Let's break down the problem step by step: ### 1. **Find Moles of Sodium Hypoiodite (NaIO)** - Molar mass of NaIO = 22.99 (Na) + 126.90 (I) + 16.00 (O) = **165.89 g/mol** - Moles = 2.14 g / 165.89 g/mol = **0.0129 mol** ### 2. **Find Initial Concentration of IO⁻** - Volume = 1.25 L - \[\text{[IO}^-\text{]}_{\text{initial}} = \frac{0.0129}{1.25} = 0.01032\,\text{M}\] ### 3. **Write the Base Dissociation Equation** \[ \text{IO}^- + \text{H}_2\text{O} \rightleftharpoons \text{HIO} + \text{OH}^- \] ### 4. **Find [OH⁻] from pH** - pOH = 14 - pH = 14 - 11.32 = 2.68 - \([\text{OH}^-] = 10^{-2.68} = 2.09 \times 10^{-3}\) M ### 5. **ICE Table** Let \( x \) = [OH⁻] at equilibrium = \(2.09 \times 10^{-3}\) M | | IO⁻ | HIO | OH⁻ | |---------|-------------|-------------|------------| | Initial | 0.01032 | 0 | 0 | | Change | -x | +x | +x | | Equil. | 0.01032-x | x | x | ### 6. **Calculate \( K_b \)** \[ K_b = \frac{[\text{HIO}][\text{OH}^-]}{[\text{IO}^-]} = \frac{(2.09 \times 10^{-3})(2.09 \times 10^{-3})}{0.01032 - 2.09 \times 10^{-3}} \] \[ 0.01032 - 0.00209 = 0.00823 \] \[ K_b = \frac{(2.09 \times 10^{-3})^2}{0.00823} \] \[ = \frac{4.37 \times 10^{-6}}{0.00823} \] \[ = 5.31 \times 10^{-4} \] ### 7. **Answer** The closest answer is: **b) \(5.4 \times 10^{-4}\)** --- **Final answer:** > **b) \(5.4 \times 10^{-4}\)**