The amino acid glycine (H2 N-CH2-COOH) can participate in the following equilibria in water: H2

step 1 Chapter 16, problem 114, chemistry is the central science. So the question gives us two equation

The amino acid glycine (H2 N-CH2-COOH) can participate in...

The amino acid glycine $\left(\mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COOH}\right)$ can participate in the following equilibria in water: $\mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons$ $\quad\quad\quad\quad\quad\quad\mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COO}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \quad K_{\mathrm{a}}=4.3 \times 10^{-3}$ $\mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons$ $\quad\quad\quad\quad\quad\quad^{+} \mathrm{H}_{3} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{OH}^{-} \quad K_{\mathrm{b}}=6.0 \times 10^{-5}$ (a) Use the values of $K_{a}$ and $K_{b}$ to estimate the equilibrium constant for the intramolecular proton transfer to form a zwitterion: $\mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COOH} \rightleftharpoons^{+} \mathrm{H}_{3} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COO}^{-}$ (b) What is the pH of a 0.050 Maqueous solution of glycine? (c) What would be the predominant form of glycine in a solution with pH 13$?$ With pH 1 ?

The amino acid glycine $\left(\mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COOH}\right)$ can participate in the following equilibria in water:
$\mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons$
$\quad\quad\quad\quad\quad\quad\mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COO}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \quad K_{\mathrm{a}}=4.3 \times 10^{-3}$
$\mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons$
$\quad\quad\quad\quad\quad\quad^{+} \mathrm{H}_{3} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{OH}^{-} \quad K_{\mathrm{b}}=6.0 \times 10^{-5}$
(a) Use the values of $K_{a}$ and $K_{b}$ to estimate the equilibrium constant for the intramolecular proton transfer to form a zwitterion:
$\mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COOH} \rightleftharpoons^{+} \mathrm{H}_{3} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{COO}^{-}$
(b) What is the pH of a 0.050 Maqueous solution of glycine?
(c) What would be the predominant form of glycine in a solution with pH 13$?$ With pH 1 ?

Key Concepts

Acid-Base Equilibria

Acid-base equilibria are foundational concepts that describe how acids and bases react with water, donating and accepting protons. These reactions are characterized by equilibrium constants, such as the acid dissociation constant (K?) for proton donation and the base dissociation constant (K_b) for proton acceptance. Understanding these equilibria is essential when analyzing reactions in aqueous solutions where multiple protonation states are possible.

Amphoteric Molecules

Amphoteric molecules are substances that can act as both acids and bases. Their ability to donate or accept protons depends on the environment and pH of the solution. This dual behavior is particularly relevant for molecules like amino acids, which contain functional groups that can undergo protonation or deprotonation under varying conditions.

Zwitterions

Zwitterions are neutral species that possess both positive and negative charges within the same molecule. In amino acids, the intramolecular proton transfer leads to a structure where one functional group is protonated and another is deprotonated, resulting in an overall neutral charge. This form is especially prevalent at the isoelectric point of the molecule.

Equilibrium Constant Estimation using K? and K_b

When a molecule can exist in multiple ionic forms through acid-base reactions, the equilibrium constant for an intramolecular proton transfer can be estimated by relating the individual dissociation constants of its functional groups. This method involves combining the equilibria of proton loss and proton gain, typically yielding a relationship that uses the ratio of K? to K_b.

Isoelectric Point and pH Dependence

The isoelectric point is the pH at which an amphoteric molecule, such as an amino acid, has an overall neutral charge because the positive and negative charges balance. At pH values near the isoelectric point, the zwitterionic form predominates. As the pH deviates significantly from the isoelectric point, one of the ionizable groups will tend to either deprotonate or protonate more fully, leading to a net charged species.

Speciation in Acidic and Basic Environments

The predominant form of an amphoteric molecule depends on the pH of the solution. In a strongly acidic environment, acidic conditions lead to protonation of basic sites, resulting in species with a net positive charge. Conversely, in a strongly basic environment, deprotonation of acidic sites occurs, leading to species with a net negative charge. This speciation governs the behavior and reactivity of such molecules in solution.

Transcript

00:01 Chapter 16, problem 114, chemistry is the central science.

00:08 So the question gives us two equations for glycines for the ka and for the kb when in equilibrium with water.

00:17 And the first question asks us to use the values of ka and kb to estimate the equilibrium constant for the intermolecular protein transfer for the zero ion.

00:30 So to do this, we know that our k equilibrium is going to equal to ka times kb.

00:37 And whenever you plug in our numbers from that is listed above, you do 4 .3 times 10 to the negative 3 times 6 .0 times 10 to the negative 5.

00:48 And when you plug that into your calculator, you get 2 .6 times 10 to the negative 7.

00:54 And so that is going to be your answer for part a.

01:00 So for part b, it's asked us what is the ph of a .05 molar aqueous solution of glycine.

01:08 And so to do you do this, we're going to have to use an ice table and a k -a value to be able to get that answer.

01:14 So the first thing we need to do is write out our equation to we would like get this.

01:17 So we're going to do is we have our h3n -c -h -2 -c -o -0 minus.

01:41 Plus our h2o goes to nh2, sorry guys, goes to h2n, c -2n -c -h -2c -o minus plus.

02:07 H -3o plus.

02:15 So now that we have this, we can now figure out what our equation is going to be, and we can use our ice table from there.

02:31 So we know we have an ice and our initial concentration was 0 .05 since we have water that does not play a role here.

02:41 And also we had a zero starting concentration off for our h3o plus.

02:53 So now what we're going to do is then right in the exchange that's going to happen.

02:59 So we're going to subtract something from the initial concentration, and then we're going to add that to the other side.

03:04 And then we're going to have the equilibrium where we add the two up.

03:08 So 0 .05 molar minus x, x, and then x.

03:14 So the next thing we have to do is actually find our ka value, especially since we have that h3 plus, we know it is a ka and not a kb.

03:22 And to be able to find our ka, we had to use the equation given above that we have to use...

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