What is the Henderson-Hasselbalch Equation?
The Henderson-Hasselbalch equation is an important equation used in biochemistry, chemistry, and pharmacology to calculate the pH of a solution containing a weak acid and its conjugate base, or a weak base and its conjugate acid. The equation is named after its developers, Lawrence J. Henderson and Karl A. Hasselbalch.
The equation is:
pH = pKa + log([A-]/[HA])
where pH is the acidity or basicity of the solution, pKa is the dissociation constant of the weak acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.
The equation is derived from the acid dissociation constant expression (Ka) of a weak acid, which is:
- Ka = [H+][A-]/[HA]
where [H+] is the concentration of hydrogen ions in the solution.
Taking the negative logarithm of both sides of the equation gives:
- -pKa = log(Ka) = log([A-]/[HA]) + log([H+])
Rearranging the equation gives the Henderson-Hasselbalch equation:
- pH = pKa + log([A-]/[HA])
The Henderson-Hasselbalch equation is useful in calculating the pH of a buffer solution, which is a solution that resists changes in pH when small amounts of acid or base are added to it. Buffers are important in biological systems, as they help maintain the pH of blood, saliva, and other bodily fluids within a narrow range, which is necessary for the proper functioning of enzymes and other biochemical processes.
Henderson-Hasselbalch Equation: Derivation
A step-by-step guide for understanding the equation is given below:
Step 1: Write the expression for the acid dissociation constant (Ka)
Ka is the equilibrium constant for the dissociation of an acid in water. It is defined as the ratio of the concentrations of the products to the concentrations of the reactants when the acid is in equilibrium with its conjugate base and hydrogen ions.
The expression for Ka is as follows:
Ka = [H+][A-]/[HA]
where [H+] is the concentration of hydrogen ions, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
Step 2: Take the negative logarithm of both sides of the equation
Taking the negative logarithm of both sides of the equation will simplify the expression and make it easier to work with.
We get:
-log Ka = -log([H+][A-]/[HA])
Step 3: Rearrange the equation
Rearrange the equation to get the logarithm of [A-]/[HA] on one side and the negative logarithm of Ka and pH on the other side:
-log Ka = -log([H+][A-]/[HA])
-log Ka = -log [H+] - log([A-]/[HA])
log([A-]/[HA]) = log Ka - log [H+]
log([A-]/[HA]) = log Ka/[H+]
Step 4: Take the antilogarithm of both sides of the equation
Taking the antilogarithm of both sides of the equation will get rid of the logarithms and give us the Henderson-Hasselbalch equation:
[A-]/[HA] = antilog (pKa - pH)
where pKa = -log Ka
Important Note: By using this equation, one can calculate the pH of a buffer solution and design buffer solutions with specific pH values.
What are the applications of the Henderson-Hasselbalch Equation?
The Henderson-Hasselbalch equation has several applications in chemistry, including:
Buffer Solutions
Buffer solutions are solutions that resist changes in pH when small amounts of acid or base are added. The Henderson-Hasselbalch equation can be used to calculate the pH of a buffer solution, which is determined by the pKa and the concentrations of the acid and its conjugate base. For example, consider a buffer solution containing acetic acid (pKa = 4.76) and its conjugate base acetate. If the concentration of acetic acid is 0.1 M and the concentration of acetate is 0.2 M, then the pH of the buffer can be calculated using the Henderson-Hasselbalch equation:
pH = 4.76 + log([0.2]/[0.1]) = 4.76 + log(2) = 4.76 + 0.30 = 5.06
Thus, the pH of the buffer solution is 5.06, which indicates that it is slightly acidic.
Acid-Base Titration Curves
The Henderson-Hasselbalch equation can also be used to predict the shape of an acid-base titration curve. An acid-base titration curve is a plot of the pH of a solution as a function of the amount of acid or base added to the solution. At the equivalence point of the titration, the moles of acid and base are equal, and the pH of the solution is determined by the pKa of the acid. For example, consider a titration of acetic acid with sodium hydroxide. At the equivalence point, the moles of acetic acid and sodium hydroxide are equal, and the pH of the solution is determined by the pKa of acetic acid (4.76). Using the Henderson-Hasselbalch equation, the pH of the solution can be calculated:
pH = 4.76 + log([A-]/[HA])
At the equivalence point, [A-] = [HA], so:
pH = 4.76 + log(1) = 4.76
Thus, at the equivalence point of the titration, the pH of the solution is 4.76.
Pharmaceutical Science
The Henderson-Hasselbalch equation is used in pharmaceutical science to calculate the pH of drug solutions and to design drug formulations. The pH of a drug solution can affect its solubility, stability, and bioavailability. For example, the pH of the stomach is acidic (pH ~ 2), which can affect the absorption of some drugs. By adjusting the pH of a drug solution, the drug's absorption and bioavailability can be optimized.
What are the limitations of the Henderson-Hasselbalch Equation?
The Henderson-Hasselbalch equation is a widely used tool for calculating the pH of a buffer solution. However, it has certain limitations that must be taken into consideration when using it for practical purposes. In this article, we will discuss the limitations of the Henderson-Hasselbalch equation in detail and provide examples to help you understand them better.
