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Chapter 13: Problem 187

A \(2.14-\mathrm{g}\) sample of sodium hypoiodite is dissolved in water to make 1.25 L of solution. The solution \(\mathrm{pH}\) is \(11.32 .\) What is \(K_{\mathrm{b}}\) for the hypoiodite ion?

Short Answer

Expert verified
The value of \(K_b\) for the hypoiodite ion is \(4.29 \times 10^{-4}\), which is calculated using the given mass of sodium hypoiodite, volume of the solution, and pH. Concentration of sodium hypoiodite and OH- ions are calculated, and finally, the Kb formula is used to determine the value.

Step by step solution

01

Identify important information

Given: Mass of sodium hypoiodite = 2.14 g and volume of the solution = 1.25 L, pH = 11.32.
02

Calculate the concentration of sodium hypoiodite

First, we need to find the molar mass of sodium hypoiodite (NaOI). The molecular weight of Na = 22.99 g/mol, O = 16.00 g/mol, and I = 126.90 g/mol. Therefore, molar mass of NaOI can be calculated as: Molar mass of NaOI = 22.99 g/mol + 16.00 g/mol + 126.90 g/mol = 165.89 g/mol Now, we will find the moles of sodium hypoiodite using the given mass and molar mass: moles of NaOI = (2.14 g) / (165.89 g/mol) = 0.0129 mol The concentration of sodium hypoiodite in the solution can be calculated by dividing the moles by the volume: [NaOI] = (0.0129 mol) / (1.25 L) = 0.01032 M
03

Calculate the concentration of OH- ions

We're given the pH of the solution which is 11.32. We can find the concentration of OH- ions using the following relationship: \(pH + pOH = 14\) \(pOH = 14 - pH\) \(pOH = 14 - 11.32 = 2.68\) To find the concentration of OH- ions, we'll use the formula: [OH-] = \(10^{-pOH}\) [OH-] = \(10^{-2.68}\) = 0.002095 M
04

Determine the Kb value for the hypoiodite ion

Since sodium hypoiodite dissociates into sodium cations (Na+) and hypoiodite anions (OI-), we can assume that the concentrations of OI- and Na+ are equal. Hence, the concentration of OH- ions is equal to the concentration of OI- ions. Now we can use the Kb formula to calculate the Kb value: \(K_b = \frac{[\text{OI}^-][\text{OH}^-]}{[\text{NaOI}]}\) \(K_b = \frac{(0.002095)^2}{0.01032}\) \(K_b = 4.29 \times 10^{-4}\) Therefore, the value of Kb for the hypoiodite ion is \(4.29 \times 10^{-4}\).

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Most popular questions from this chapter

Arrange the following 0.10 \(M\) solutions in order from most acidic to most basic. See Appendix 5 for \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) values. $$\mathrm{CaBr}_{2}, \quad \mathrm{KNO}_{2}, \quad \mathrm{HClO}_{4}, \quad \mathrm{HNO}_{2}, \quad \mathrm{HONH}_{3} \mathrm{ClO}_{4}$$

Calculate the \(\mathrm{pH}\) of a \(0.050-M\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\) solution \(\left(K_{\mathrm{b}}=\right.\) \(\left.1.3 \times 10^{-3}\right)\)

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Rank the following 0.10 \(M\) solutions in order of increasing pH. a. HI, HF, NaF, NaI b. \(\mathrm{NH}_{4} \mathrm{Br}, \mathrm{HBr}, \mathrm{KBr}, \mathrm{NH}_{3}\) c. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}, \mathrm{NaNO}_{3}, \mathrm{NaOH}, \mathrm{HOC}_{6} \mathrm{H}_{5}, \mathrm{KOC}_{6} \mathrm{H}_{5}\) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{HNO}_{3}\)
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