Equilibrium Question 11
Question 11 - 31 January - Shift 2
At $298 K$, the solubility of silver chloride in water is $1.434 \times 10^{-3} g L^{-1}$. The value of $-\log K _{sp}$ for silver chloride is
(Given mass of $Ag$ is $107.9 g mol^{-1}$ and mass of $Cl$ is $35.5 g mol^{-1}$ )
Show Answer
Answer: (10)
Solution:
Formula: Solubility product
$ \begin{aligned} & AgCl(s) \to Ag^{+}(\text{aq. })+Cl^{-} \text{(aq.) } \\ & K _{sp}=S^{2}=(\frac{1.43}{143.4} \times 10^{-3})^{2}=10^{-10} \\ & -\log K _{sp}=10 \end{aligned} $
