Rank the compounds below in order of increasing acidity. " COH HO HO O2N Relevant Information: Kw= [H,O"] [OH]=1x 10-at 25 °C !! Strong Weak pH + pOH = 14 at 25 °C : pX =-log X Acid pH = -logC, 12 pH = -log(C,K] pK. + pK, - 14 at 25 °C for conjugate pairs Base POH = -logC, 1/2 %3D pOH = -log(C,K pH = pK, +log 4] [HA] !3! TA SA SA He 14.003 10 Ne 10. 12 4.0 40 20.1 is IN 4341012 13 AI SI 263 28 30 30 145 103 12 Na Mg 22 2431 CC Ar 20 Ca Sc TI 33 Cr Ma Fe 54 24 14 Co NI Cu Za Ga Ge As Se Br Kr 19.10 40.0 44447.041200 4 N11 93 s 4339 72 2 4 T NT o

Organic Chemistry: A Guided Inquiry
2nd Edition
ISBN:9780618974122
Author:Andrei Straumanis
Publisher:Andrei Straumanis
Chapter4: Polar Bonds, Polar Reactions
Section: Chapter Questions
Problem 24E: Summarize the relationship between pKa and base strength by completing the followingsentences: a....
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least acidic I < || < III most acidic
least acidic II < I < III most acidic
least acidic III < || < I most acidic
least acidic III < | < || most acidic
Transcribed Image Text:least acidic I < || < III most acidic least acidic II < I < III most acidic least acidic III < || < I most acidic least acidic III < | < || most acidic
Rank the compounds below in order of increasing acidity.
CO
LOH
HO
O2N
II
Relevant Information:
Kw = [H,O"] [OH) =1x 10-"at 25 °C
Strong
Weak
pH + pOH = 14 at 25 °C : pX =-log X
12
Acid
pH = -logC,
pH = -log(C,K]
pK. + pks = 14 at 25 °C for conjugate pairs
Base
pOH = -logC,
pOH = -log[C K)
12
[4]
HA]
pH = pK, + log:
!3!
6A
TA
SA
H
1.008
He
14.005
10
a LI Be
4341.012
IN
10.1 1201 14.0 14.00 1a0 20 1E
Ne
12
Na Mg
229 2431
(14
AI SI P
15
17
CIAr
269 28.09 30 320 334 3
20
Ca Sc
19.10 40.0 44447. 0o.412.00 34 N11 93 354139 s72 T2 1492 1 H 0
(34
Cr Ma Fe
114
4K
Co NI Cu Za Ga Ge As
Se
Br Kr
Transcribed Image Text:Rank the compounds below in order of increasing acidity. CO LOH HO O2N II Relevant Information: Kw = [H,O"] [OH) =1x 10-"at 25 °C Strong Weak pH + pOH = 14 at 25 °C : pX =-log X 12 Acid pH = -logC, pH = -log(C,K] pK. + pks = 14 at 25 °C for conjugate pairs Base pOH = -logC, pOH = -log[C K) 12 [4] HA] pH = pK, + log: !3! 6A TA SA H 1.008 He 14.005 10 a LI Be 4341.012 IN 10.1 1201 14.0 14.00 1a0 20 1E Ne 12 Na Mg 229 2431 (14 AI SI P 15 17 CIAr 269 28.09 30 320 334 3 20 Ca Sc 19.10 40.0 44447. 0o.412.00 34 N11 93 354139 s72 T2 1492 1 H 0 (34 Cr Ma Fe 114 4K Co NI Cu Za Ga Ge As Se Br Kr
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