VIPSolutionsAct like a helpful tutor and exlain me : Give step-by-step solution with explanation and final answer:1. (613.3) Relationship between the Tiraton Curve and the Acid-Base Properties of Giycing: A researcher titrated a 100 mL solution of 0.1 M glycine at pH 1.72 with 2 M NaOH solution. She then monitored the pH and plotted the rests in the graph shown, The key points i the tration are designated 10 V. For each o th following statements, anti ths appropriate key point in th fiztion.. A. The pis equal tothe pK 2 of the carbon! group 5.The pH is squal to ths pA. of the protonated LE — P amino group. H C-Th predominant gine species ol seo ] “HsN— CH. Coo. H 0. The predominant icine species is N ! “HIN—CH—C00-. i . Glycine exists as 50:50 murs of sor i “HN—CH2— COOH and + HN ~—CH:—coo= P1 © i F. The average net charge of glycine is + ¥ (half. t G. Half of the amino groups are ionized. “ { H. The average net charge of glycine is 0. 234 i 1. The average net charge of glycine is ~1. 2 J. This isthe isoetscric point for cine. K. Glycine exibit ts maximum buffering o capacity in these regions. Note that statement (K) 0 © © 0 applies to more than one key point in the titration OH” (ecuivalents)
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Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:
1. (613.3) Relationship between the Tiraton Curve and the Acid-Base Properties of Giycing: A
researcher titrated a 100 mL solution of 0.1 M glycine at pH 1.72 with 2 M NaOH solution. She then
monitored the pH and plotted the rests in the graph shown, The key points i the tration are
designated 10 V. For each o th following statements, anti ths appropriate key point in th fiztion..
A. The pis equal tothe pK 2 of the carbon! group
5.The pH is squal to ths pA. of the protonated LE — P
amino group. H
C-Th predominant gine species ol seo ]
“HsN— CH. Coo. H
0. The predominant icine species is N !
“HIN—CH—C00-. i
. Glycine exists as 50:50 murs of sor i
“HN—CH2— COOH and + HN ~—CH:—coo= P1 © i
F. The average net charge of glycine is + ¥ (half. t
G. Half of the amino groups are ionized. “ {
H. The average net charge of glycine is 0. 234 i
1. The average net charge of glycine is ~1. 2
J. This isthe isoetscric point for cine.
K. Glycine exibit ts maximum buffering o
capacity in these regions. Note that statement (K) 0 © © 0
applies to more than one key point in the titration OH” (ecuivalents)
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Let's tackle each statement step by step, referencing the titration curve of glycine:
---
## **Understanding the Curve**
- **Glycine** has two ionizable groups:
- Carboxyl group (\(-COOH\)), pK₁ ≈ 2.34
- Amino group (\(-NH_3^+\)), pK₂ ≈ 9.60
- **Titration Curve Key Points:**
- **I:** Start (all in fully protonated form)
- **II:** pH = pK₁ (carboxyl group half-neutralized)
- **III:** Isoelectric point (zwitterion dominant)
- **IV:** pH = pK₂ (amino group half-neutralized)
- **V:** End (all in deprotonated form)
---
## **Step-by-Step Solution**
### **A. The pH is equal to the pKₐ of the carboxyl group.**
- **pK₁ = 2.34**
- **Corresponds to point II.**
### **B. The pH is equal to the pKₐ of the protonated amino group.**
- **pK₂ = 9.60**
- **Corresponds to point IV.**
### **C. The predominant glycine species is \( \text{H}_3\text{N}^+–\text{CH}_2–\text{COOH} \).**
- This is the fully protonated form (low pH, before first equivalence).
- **Corresponds to point I.**
### **D. The predominant glycine species is \( \text{H}_3\text{N}^+–\text{CH}_2–\text{COO}^- \).**
- This is the zwitterion (between pK₁ and pK₂, at isoelectric point).
- **Corresponds to point III.**
### **E. Glycine exists as a 50:50 mixture of \( \text{H}_3\text{N}^+–\text{CH}_2–\text{COOH} \) and \( \text{H}_3\text{N}^+–\text{CH}_2–\text{COO}^- \).**
- This occurs at pH = pK₁, where half the carboxyl groups are deprotonated.
- **Corresponds to point II.**
### **F. The average net charge of glycine is ½ (half).**
- At point II, half the molecules have a net charge of +1, half have , so the average is **+.5**.
- **Corresponds to point II.**
### **G. Half of the amino groups are ionized.**
- This occurs at pH = pK₂, where half the amino groups have lost their proton.
- **Corresponds to point IV.**
### **H. The average net charge of glycine is .**
- This is at the isoelectric point (zwitterion form).
- **Corresponds to point III.**
### **I. The average net charge of glycine is –1.**
- At high pH (all groups deprotonated).
- **Corresponds to point V.**
### **J. This is the isoelectric point for glycine.**
- Where glycine has no net charge (zwitterion).
