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yKpoI14uk [10]
3 years ago
11

The concentration of iodide ions in a saturated solution of lead (ii) iodide is __________ m. the solubility product constant of

pbi2 is 1.4 x 10-8.
Chemistry
1 answer:
Allisa [31]3 years ago
3 0
When PbI₂ dissolves, it dissociates as follows
PbI₂ --> Pb²⁺ + 2I⁻
Molar solubility is the number of moles of salt that can be dissolved in 1 L of solution 
If molar solubility of PbI₂ is x , then molar solubility of Pb²⁺ is x and I⁻ is 2x .
ksp is solubility product constant. 
ksp = [Pb²⁺][I⁻]²
ksp = [x][2x]²
ksp = 4x³
4x³ = 1.4 x 10⁻⁸
x³ = 0.35 x 10⁻⁸
x = 1.51 x 10⁻³ M
since molar solubility of I⁻ is 2x , then molar solubility of I⁻ is 3.03 x 10⁻³ M 
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Answer:first D. 88L

Second A 2*10^24

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trapecia [35]

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10.945 x 10^-4

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Balanced equation:

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it takes 2 moles HCL for each mole Mn(OH)2

Next find the molarity of the Mn(OH)2 solution

= (1 mole Mn(OH)2 / 2 mole HCl)  X (0.0020 mole HCl / 1000ml) X (4.86 ml)    

= 4.86 x 10^-3 mole  

this is now dissolved in (70 + 4.86)  =  74.86 ml or 0.07486 L

thus [Mn(OH)2]  =  4.86 x 10^-3 mole / 0.07486 L  =  0.064921 M

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