Integrative Problems equire the integration of multiple concepts to fi.docx
Download as DOCX, PDF0 likes11 views
The document outlines the calculation of the equilibrium constant (Kp) for the hypoiodite ion from a sodium hypoiodite solution. It details the steps taken to derive molarity, pH, and the reaction that establishes Kb. The final computed value for the Kb of the hypoiodite ion is 5.3 x 10^-4.
1 of 2
Download to read offline
![Integrative Problems equire the integration of multiple concepts to find the solutions. 187. A
2.14-g sample of sodium hypoiodite is dissolved in water make 1.25 L of solution. The solution
pH is 11.32. What is Kp for the hypoiodite ion?
Solution
molarity = W/G.M.Wt * volume of solution in L
= 2.14/165.893*1.25 = 0.0103M
NaIO(aq) ---------> Na^+ (aq) + IO^- (aq)
0.0103M 0.0103M
PH = 11.32
POH = 14-PH
= 14-11.32
=2.68
POH = 2.68
-log[OH^-] = 2.68
[OH^-] = 10^-2.68 = 0.00209M
IO^- (aq) + H2O ----------> HIO(aq) + OH^-
I 0.0103 0 0
C -0.00209 0.00209 0.00209
E 0.00821 0.00209 0.00209](https://image.slidesharecdn.com/integrativeproblemsequiretheintegrationofmultipleconceptstofi-230204011443-138f2e4b/85/Integrative-Problems-equire-the-integration-of-multiple-concepts-to-fi-docx-1-320.jpg)
![Kb = [HIO][OH^-]/[IO^-]
= 0.00209*0.00209/0.00821 = 0.00053
Kb of hypoiodite ion = 5.3*10^-4 >>>>answer](https://image.slidesharecdn.com/integrativeproblemsequiretheintegrationofmultipleconceptstofi-230204011443-138f2e4b/85/Integrative-Problems-equire-the-integration-of-multiple-concepts-to-fi-docx-2-320.jpg)
Integrative Problems equire the integration of multiple concepts to fi.docx
- 1. Integrative Problems equire the integration of multiple concepts to find the solutions. 187. A 2.14-g sample of sodium hypoiodite is dissolved in water make 1.25 L of solution. The solution pH is 11.32. What is Kp for the hypoiodite ion? Solution molarity = W/G.M.Wt * volume of solution in L = 2.14/165.893*1.25 = 0.0103M NaIO(aq) ---------> Na^+ (aq) + IO^- (aq) 0.0103M 0.0103M PH = 11.32 POH = 14-PH = 14-11.32 =2.68 POH = 2.68 -log[OH^-] = 2.68 [OH^-] = 10^-2.68 = 0.00209M IO^- (aq) + H2O ----------> HIO(aq) + OH^- I 0.0103 0 0 C -0.00209 0.00209 0.00209 E 0.00821 0.00209 0.00209
- 2. Kb = [HIO][OH^-]/[IO^-] = 0.00209*0.00209/0.00821 = 0.00053 Kb of hypoiodite ion = 5.3*10^-4 >>>>answer