TABLE 15.5Acid lonization Constants (Ka) for Some Monoprotic Weak Acids at 25 °CAcidFormulaStructural Formulalonization ReactionK.HCIO2(aq) + H20(0)-2Chlorous acidHCI021.1 x 10H3o(a) CIO2 (aq)HNO2(aq) +H200)-Nitrous acidHNO24.6 x 10H30(aq) NO2(aq)HF(aq) H200)-Hydrofluoric acidHFH-F3.5 x 10H3o (a) F(aq)0нсног(aq)H20(1)+Formic acidнсног1.8 x 104H30(a) CHO2 (aq)HC7H502(aq) H200)Benzoic acidHC7H502CH6.5 x 10H30(aq) C7H502 (aq)C-C0C2302(aq) H20U)Acetic acidHC2H3021.8 x 10H-0-0-CH3H30() C2H302 (aq)

Question
Asked Feb 24, 2019

(7): Calculate the pH and percent deprotonation of 0.10 M acetic acid CH3COOH (aq)
(use Table 15.5 to find the Ka of this weak acid)

TABLE 15.5
Acid lonization Constants (Ka) for Some Monoprotic Weak Acids at 25 °C
Acid
Formula
Structural Formula
lonization Reaction
K.
HCIO2(aq) + H20(0)
-2
Chlorous acid
HCI02
1.1 x 10
H3o(a) CIO2 (aq)
HNO2(aq) +H200)-
Nitrous acid
HNO2
4.6 x 10
H30(aq) NO2(aq)
HF(aq) H200)-
Hydrofluoric acid
HF
H-F
3.5 x 10
H3o (a) F(aq)
0
нсног(aq)
H2
0(1)
+
Formic acid
нсног
1.8 x 104
H30(a) CHO2 (aq)
HC7H502(aq) H200)
Benzoic acid
HC7H502
CH
6.5 x 10
H30(aq) C7H502 (aq)
C-C
0
C2302(aq) H20U)
Acetic acid
HC2H302
1.8 x 10
H-0-0-CH3
H30() C2H302 (aq)
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TABLE 15.5 Acid lonization Constants (Ka) for Some Monoprotic Weak Acids at 25 °C Acid Formula Structural Formula lonization Reaction K. HCIO2(aq) + H20(0) -2 Chlorous acid HCI02 1.1 x 10 H3o(a) CIO2 (aq) HNO2(aq) +H200)- Nitrous acid HNO2 4.6 x 10 H30(aq) NO2(aq) HF(aq) H200)- Hydrofluoric acid HF H-F 3.5 x 10 H3o (a) F(aq) 0 нсног(aq) H2 0(1) + Formic acid нсног 1.8 x 104 H30(a) CHO2 (aq) HC7H502(aq) H200) Benzoic acid HC7H502 CH 6.5 x 10 H30(aq) C7H502 (aq) C-C 0 C2302(aq) H20U) Acetic acid HC2H302 1.8 x 10 H-0-0-CH3 H30() C2H302 (aq)

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check_circleExpert Solution
Step 1

Calculation for finding out hydrogen ions:

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Step 2

Assuming that x is small compared to 0.10, we neglect it in the denominator:

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Step 3

Calculation of pH

 The pH can be calculated as fo...

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