Henderson-Hasselbalch Equation: Videos & Practice Problems

Understanding the pH of a buffer solution is crucial in chemistry, and the Henderson-Hasselbalch equation provides a simplified method for calculating it. This equation is particularly useful when dealing with a buffer composed of a conjugate acid-base pair. The pH can be determined using the equation:

pH=pK_a+log([conjugate base]/[weak acid])

if the acid dissociation constant (K_a) is given, or:

pH=pK_b+log([conjugate acid]/[weak base])

if the base dissociation constant (K_b) is provided. The brackets denote the molarity or moles of the components. An ideal buffer has equal concentrations of weak acid and its conjugate base, or weak base and its conjugate acid, which simplifies the equation to:

pH=pK_a

or:

pH=pK_b Buffers are most effective within a pH range of pKa ± 1, where they best resist changes in pH. This knowledge is key in preparing buffer solutions that maintain a stable pH in various chemical applications.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a valuable tool for calculating the pH of a buffer solution, specifically those composed of a conjugate acid-base pair, without the need for an ICE chart. This equation can be expressed in two forms, depending on whether the acid dissociation constant (K_a) or the base dissociation constant (K_b) is provided.

When given K_a, the equation is:

pH = pK_a + log( [A^-] / [HA] )

In this formula, [A^-] represents the concentration of the conjugate base, and [HA] represents the concentration of the weak acid.

If K_b is provided, the equation changes to:

pH = pK_b + log( [HA] / [B] )

Here, [HA] is the concentration of the conjugate acid, and [B] is the concentration of the weak base.

It is important to note that the brackets in these equations indicate concentration, which can be expressed in terms of molarity (moles per liter) or simply in moles. Remember that moles can be calculated using the formula:

moles = liters × molarity

This relationship is crucial when applying the Henderson-Hasselbalch equations to ensure accurate calculations of pH in buffer solutions.

Buffer pH Calculation Example

To calculate the pH of a buffer solution containing 2.0 M nitrous acid (HNO₂) and 1.48 M lithium nitrite (LiNO₂), we can utilize the Henderson-Hasselbalch equation, which is particularly useful for buffer solutions. The equation is expressed as:

pH = pKa + log₁₀([A^-]/[HA])

In this case, HNO₂ is the weak acid (HA), and LiNO₂ provides the conjugate base (A^-), which is nitrite ion (NO₂^-). Given that the acid dissociation constant (K_a) for nitrous acid is 4.6 × 10^-4, we first need to calculate the pK_a:

pK_a = -log₁₀(K_a) = -log₁₀(4.6 × 10^-4)

Calculating this gives us:

pK_a ≈ 3.34

Next, we substitute the values into the Henderson-Hasselbalch equation:

pH = 3.34 + log₁₀(1.48/2.0)

Calculating the log term:

log₁₀(1.48/2.0) ≈ log₁₀(0.74) ≈ -0.128

Now, we can find the pH:

pH ≈ 3.34 - 0.128 ≈ 3.21

Thus, the pH of the buffer solution is approximately 3.21. This calculation demonstrates how the combination of a weak acid and its conjugate base can effectively resist changes in pH, a key characteristic of buffer solutions.

The K_b of C₆H₅NH₂ (aniline) is 3.9 × 10⁻¹⁰. Determine pH of a buffer solution made up of 500 mL of 1.4 M C₆H₅NH₂ and 230 mL of 2.3 M C₆H₅NH₃⁺.

• A. 4.81
• B. 9.62
• C. 9.29
• D. 4.71

Answer: (D)

Determine the buffer component concentration ratio (CB/WA) for a buffer with a pH of 4.7. K_a of boric acid (H₃BO₃) is 5.4 × 10⁻¹⁰.

• A. 4.568 : 1
• B. 2.706 × 10⁻⁵ : 1
• C. 1 : 4.568
• D. 1 : 2.706 × 10⁻⁵

Answer: (B)

Calculate mass of NaN₃ that needs be added to 1.8 L of 0.35 M HN₃ in order to make a buffer with a pH of 6.5. K_a of hydrazoic acid is 1.9 × 10⁻⁵.

• A. 1.4 g NaN₃
• B. 1.0 × 10⁻² g NaN₃
• C. 2.5 × 10³ g NaN₃
• D. 3.55 g NaN₃

Answer: (C)

Buffer Range Calculation

Buffers play a crucial role in maintaining pH stability in various chemical and biological systems. Their effectiveness is confined to a specific pH range, which can be calculated using the relationship: pH = pK_a ± 1. This indicates that a buffer is most effective at resisting pH changes within one unit above or below its pK_a.

For a buffer to function optimally, the concentrations of the weak acid and its conjugate base must be equal. This condition ensures that the pH of the buffer is equal to the pK_a of the weak acid, providing the best resistance to pH fluctuations. The Henderson-Hasselbalch equation is instrumental in this context, expressed as:

pH = pK_a + log₁₀([A^-]/[HA])

Here, [A^-] represents the concentration of the conjugate base, and [HA] is the concentration of the weak acid. Alternatively, if the base dissociation constant (K_B) is provided, the equation can be rearranged to:

pH = pK_B + log₁₀([HA]/[B])

In a scenario where the concentrations of the weak acid and conjugate base are equal, such as [HA] = 0.40 M and [A^-] = 0.40 M, the equation simplifies significantly. The log of 1 (since 0.40/0.40 = 1) equals 0, leading to:

pH = pK_a + 0

Thus, when the concentrations of the conjugate base and weak acid are equal, the buffer's pH directly corresponds to the pK_a. This principle also applies when considering the conjugate acid and weak base, reinforcing that an ideal buffer is achieved when their concentrations are balanced.

