The Ksp for cerium iodate, Ce(IO3)3 is 3.2 e-10. What is the molar solubility of Ce ion in pure water?
= .0019 M
B) A 0.031 M sodium iodate is added as a common ion. What is the concentration of Ce ion with the common ion present?
C) What is the ratio of the Ce ion with the common ion added to the Ce ion in pure water?
Ce ion with common ion / Ce ion in pure water
how do i go about starting this problem for B and C?
Ksp = (Ce^+3)(IO3^-)^3
For B, enter IO3^- as 0.031 (from NaIO3). Technically, if S is solubility of Ce(IO3)3, then 3S is concn of IO3^- so in the NaIO3 soln, concn is 3S + 0.031; however, in general, 3S is so small in comparison to the common ions that are added that 3S can be neglected. It is good to leave them in, neglect them, solve the problem for Ce^+3, then check the assumption and see if it is valid. In my many years of doing this I have found one (just 1) case in which the 3S could NOT be neglected.
For C, just right the ratio of one over the other and you have it.
To solve parts B and C, we can use the concept of common ion effect and apply it to the equilibrium of the dissolution of cerium iodate in water.
B) To find the concentration of Ce ion when a 0.031 M sodium iodate is added as a common ion, we need to consider the common ion effect. The common ion effect states that the solubility of a salt is reduced when a common ion (from another salt) is added to the solution.
In this case, when sodium iodate (NaIO3) is added, it provides additional iodate ions (IO3-) to the solution. These iodate ions are common to both cerium iodate (Ce(IO3)3) and sodium iodate (NaIO3).
We can set up an equilibrium expression for the dissolution of cerium iodate as follows:
Ce(IO3)3(s) ⇌ Ce3+(aq) + 3 IO3-(aq)
The Ksp expression for cerium iodate is:
Ksp = [Ce3+][IO3-]^3
Now, since sodium iodate is added, we need to consider the concentration of iodate ions ([IO3-]) in the solution. Assuming the dissolution process reaches equilibrium, the concentration of iodate ions can be represented as: [IO3-] = [added iodate] + [from cerium iodate]
[IO3-] = [0.031 M] + [x] where x is the molar solubility of Ce ion in the presence of the common ion.
We can substitute this value of [IO3-] into the Ksp expression:
Ksp = [Ce3+][[0.031 M] + [x]]^3
Solving this equation for x will give us the concentration of Ce ion with the common ion present.
C) To find the ratio of Ce ion with the common ion added to the Ce ion in pure water, we can compare the concentrations of Ce ion in both cases.
The ratio can be calculated as:
Ratio = [Ce ion with common ion] / [Ce ion in pure water]
Substitute the concentrations of Ce ions found in parts B and A to calculate the ratio.
Note: The molar solubility of Ce ion in pure water is given as 0.0019 M.
To solve part B and C of this problem, you need to use the concept of common ion effect. Let's break it down step by step:
Part B:
1. Start by writing the balanced equation for the dissolution of cerium iodate in water:
Ce(IO3)3(s) ⇌ Ce3+(aq) + 3 IO3-(aq)
2. Since a 0.031 M sodium iodate is added as a common ion, it will provide an additional source of IO3- ions. Therefore, the equilibrium will shift to the left to counteract the increase in concentration of IO3- ions.
The reaction can now be rewritten as:
Ce(IO3)3(s) + 3NaIO3(aq) ⇌ Ce3+(aq) + 3 IO3-(aq) + 3Na+(aq)
3. Use the given Ksp to calculate the initial molar solubility of Ce ion in the presence of the common ion.
Ce3+ concentration = (Ksp / [IO3-]^3)^0.5
= (3.2 x 10^-10 / (0.031 M)^3)^0.5
4. Calculate the concentration of Ce ion with the common ion present by subtracting the initial concentration of the common ion from the calculated Ce3+ concentration.
Ce3+ concentration with common ion = (Ce3+ concentration) - (initial concentration of Na+)
Part C:
1. Calculate the ratio of Ce ion with the common ion added to the Ce ion in pure water.
Ratio = (Ce3+ concentration with common ion) / (Ce3+ concentration in pure water)
2. Substitute the values calculated from Part B to find the ratio.
By following these steps, you should be able to solve parts B and C of the problem.