Question

The molar solubility of Pb(IO3)2 in a 0.10 M NaIO3 solution is 2.4 10-11 mol/L. What is Ksp for Pb(IO3)2?

Answer 1

Pb(IO3)2 ==> Pb^2+ + 2IO3^-
...x..........x........2x

...............NaIO3 ==> Na^+ + IO3^-
initial.........0.1M......0.......0
change..........-0.1....+0.1......+0.1M
equil..............0......0.1......0.1

Ksp = (Pb^2+)(IO3^-)^2
(Pb^2+) = 2.4E-11 from the problem.
(IO3^-) = 2x from the Pb(IO3)2 and 0.1M from the NaIO3. The total (IO3^-) = 2x + 0.1 and we assume that x is so small that we can neglect it.
Thus, Ksp = (2.4E-11)(0.1)^2 and solve for Ksp.

Answer 2

The molar solubility of Pb(IO3)2 in a 0.10 M NaIO3 solution is 2.4x 10-11 mol/ L. What is the Ksp for Pb(IO3)2?

Answer 3

ph of naio3 is 4 as concentration is 0.0001M and (h+) conacentration is ten to the power of minus four.
by writings balanced equations we get,
enot=0
e cell=-0.0591*log10^4
e cell=-0.0591*4
:e cell=0.06*4=0.24
thank you