El aminoácido glicina (H2 N-CH2-COOH) participa en los equilibrios siguientes en agua: H2

step 1 Problem 130 Part A according to Hays law, we can write this. Now we need to add this equation. S

El aminoácido glicina (H2 N-CH2-COOH) participa en los...

El aminoácido glicina $\left(\mathrm{H}_2 \mathrm{~N}-\mathrm{CH}_2-\mathrm{COOH}\right)$ participa en los equilibrios siguientes en agua: $$ \begin{aligned} & \mathrm{H}_2 \mathrm{~N}-\mathrm{CH}_2-\mathrm{COOH}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \\ & \mathrm{H}_2 \mathrm{~N}-\mathrm{CH}_2-\mathrm{COO}^{-}+\mathrm{H}_3 \mathrm{O}^{+} \quad \mathrm{K}_{\mathrm{a}}=4.3 \times 10^{-3} \\ & \mathrm{H}_2 \mathrm{~N}-\mathrm{CH}_2-\mathrm{COOH}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \\ & { }^{+} \mathrm{H}_3 \mathrm{~N}-\mathrm{CH}_2-\mathrm{COOH}+\mathrm{OH}^{-} \quad K_b=6.0 \times 10^{-5} \end{aligned} $$ (a) Use los valores de $K_a$ y $K_b$ para estimar la constante de equilibrio de la transferencia protónica intramolecular para formar un ion dipolo: $$ \mathrm{H}_2 \mathrm{~N}-\mathrm{CH}_2-\mathrm{COOH} \rightleftharpoons{ }^{+} \mathrm{H}_3 \mathrm{~N}-\mathrm{CH}_2-\mathrm{COO}^{-} $$ ¿Qué suposiciones es necesario hacer? (b) ¿Cuál es el pH de una disolución acuosa $0.050 \mathrm{M}$ de glicina? (c) ¿Cuál seria la forma predominante de la glicina en una disolución de $\mathrm{pH} 13$ ? ¿Y $\mathrm{Y}$ de $\mathrm{pH} 1$ ?

El aminoácido glicina $\left(\mathrm{H}_2 \mathrm{~N}-\mathrm{CH}_2-\mathrm{COOH}\right)$ participa en los equilibrios siguientes en agua:
$$
\begin{aligned}
& \mathrm{H}_2 \mathrm{~N}-\mathrm{CH}_2-\mathrm{COOH}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \\
& \mathrm{H}_2 \mathrm{~N}-\mathrm{CH}_2-\mathrm{COO}^{-}+\mathrm{H}_3 \mathrm{O}^{+} \quad \mathrm{K}_{\mathrm{a}}=4.3 \times 10^{-3} \\
& \mathrm{H}_2 \mathrm{~N}-\mathrm{CH}_2-\mathrm{COOH}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \\
& { }^{+} \mathrm{H}_3 \mathrm{~N}-\mathrm{CH}_2-\mathrm{COOH}+\mathrm{OH}^{-} \quad K_b=6.0 \times 10^{-5}
\end{aligned}
$$
(a) Use los valores de $K_a$ y $K_b$ para estimar la constante de equilibrio de la transferencia protónica intramolecular para formar un ion dipolo:
$$
\mathrm{H}_2 \mathrm{~N}-\mathrm{CH}_2-\mathrm{COOH} \rightleftharpoons{ }^{+} \mathrm{H}_3 \mathrm{~N}-\mathrm{CH}_2-\mathrm{COO}^{-}
$$
¿Qué suposiciones es necesario hacer? (b) ¿Cuál es el pH de una disolución acuosa $0.050 \mathrm{M}$ de glicina? (c) ¿Cuál seria la forma predominante de la glicina en una disolución de $\mathrm{pH} 13$ ? ¿Y $\mathrm{Y}$ de $\mathrm{pH} 1$ ?

Key Concepts

pH Dependence of Ionic Species

The ionic state of an amphoteric molecule such as an amino acid is highly dependent on the pH of the solution. At intermediate pH values near the isoelectric point, the zwitterion is predominant, whereas in strongly acidic or basic conditions the molecule exists mainly in its protonated or deprotonated forms, respectively. Understanding how pH shifts equilibria among different ionic species is crucial in predicting the behavior of such molecules in solution.

Equilibrium Constant Estimation Using Acid and Base Constants

Estimation of an equilibrium constant for an intramolecular proton transfer can sometimes be accomplished by relating the known acid dissociation constant (Ka) for deprotonation of the carboxyl group and the base dissociation constant (Kb) for deprotonation of the ammonium group. Under appropriate assumptions—such as negligible changes in solvent activity and ideal behavior—the equilibrium constant for the internal proton transfer can be approximated as the ratio of Ka to Kb.

Isoelectric Point and Predominant Forms

The isoelectric point (pI) is the pH at which an amino acid or small peptide carries no net electrical charge. At this pH, the concentrations of the positively and negatively charged forms balance, making the zwitterionic form predominant. In environments with pH values significantly above or below the pI, the equilibrium shifts so that either deprotonated (anionic) or protonated (cationic) forms dominate, respectively.

Amino Acid Amphoterism

Amino acids can act as both acids and bases because they contain both a carboxylic acid group and an amino group. In aqueous solutions, this property gives rise to several equilibria, including fully protonated forms, deprotonated forms, and zwitterions (dipolar ions) where internal proton transfer leads to a molecule bearing both positive and negative charges but having an overall neutral charge.

Intramolecular Proton Transfer Equilibrium

This concept refers to the process in which a proton is transferred within the same molecule, converting one functional group into another state without net exchange with the solvent. For amino acids, this often involves the transfer of a proton from the carboxyl group to the amino group, generating a zwitterionic species. Studying the equilibrium of such intramolecular proton transfers requires assumptions about the negligible effect of the solvent once the proton is transferred and that the process is rapid relative to other equilibria.

Transcript

00:03 Problem 130 part a according to hays law, we can write this.

00:32 Now we need to add this equation.

00:38 So after adding, we get n .s2, double 1 o 'h equilibrium with ns3 positive, ch2, negative.

01:05 For this reaction is we simply multiply these all value of individual state of rate constant so that is k a multiply by k b divided by k w now k for this reaction is equal to k a multiply by k b divided by k w now put the values 4 .33 into 10 to the power minus 3 multiply by 6 into 10 to the power minus 5 divided by 1 into 10 to the power minus 14 that is equals to 2 .6 into 10 to the power 7 so this is a estimated value of k for intermole proton transfer transfer from zuterion so this is answer for part a now we can clearly see here the larger value of k indicate that formation of juter ion is favorable okay the assumption is that the same among and this molecule molecule is acting like an acid and the base is this molecule sorry this molecule is acting like acting like both acid acting like both acid and a base simultaneously now in part b since glycerine glycine exists as the zuterion in aqua's solution the ph is determined by the equilibrium below that this molecule this glycine molecule combined with h2o and it will be equilibrium with neutral glycine molecule plus h2o now we can write ca for this reaction that is concentration of neutral glycine into concentration of hydrogenium ion divided by by concentration of ionized glycine molecule that is equals to kw divided by kb 1 into 10 to the power minus 14 divided by 6 into 10 to the power minus 5 that is equals to 1 .67 into 10 to the power minus 10 now concentration now we can write as x equals to concentration of hydronium ion that is equals to concentration of glycline so this is juter form okay and this is ionized form of glycine and concentration of juter ion form glycine equals to 0 .050 minus x now we can write 1 .67 into 10 to the power minus 10 equals to x squared divided by 0 .0 .0 .0 minus x...

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