K = A B Q = A B Kw = [H3O*][OH ]= 1.00 x 10-14 @ 25°C %3D %3D pH = -log10([H3O*) pOH = -log10([OH-]) %3D [H3O*] = 10-pH [OH] = 10-POH pH + pОН — 14 %3D [H3O"|[A] HA [HB"||OH| %3D 1.00 x 10-14 Ka K.K %3D [B] pKa = -log(Ka) pK, = -log(K») pKa+ pK = 14 Henderson Hasselbach: pH = pK. + log (HA) 497. mL of a 0.14 MX solution is titrated to the equivalence point with a 0.14. M HCI solution. The base dissociation constant for X is K, = 1.68 x 10-10

Introduction to General, Organic and Biochemistry
11th Edition
ISBN:9781285869759
Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Chapter16: Amines
Section: Chapter Questions
Problem 16.27P: 16-27 Guanidine, p/Ca 13.6, is a very strong base, almost as basic as hydroxide ion. NH NH2+ II II...
icon
Related questions
icon
Concept explainers
Question
K
[A]«[B]*
Q = AB
Kw = [H3O*][OH¯]=1.00 × 10-14 @25°C
pH = –log10([H3O+))
pOH = –log10([OH-])
[H3O*]
[OH ] = 10¬POH
pH + pOH = 14
10-pH
[H3O¯][A¯]
HA
[HB"]|OH¯]
[B]
Ка
Ki =
KaK = 1.00 x 10-14
pKa = -log(Ka)
pK, = –log(K)
pKa + pK, = 14
%3D
Henderson Hasselbach: pH = pK. + log ( )
HA
497. mL of a 0.14 M X solution is titrated to the equivalence point with a 0.14. M HCI solution.
The base dissociation constant for X is K = 1.68 x 101
What is the pH at the equivalence point?
Please express your answer to the 3rd decimal place.
Transcribed Image Text:K [A]«[B]* Q = AB Kw = [H3O*][OH¯]=1.00 × 10-14 @25°C pH = –log10([H3O+)) pOH = –log10([OH-]) [H3O*] [OH ] = 10¬POH pH + pOH = 14 10-pH [H3O¯][A¯] HA [HB"]|OH¯] [B] Ка Ki = KaK = 1.00 x 10-14 pKa = -log(Ka) pK, = –log(K) pKa + pK, = 14 %3D Henderson Hasselbach: pH = pK. + log ( ) HA 497. mL of a 0.14 M X solution is titrated to the equivalence point with a 0.14. M HCI solution. The base dissociation constant for X is K = 1.68 x 101 What is the pH at the equivalence point? Please express your answer to the 3rd decimal place.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

Blurred answer
Knowledge Booster
Ionic Equilibrium
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Introduction to General, Organic and Biochemistry
Introduction to General, Organic and Biochemistry
Chemistry
ISBN:
9781285869759
Author:
Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher:
Cengage Learning