1. Understand the Chemistry
* Sodium hypoiodite (NaIO) is a salt that dissolves in water to produce sodium ions (Na+) and hypoiodite ions (IO-).
* Hypoiodite ions (IO-) are the conjugate base of hypoiodous acid (HIO). They can react with water in an equilibrium reaction to produce hydroxide ions (OH-):
```
IO-(aq) + H2O(l) ⇌ HIO(aq) + OH-(aq)
```
* Kb is the base dissociation constant for this equilibrium. A higher Kb value indicates a stronger base.
2. Set Up the ICE Table
We'll use an ICE (Initial, Change, Equilibrium) table to find the equilibrium concentrations:
| | IO- | HIO | OH- |
|-------------|---------|---------|---------|
| Initial | x | 0 | 0 |
| Change | -y | +y | +y |
| Equilibrium | x - y | y | y |
* x is the initial concentration of IO-
* y is the change in concentration at equilibrium
3. Calculate the Initial Concentration of IO-
* Molar mass of NaIO: 149.89 g/mol
* Moles of NaIO: (2.14 g) / (149.89 g/mol) = 0.0143 mol
* Initial concentration of IO-: (0.0143 mol) / (1.25 L) = 0.0114 M
4. Calculate the Hydroxide Ion Concentration
* pH + pOH = 14
* pOH = 14 - 11.32 = 2.68
* [OH-] = 10^-2.68 = 2.1 x 10^-3 M
5. Use the Equilibrium Concentrations to Find Kb
* [OH-] = y = 2.1 x 10^-3 M
* [HIO] = y = 2.1 x 10^-3 M
* [IO-] = x - y = 0.0114 M - 2.1 x 10^-3 M = 0.0093 M
Kb = ([HIO][OH-]) / [IO-]
Kb = (2.1 x 10^-3 M)(2.1 x 10^-3 M) / (0.0093 M)
Kb = 4.7 x 10^-4
Therefore, the Kb for the hypoiodite ion (IO-) is approximately 4.7 x 10^-4.
