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
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4b90110e-d69f-4502-ad47-61daee248c92%2F65ed3c23-79ee-4198-8435-ee78675f6069%2F022kjy_processed.jpeg&w=3840&q=75)
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