Answer:
Explanation:
Hello,
In this case, the dissociation reaction is:
For which the equilibrium expression is:
![Ksp=[Pb^{2+}][I^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BPb%5E%7B2%2B%7D%5D%5BI%5E-%5D%5E2)
Thus, since the saturated solution is 0.064g/100 mL at 20 °C we need to compute the molar solubility by using its molar mass (461.2 g/mol)
In such a way, since the mole ratio between lead (II) iodide to lead (II) and iodide ions is 1:1 and 1:2 respectively, the concentration of each ion turns out:
![[Pb^{2+}]=1.39x10^{-3}M](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%3D1.39x10%5E%7B-3%7DM)
![[I^-]=1.39x10^{-3}M*2=2.78x10^{-3}M](https://tex.z-dn.net/?f=%5BI%5E-%5D%3D1.39x10%5E%7B-3%7DM%2A2%3D2.78x10%5E%7B-3%7DM)
Thereby, the solubility product results:
Regards.
Answer: 2.17mol/L
Explanation:
C=n/V
n: 0.217mol
V: 100.0mL
1. Convert mL to L (100.0mL/1000 = 0.01L)
2. Put numbers into equation (C= 0.217/0.01L)
3. Molarity is 2.17mol/L
Answer:
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Explanation:
