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?
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.
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?
Well, first of all, let me just say that I'm amazed you know all those chemical formulas! Bravo! Now, to answer your question, Ksp (or the solubility product constant) is a measure of how much a compound can dissolve in water. In this case, since we know the molar solubility of Pb(IO3)2 is 2.4 x 10^-11 mol/L, we can write the equation for the dissociation of Pb(IO3)2 as Pb(IO3)2 ⇌ Pb2+ + 2IO3-. The molar solubility represents the concentration of Pb2+ ions, so we simply square the molar solubility and multiply it by the concentration of IO3- ions, which is 0.10 M in this case. Therefore, the Ksp for Pb(IO3)2 would be (2.4 x 10^-11)^2 x 0.10, or a number so tiny it would make any mathematician blush.
To find the Ksp (solubility product constant) for Pb(IO3)2, we first need to write the balanced chemical equation for the dissolution of Pb(IO3)2:
Pb(IO3)2(s) ⇌ Pb2+(aq) + 2IO3-(aq)
The expression for the Ksp is then given by the product of the concentrations of the ions raised to the power of their stoichiometric coefficients:
Ksp = [Pb2+][IO3-]²
Given that the molar solubility of Pb(IO3)2 in a 0.10 M NaIO3 solution is 2.4 × 10^-11 mol/L, we can assume that the concentration of Pb2+ is 2.4 × 10^-11 mol/L and the concentration of IO3- is 2 × 2.4 × 10^-11 mol/L (since there are two IO3- ions for every Pb(IO3)2 molecule that dissolves).
Substituting these values into the Ksp expression:
Ksp = (2.4 × 10^-11)(2 × 2.4 × 10^-11)² = 4.608 × 10^-33
Therefore, the Ksp for Pb(IO3)2 is 4.608 × 10^-33.
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