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The amino acid histidine has ionizable groups with \(\mathrm{pK}_{\mathrm{a}}\) values of \(1.8,6.0\), and \(9.2\), as shown (His m imidazole group). A bobohemist makes up \(100 \mathrm{~mL}\) of a \(0.10 \mathrm{~m}\) solution of histidine at a pH of \(5.40\). She then adds \(40 \mathrm{~mL}\) of \(0.10 \mathrm{~m}\) HCl. What is the pll of the resulting solution?

Short Answer

Expert verified
The pH of the resulting solution is approximately 6.18.

Step by step solution

01

Identify Ionizable Groups and Their pKa

Histidine has three ionizable groups with the following pKa values: the carboxyl group with \( \text{pK}_\text{a} = 1.8 \), the imidazole group with \( \text{pK}_\text{a} = 6.0 \), and the amino group with \( \text{pK}_\text{a} = 9.2 \). At pH 5.40, the main concern is the transition between the carboxyl and imidazole groups.
02

Determine Initial Protonation States

Given the pH of 5.40, the carboxyl group (\( \text{pK}_\text{a} = 1.8 \)) is largely deprotonated, and the imidazole group (\( \text{pK}_\text{a} = 6.0 \)) is partially protonated. The amino acid initially acts as a buffer close to the pKa of the imidazole group.
03

Calculate Moles of Histidine and HCl

Calculate the initial moles of histidine: \(0.10 \, \text{mol/L} \times 0.10 \, \text{L} = 0.01 \, \text{mol}\). Calculate the moles of HCl added: \(0.10 \, \text{mol/L} \times 0.04 \, \text{L} = 0.004 \, \text{mol}\). The additional moles of HCl will shift the pH of the buffer system.
04

Shift in Buffer System (Henderson-Hasselbalch Equation)

With the addition of HCl, more histidine will be in the protonated form. Use the Henderson-Hasselbalch equation to find the new pH: \[ \text{pH} = \text{pK}_\text{a} + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \]. Let \([\text{base}] = 0.01 - 0.004 = 0.006 \, \text{mol}\) and \([\text{acid}] = 0.004 \, \text{mol}\).
05

Solve for New pH

Using \( \text{pK}_\text{a} = 6.0 \), substitute into the Henderson-Hasselbalch equation: \[ \text{pH} = 6.0 + \log \left( \frac{0.006}{0.004} \right) \]. Thus, \[ \text{pH} = 6.0 + \log(1.5) \approx 6.18 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Amino Acids
Amino acids are the building blocks of proteins. These small molecules possess two key feature groups: a basic amino group (鈥揘H鈧) and an acidic carboxyl group (鈥揅OOH). What makes each amino acid unique is its side chain, which can vary widely in structure and properties. The side chains determine the role they play in living organisms. For instance, histidine, an amino acid discussed in the solution, contains an imidazole group as part of its side chain.
Apart from the basic functionality provided by its amino and carboxyl groups, this imidazole group also contributes to ionization processes. Histidine is particularly notable for its role in enzyme catalysis and as a precursor in the biosynthesis of histamine. It's a key player in the regulation of pH in biological systems due to its unique side chain, making it essential in buffer systems as well.
Buffer Systems
Buffer systems are crucial in biological contexts because they help maintain a stable pH, which is vital for the proper functioning of biological molecules and systems. A buffer consists of a weak acid and its conjugate base. It resists pH changes upon the addition of small amounts of acids or bases.
Histidine, owing to its ionizable groups, acts as an excellent buffer at physiological pH. The presence of a weak acid (such as the protonated form of histidine) and its conjugate base helps in absorbing excess H鈦 or OH鈦 ions.
This buffering action is particularly important in biological systems where enzymes and biochemical reactions are sensitive to pH changes. For example, our blood relies on buffer systems, such as the bicarbonate buffer, to stabilize its pH around 7.4. Understanding buffer systems allows scientists and medical professionals to comprehend how organisms maintain homeostasis.
Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is a useful mathematical tool that allows us to calculate the pH of a buffer solution. It provides a way to estimate the pH when the concentration of an acid and its conjugate base is known. The equation is given by:
\[ \text{pH} = \text{pK}_\text{a} + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \]
This equation is derived from the acid dissociation constant expression and is particularly useful for understanding how buffer systems work. It lets us predict the change in pH when an acid or base is added, making it a powerful tool in both laboratory settings and natural biological processes.
In our exercise, this equation was used to determine the new pH after adding HCl to a histidine solution. By substituting the concentration of the deprotonated and protonated forms of histidine into the equation, we calculate the shift in pH, illustrating practical application in biochemical processes.
pKa Values
In biochemistry, the pK鈧 value is a critical concept as it indicates the strength of an acid. It is the negative logarithm of the acid dissociation constant (K鈧) and signifies the pH at which a particular group is half dissociated (half protonated and half deprotonated).
This value is intrinsic to an acid and its conjugate base; the lower the pK鈧, the stronger the acid. For example, histidine has three ionizable groups with different pK鈧 values 鈥 each corresponding to its carboxyl group, imidazole group, and amino group.
In the given problem, the imidazole group with a pK鈧 of 6.0 is of primary interest. Understanding these values helps predict the behavior of amino acids in different pH environments and guides predictions about how and when the molecules will donate or accept protons, a vital factor in buffer capacity and enzyme activity.

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Most popular questions from this chapter

Calculation of \(\mathrm{p} K_{\mathrm{a}}\) An unknown compound, \(\mathrm{X}\), is thought to have a carboxyl group with a \(\mathrm{p} K_{\mathrm{n}}\) of \(2.0\) and another ionizable group with a \(\mathrm{p} K_{\mathrm{n}}\) between 5 and 8 . When \(75 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{NaOH}\) is added to \(100 \mathrm{~mL}\) of a \(0.1 \mathrm{~m}\) solution of \(\mathrm{X}\) at \(\mathrm{pH} 2.0\), the \(\mathrm{pH}\) increases to \(6.72\). Calculate the \(\mathrm{p} K_{\mathrm{a}}\) of the second ionizable group of \(X\).

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Calculation of Molar Ratios of Conjugate Base to Weak Acid from pll For a weak acid with a pK of \(6.00\), calculate the ratio of conjugate base to acid at a pH of \(5.00\).

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pH and Drug Absorption Asp?rin is a weak acid with a \(p K_{n}\) of \(3.5\) (the ionizable \(H\) is shown in red): Aspirin is absorbed into the blood through the cells lining the stomach and the small intestine. Absorption requires passage through the plasma membrane. The polarity of the molecule determines the absorption rate: charged and highly polar molecules pass slowly, whereas neutral hydrophobic molecules pass rapidly. The \(\mathrm{pH}\) of the stomach contents is about \(1.5\), and the \(\mathrm{pH}\) of the contents of the small intestine is about 6. Rased on this information, is more aspirin absorbed into the bloodstream from the stomach or from the small intestine? Clearly justify your choice

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