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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

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

Chemistry
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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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.
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