1. Ideal buffer behavior assumption
The Henderson-Hasselbalch equation assumes that the buffer is ideal and behaves perfectly. However, in real-life situations, buffers may not behave ideally due to various factors such as temperature, ionic strength, and the presence of other ions or molecules in the solution. This assumption can result in inaccuracies when using the equation to calculate the pH of a buffer solution.
Example: Suppose you want to calculate the pH of a buffer solution at a different temperature than the one at which it was prepared. The Henderson-Hasselbalch equation may not be accurate in this case since temperature affects the equilibrium constant of the buffer system.
2. Limited to weak acids and bases
The Henderson-Hasselbalch equation only works well for weak acids and bases. Strong acids and bases, which are completely dissociated in solution, cannot be treated as buffers and therefore cannot be analyzed using the Henderson-Hasselbalch equation. In the case of strong acids and bases, the pH of the solution can be directly calculated using the concentration of the acid or base and the dissociation constant.
Example: Hydrochloric acid (HCl) is a strong acid and cannot be analyzed using the Henderson-Hasselbalch equation. The pH of a hydrochloric acid solution can be directly calculated using the concentration of HCl and the dissociation constant.
3. Assumption of equal amounts of acid and conjugate base
The Henderson-Hasselbalch equation assumes that the acid and its conjugate base are present in equal amounts. However, in practice, the concentrations of the acid and its conjugate base may not be equal, which can result in inaccuracies when using the equation to calculate the pH of a buffer solution.
Example: Suppose you have a buffer solution in which the concentration of the acid is much higher than that of the conjugate base. In this case, the Henderson-Hasselbalch equation may not be accurate in predicting the pH of the solution since the concentration ratio is not equal.
Henderson-Hasselbalch Equation: Solved Examples
Question 1: Calculate the pH of a buffer solution that contains 0.1 M acetic acid and 0.2 M sodium acetate. The pKa of acetic acid is 4.76.
Solution:
pH = pKa + log([A-]/[HA])
pH = 4.76 + log(0.2/0.1)
pH = 4.76 + 0.301
pH = 5.06
Therefore, the pH of the buffer solution is 5.06.
Question 2: What is the pH of a buffer solution made by mixing 100 mL of 0.1 M acetic acid and 50 mL of 0.2 M sodium acetate? The pKa of acetic acid is 4.76.
Solution:
First, we need to calculate the moles of acetic acid and sodium acetate:
moles of acetic acid = 0.1 M x 0.1 L = 0.01 moles
moles of sodium acetate = 0.2 M x 0.05 L = 0.01 moles
Next, we can calculate the total volume of the buffer solution:
total volume = 100 mL + 50 mL = 0.15 L
Now we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = 4.76 + log(0.01/0.01)
pH = 4.76
Therefore, the pH of the buffer solution is 4.76.
Question 3: Calculate the pH of a buffer solution made by dissolving 0.1 M NH4Cl and 0.15 M NH3. The pKa of NH4+ is 9.24.
Solution:
pH = pKa + log([A-]/[HA])
pH = 9.24 + log(0.15/0.1)
pH = 9.24 + 0.176
pH = 9.42
Therefore, the pH of the buffer solution is 9.42.
Question 4: What is the pH of a buffer solution made by mixing 50 mL of 0.1 M NaH2PO4 and 50 mL of 0.1 M Na2HPO4? The pKa of HPO42- is 7.21.
Solution:
First, we need to calculate the moles of NaH2PO4 and Na2HPO4:
moles of NaH2PO4 = 0.1 M x 0.05 L = 0.005 moles
moles of Na2HPO4 = 0.1 M x 0.05 L = 0.005 moles
Next, we can calculate the total volume of the buffer solution:
total volume = 50 mL + 50 mL = 0.1 L
Now we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = 7.21 + log(0.005/0.005)
pH = 7.21
Therefore, the pH of the buffer solution is 7.21.
Question 5: Calculate the pH of a buffer solution made by dissolving 0.2 M lactic acid and 0.1 M sodium lactate. The pKa of lactic acid is 3.86.
Solution:
pH = pKa + log([A-]/[HA])
pH = 3.86 + log(0.1/0.2)
pH = 3.86 + (-0.301)
pH = 3.56
Therefore, the pH of the buffer solution is 3.56.
Question 6: What is the pH of a buffer solution made by mixing 25 mL of 0.2 M HNO2 and 75 mL of 0.1 M NaNO2? The pKa of HNO2 is 3.35.
Solution:
First, we need to calculate the moles of HNO2 and NaNO2:
moles of HNO2 = 0.2 M x 0.025 L = 0.005 moles
moles of NaNO2 = 0.1 M x 0.075 L = 0.0075 moles
Next, we can calculate the total volume of the buffer solution:
total volume = 25 mL + 75 mL = 0.1 L
Now we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = 3.35 + log(0.0075/0.005)
pH = 3.35 + 0.079
pH = 3.43
Therefore, the pH of the buffer solution is 3.43
Question 7: Calculate the pH of a buffer solution made by dissolving 0.15 M formic acid and 0.1 M sodium formate. The pKa of formic acid is 3.75.
Solution:
pH = pKa + log([A-]/[HA])
pH = 3.75 + log(0.1/0.15)
pH = 3.75 + (-0.176)
pH = 3.57
Therefore, the pH of the buffer solution is 3.57.