ül11. For the weak base piperidine, CSH10NH, K, = 1,30 x 10-3. What is the equilibriumconcentration of all species, the pH, % ionization, and pOH of a O.545 M solution ofpiperidine?108CらHioNI4 +H20 cg Itio NH2 t OH[coHiO NH2 ]•COH] = •0260 m[ cgHIO NH] =*Sしらーメ+ x tx* Sしら -入メPOH= 1.5851.30X10sPIt12,4115%3D•SIS - X4.lon= - 0260XIC0 = い、7%。※+1.30×1 -703×IG"ニ0×ニ,026012. What is the pH of a 0.459 M solution of NaCIO. Ka = 2.87 x 10-8(e) TH Skb=CHClot3 COHT kb= \xlo4NacIONat+ CIO%3DNeutralbasic2.87x10

11) Reaction- C5H10NH + H2O → C5H10NH2+ + OH- Your steps and calculation for question number 11 are correct. Answers- Equilibrium concentrations [ C5H10NH ] = 0.519 M [C5H10NH2+] = 0.0260 M [OH-] = 0.0260 M pH = 12.41 pOH = 1.59 % ionization = 4.77 %12) NaClO is weak base. Ka = 2.87 x 10-3 Kb = 1.0 x 10-14KaKb = 1.0 x 10-142.87 x 10-8Kb = 3.48 x 10-7 Reaction- NaClO → Na+ + ClO- ClO- HClO OH- I 0.459 0 0 C -X +X +X E 0.459 - X X X Kb = [HClO] [OH-][ClO-]3.48 x 10-7 = X20.459 - X5 % rule is applicable , hence ignore the X in denominator3.48 x 10-7 = X20.459 X= 3.9966 x 10-4 X = [OH-] = 3.9966 x 10-4 M pOH = -log [OH-] pOH = -log (3.9966 x 10-4) pOH = 3.40 pH + pOH = 14 pH = 14 - pOH pH = 10.60

Science

Chemistry

11. For the weak base piperidine, CsHioNH, K. = 1.30 x 10-3. What is the equilibrium concentration of all species, the pH, % ionization, and pOH of a 0.545 M solution of piperidine? Cg HIONIH + H20 = CSItio NH2+ OH- 10 エSし1ら [COHIO NH2 I•COH] [ CgHIO NH] = = •0260 m ーメ + x tx • SLIS - X POH= I.S8らメ 1.30 X10s PIt %3D • SIS - X -6260 = いい17%。入IO0 文+1.30x10ー0るX10 ×ニ0260 ーい 12. What is the pH of a 0.459 M solution of NaCIO. Ka = 2.87 x 10-8

11. For the weak base piperidine, CsHioNH, K. = 1.30 x 10-3. What is the equilibrium concentration of all species, the pH, % ionization, and pOH of a 0.545 M solution of piperidine? Cg HIONIH + H20 = CSItio NH2+ OH- 10 エSし1ら [COHIO NH2 I•COH] [ CgHIO NH] = = •0260 m ーメ + x tx • SLIS - X POH= I.S8らメ 1.30 X10s PIt %3D • SIS - X -6260 = いい17%。入IO0 文+1.30x10ー0るX10 ×ニ0260 ーい 12. What is the pH of a 0.459 M solution of NaCIO. Ka = 2.87 x 10-8

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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