How to write the Ksp Expression? - write an ksp of pb3(aso4)2,
The chemical equation is: Pb (NO3) 2 + 2NaI -----> PbI2 + 2NaNO3
The total ionic equation: Pb 2no3 + 2 Na + + 2I + PbI2 -----> 2Na + 2no3
net ionic equation is: Pb + 2I -----> PbI2
Is it the right to express Ksp?
Ks = [2 x I] 2 [Pb]
I do not know if it is double or iodine does not
Write An Ksp Of Pb3(aso4)2, How To Write The Ksp Expression?
4 comments:
Complete Ionic:
Pb 2 + + 2 NO3-+ 2 Na + + 2 I->> PbI2 (s) + 2 Na + + 2 NO3 --
net ionic:
Pb 2 + + 2 I->> PbI2
Ksp = [Pb 2 +] [I-] ^ 2
10 KS = 7.1 x ^ -9
If you are the molar solubility:
Let x = mol / L Pb 2 +, to solve. That gives x mol / L Pb 2 + and 2x mol / LI --
7.1 x 10 ^ -9 = (x) (2x) ^ 2 = 4x ^ 3
molar solubility = 0.0012 M
Complete Ionic:
Pb 2 + + 2 NO3-+ 2 Na + + 2 I->> PbI2 (s) + 2 Na + + 2 NO3 --
net ionic:
Pb 2 + + 2 I->> PbI2
Ksp = [Pb 2 +] [I-] ^ 2
10 KS = 7.1 x ^ -9
If you are the molar solubility:
Let x = mol / L Pb 2 +, to solve. That gives x mol / L Pb 2 + and 2x mol / LI --
7.1 x 10 ^ -9 = (x) (2x) ^ 2 = 4x ^ 3
molar solubility = 0.0012 M
Hello!
Its expression is correct. But do not forget to show the charge of the ions, namely, self-or Pb + 2. Yes iodide ions have to be multiplied by 2, their concentration is two times.
they do not multiply by two of iodine, which indicated that 2 moles of iodine by the exhibitor.
Ks = [I] 2 [Pb]
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