Question

Consider the ionic solid Pb(OH)2. When dissolved in pure water, its solubility is found to be 6.7×10-6 M.

Give the formula of the anion in this compound.

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[For example, Sr2+ would be "Sr^2+"; or PO43- would be"PO4^3-".] )

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What is the concentration of this anion in a saturated solution of Pb(OH)2?

Determine the value of Ksp for Pb(OH)2.

Tries 0/10

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Answer 1

Consider reaction Pb(OH)_{2 (s)}  Pb ²⁺_{(aq )} + 2 OH ^-_(aq)  

For above reaction, K sp = [ Pb ²⁺] [ OH ^-] ²  

If S is the solubility of Pb(OH)₂ in mol / L then [ Pb ²⁺] = S mol / L and [ OH ^-]= 2 S mol / L .

Therefore, K sp = S ( 2 S) ²

K sp = 4 S ³

We have, S = 6.7 10 ^-06 M

K sp = 4 ( 6.7 10 ^-06 )  ³

K sp = 1.2 10 ^-15.

Concentration of anion in saturated solution of Pb(OH)₂ = 2 S = 2 ( 6.7 10 ^-06 M ) = 1.34 10 ^-05 M

ANSWER : Formula of anion : OH^1-

Concentration of anion in saturated solution of Pb(OH)₂ : 1.3   10 ^-05M

Ksp of Pb(OH)₂ = 1.2 10^-15