- **Corresponds to point III.**
### **K. Glycine exhibits its maximum buffering capacity in these regions.**
- Maximum buffering occurs near pK₁ and pK₂ (where slope is flattest).
- **Corresponds to points II and IV.**
---
## **Summary Table**
| Statement | Key Point on Curve |
|-----------|-------------------|
| A | II |
| B | IV |
| C | I |
| D | III |
| E | II |
| F | II |
| G | IV |
| H | III |
| I | V |
| J | III |
| K | II and IV |
---
### **Final Answers**
- **A:** II
- **B:** IV
- **C:** I
- **D:** III
- **E:** II
- **F:** II
- **G:** IV
- **H:** III
- **I:** V
- **J:** III
- **K:** II and IV
Let me know if you have questions on any specific step!
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# Detailed Explanation of Glycine Titration Curve and Its Key Points
Glycine is an amino acid with two ionizable groups: the carboxyl group (\(-COOH\)) and the amino group (\(-NH_3^+\)). Understanding the relationship between its structure and the pH during titration is crucial for interpreting the titration curve.
---
## **Understanding Glycine's Ionizable Groups**
### **Ionization Constants**
- **pK₁** (Carboxyl group): Approximately 2.34
- **pK₂** (Amino group): Approximately 9.60
---
## **Titration Curve Key Points**
The titration curve for glycine can be divided into several key points:
1. **Point I**: Fully protonated form (\(\text{H}_3\text{N}^+–\text{CH}_2–\text{COOH}\))
2. **Point II**: pH = pK₁ (Half of the carboxyl groups are deprotonated)
3. **Point III**: Isoelectric point (Zwitterion form dominant)
4. **Point IV**: pH = pK₂ (Half of the amino groups are deprotonated)
5. **Point V**: Fully deprotonated form (\(\text{H}_2\text{N}–\text{CH}_2–\text{COO}^-\))
---
## **Step-by-Step Analysis of Statements**
### **A. The pH is equal to the pKₐ of the carboxyl group.**
- **Explanation**: At point II, glycine is at its pK₁, meaning that the carboxyl group is half neutralized.
- **Key Point**: **II**
### **B. The pH is equal to the pKₐ of the protonated amino group.**
- **Explanation**: At point IV, the pH equals pK₂, indicating that half of the amino groups have lost their protons.
- **Key Point**: **IV**
### **C. The predominant glycine species is \( \text{H}_3\text{N}^+–\text{CH}_2–\text{COOH} \).**
- **Explanation**: This occurs at low pH, where glycine exists mainly in its fully protonated form.
- **Key Point**: **I**
### **D. The predominant glycine species is \( \text{H}_3\text{N}^+–\text{CH}_2–\text{COO}^- \).**
- **Explanation**: The zwitterionic form dominates between pK₁ and pK₂.
- **Key Point**: **III**
### **E. Glycine exists as a 50:50 mixture of \( \text{H}_3\text{N}^+–\text{CH}_2–\text{COOH} \) and \( \text{H}_3\text{N}^+–\text{CH}_2–\text{COO}^- \).**
- **Explanation**: This occurs at pH = pK₁, where half of the carboxyl groups are deprotonated.
- **Key Point**: **II**
### **F. The average net charge of glycine is +½ (half).**
- **Explanation**: At point II, the average charge is +½, corresponding to the balance of charged species.
- **Key Point**: **II**
### **G. Half of the amino groups are ionized.**
- **Explanation**: This is true at pH = pK₂, where half of the amino groups have lost protons.
- **Key Point**: **IV**
### **H. The average net charge of glycine is 0.**
- **Explanation**: This is at the isoelectric point (point III), where glycine has no net charge.
- **Key Point**: **III**
### **I. The average net charge of glycine is -1.**
- **Explanation**: At high pH (point V), all groups are deprotonated, resulting in a net charge of -1.
- **Key Point**: **V**
### **J. This is the isoelectric point for glycine.**
- **Explanation**: At point III, glycine has no net charge, which defines the isoelectric point.
- **Key Point**: **III**
### **K. Glycine exhibits its maximum buffering capacity in these regions.**
- **Explanation**: Maximum buffering occurs near pK₁ and pK₂, where the pH is stable despite added acids or bases.
- **Key Points**: **II and IV**
---
## **Summary Table of Statements and Corresponding Key Points**
| Statement | Key Point on Curve |
|-----------|-------------------|
| A | II |
| B | IV |
| C | I |
| D | III |
| E | II |
| F | II |
| G | IV |
| H | III |
| I | V |
| J | III |
| K | II and IV |
---
## **Final Answers**
- **A:** II
- **B:** IV
- **C:** I
- **D:** III
- **E:** II
- **F:** II
- **G:** IV
- **H:** III
- **I:** V
- **J:** III
- **K:** II and IV
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