9. Titration of valine by a strong base, for example NaOH, reveals two pk's. The titration reaction occurring at pK2 (pK2 = 9.62) is: а) —СООН + ОН ->—СОО" + Н20. b) -COOH + -NH2 -> -COO" +–NH2*. c) -coO +-NH2* -> -COOH + –NH2. d) –NH3* + OH" -> –NH2 + H2O. e) –NH2 + OH" -> –NH" + H2O. 10.In a highly basic solution, pH = 13, the dominant form of glycine is: а) NH2—CH2—соон. b) NH2-CH2-COO". c) NH2-CH3*-cOO". d) NH3*—CH2—соон. e) NH3*-CH2–COO".

Biochemistry
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Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
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Chapter1: Biochemistry: An Evolving Science
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9. Titration of valine by a strong base, for example NaOH, reveals two pK's. The titration reaction occurring at
pK2 (pK2 = 9.62) is:
a) —СООН + Он -> —соо" + Н20.
b) -COOH + -NH2 -> -COO" +-NH2*.
c) -co0 +-NH2* ->-COOH +–NH2.
d) –NH3* + OH" -> –NH2 + H2O.
e) –NH2 + OH -> –NH + H2O.
10.In a highly basic solution, pH = 13, the dominant form of glycine is:
а) NH2—CH2—СООН.
b) NH2-CH2-coO".
c) NH2-CH3*-coo.
d) NH3*-CH2-COOH.
e) NH3*-CH2-coo".
Transcribed Image Text:9. Titration of valine by a strong base, for example NaOH, reveals two pK's. The titration reaction occurring at pK2 (pK2 = 9.62) is: a) —СООН + Он -> —соо" + Н20. b) -COOH + -NH2 -> -COO" +-NH2*. c) -co0 +-NH2* ->-COOH +–NH2. d) –NH3* + OH" -> –NH2 + H2O. e) –NH2 + OH -> –NH + H2O. 10.In a highly basic solution, pH = 13, the dominant form of glycine is: а) NH2—CH2—СООН. b) NH2-CH2-coO". c) NH2-CH3*-coo. d) NH3*-CH2-COOH. e) NH3*-CH2-coo".
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