Asked by Anonymous

The Ksp of mercury II hydroxide is 3.60 x 10^-26. Calculte the solubility of this compound in g/L
Answered by DrBob222
...........Hg(OH)2 ==> Hg^2+ + 2OH^-
I..........solid.......0........0
C............-x........x........2x
E..........solid.......x........2x

x = solubility
Ksp = (Hg^2+)(OH^-)^2
Substitute the E line and solve for x = solubility in mols/L, then convert to g/L.
Answered by Anonymous
1.09*10^-74
Answered by Anonymous
4.86*10^-7 is the real answer
Answered by teal
Actually both are wrong the answer is 1.69*10 to the power -7. Trust me I got the answer write on sapling
Answered by Justin
4.86*10^7 is correct. I also go it right on Sapling. x=2.09*10^-9 multiply by Molar Mass 234.605
Answered by Liana
2.08*10^-9
Answered by Adam
4.88*10^7
Answered by anon
Hg(OH)2 dissociates in water to a small extent.

Hg(OH)2(s) ↽−−⇀ Hg2+(aq) + 2OH−(aq)
initial excess 0 0
change −𝑥 +𝑥 +2𝑥
equilibrium 𝑥 𝑥
The equilibrium constant expression for 𝐾sp is

𝐾sp=[Hg2+][OH−]2=(𝑥)(2𝑥)2=4𝑥3

Substitute the value of 𝐾sp ( 3.60×10−26 ) into the expression and solve for the molar solubility, 𝑥 .

3.60×10−26=4𝑥3

𝑥3=3.60×10−264

x=3.60×10−264⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√3=2.08×10−9 M

Convert the solubility from molarity (moles per liter) to grams per liter using the molar mass of Hg(OH)2.

2.08×10−9 mol1 L×234.61 g1 mol=4.88×10−7 gL
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