In summary, the effectiveness of a buffer is maximized within the pH range of pH = pK_a ± 1, and it is most effective when the concentrations of the acid and base components are equal, simplifying the Henderson-Hasselbalch equation to pH = pK_a or pH = pK_B, depending on the context.

Buffering Range Example

To determine the buffering range of a solution containing lactic acid and its conjugate base, sodium lactate, we first need to calculate the pKa of lactic acid. The acid dissociation constant (Ka) for lactic acid is given as \(1.4 \times 10^{-4}\). The pKa can be calculated using the formula:

\( \text{pKa} = -\log(\text{Ka}) \)

Substituting the value of Ka, we find:

\( \text{pKa} = -\log(1.4 \times 10^{-4}) \approx 3.85 \)

The buffering range, which indicates the pH range where the buffer is most effective, is determined by the equation:

\( \text{pH} = \text{pKa} \pm 1 \)

Applying this to our calculated pKa, we have:

\( \text{pH} = 3.85 \pm 1 \)

This results in a buffering range of:

\( 3.85 - 1 \) to \( 3.85 + 1 \), or \( 2.85 \, \text{pH} \) to \( 4.85 \, \text{pH} \).

Thus, the effective buffering range for this solution is between 2.85 pH and 4.85 pH, where the buffer will maintain a relatively stable pH in the presence of added acids or bases.

Which of the following weak acid-conjugate base combinations would result in an ideal buffer solution with a pH of 9.4?
a) formic acid (HCHO₂) and sodium formate (K_a = 1.8 x 10^-4)
b) benzoic acid (HC₇H₅O₂) and potassium benzoate (K_a = 6.5 x 10^-5)
c) hydrocyanic acid (HCN) and lithium cyanide (K_a = 4.9 x 10^-10)
d) iodic acid (HIO₃) and sodium iodate (K_a = 1.7 x 10^-1)

• A. formic acid (HCHO₂) and sodium formate (K_a = 1.8 x 10^-4)
• B. benzoic acid (HC₇H₅O₂) and potassium benzoate (K_a = 6.5 x 10^-5)
• C. hydrocyanic acid (HCN) and lithium cyanide (K_a = 4.9 x 10^-10)
• D. iodic acid (HIO₃) and sodium iodate (K_a = 1.7 x 10^-4)

Answer: (C)

Here’s what students ask on this topic:

What is the Henderson-Hasselbalch equation?

The Henderson-Hasselbalch equation is a formula that relates the pH (a measure of acidity) of a solution to the pKa (the acid dissociation constant) of the acid and the ratio of the concentrations of its conjugate base and acid forms. It is expressed as:

$\mathrm{pH}=\mathrm{pKa}+\log \left(\frac{[A-]}{\left[\mathrm{HA}\right]}\right)$

In this equation, [A^-] represents the concentration of the conjugate base, and [HA] represents the concentration of the acid. The pKa is a constant that is specific to each acid and indicates the pH at which the acid is half dissociated. This equation is particularly useful in the field of biochemistry for understanding buffer systems, which are solutions that resist changes in pH when small amounts of acid or base are added. It helps predict the pH of a buffer solution and design buffer systems with desired pH levels.

How do you use the Henderson-Hasselbalch equation?

The Henderson-Hasselbalch equation is used to estimate the pH of a buffer solution. The formula is:

$\mathrm{pH}=\mathrm{pK}{a}_{}+\log \left(\frac{{\mathrm{[A]}}_{-}}{{\mathrm{[HA]}}_{}}\right)$

Here, pH is the measure of the acidity or basicity of the solution, pK_a is the acid dissociation constant (a measure of the strength of the acid), [A^-] is the concentration of the base form (the conjugate base of the acid), and [HA] is the concentration of the acid form.

To use the equation, follow these steps:

1. Determine the pK_a of the weak acid component of the buffer system. This value is often provided in tables or can be calculated from the K_a value.
2. Measure or calculate the molar concentrations of the conjugate base ([A^-]) and the acid ([HA]) in the solution.
3. Plug these values into the Henderson-Hasselbalch equation.
4. Calculate the log of the concentration ratio ([A^-]/[HA]).
5. Add this log value to the pK_a to find the pH of the buffer solution.

This equation is particularly useful when preparing buffer solutions with a specific target pH or when analyzing the pH changes in buffer systems during tit

According to the Henderson-Hasselbalch equation, when is the pH equal to the pKa?

The Henderson-Hasselbalch equation is a mathematical relationship used to describe the pH as a measure of acidity in biological and chemical systems. The equation is as follows:

$\mathrm{pH}=\mathrm{pK_a}+\log \left(\frac{[A^-]}{\left[\mathrm{HA}\right]}\right)$

Here, pH is the measure of acidity, pK_a is the acid dissociation constant (a measure of the strength of an acid), [A^-] is the concentration of the base form (conjugate base), and [HA] is the concentration of the acid form.

According to this equation, the pH of a solution is equal to the pK_a when the concentrations of the acid form ([HA]) and its conjugate base ([A^-]) are equal. In other words, when [A^-] = [HA], the log([A^-]/[HA]) becomes log(1), which is equal to 0. Therefore, under these conditions, the pH is directly equal to the pK_a. This particular point is important because it represents the point at which the acid is half dissociated, and the solution acts as a buffer, resisting changes in pH.