Henderson-Hasselbalch Equation - Online Tutor, Practice Problems & Exam Prep

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

Now, the Henderson Hasselbeck equation allows you to calculate the pH of a buffer without having to use an ice chart. Here we're going to say it only applies to buffers composed of a conjugate acid base pair. We can look at the Henderson Hasselbeck equation as two different formulas. The formula that you use is based on if they give you the K_A or K_B of your buffer solution.

If they give you the K_A of your buffer solution, you can say that

pH = pK A + log ( conjugate base weak acid )

Now if they give you K_B then you could use this formula.

pH = pK B + log ( conjugate acid weak base )

Now when it comes to the sign the brackets, we know that it's tensamine, molarity or concentration for the Henderson Hasselbeck equations.

It could also be used for moles. OK, so just remember the unit that can go within these brackets can be either molarity or moles. Remember that moles itself equals liters times molarity, so keep that in mind when you utilize the Henderson Hasselbeck equations.

Henderson-Hasselbalch Equation Example

Calculate the pH of a solution containing 2.0 molar of nitrous acid and 1.48 molar of lithium nitrate. Here we're told that the K_A of our weak acid is 4.6×10-4. Now here, because the K value is less than one, we know that nitrous acid is a weak acid.

Lithium nitrite looks similar to nitrous acid except it has one less H⁺ ion. Because of this, this has to be the conjugate base. So we have a weak acid. We have a conjugate base. This is the pairing that helps to make a buffer. So we're going to use the Henderson-Hasselbalch equation to calculate the pH of this buffer.

Here we're going to say pH equals. Now because they give us K_A, we can say pH equals pK_A plus log of conjugate base over weak acid. Here pK_A remember is just negative log of K_A. So now of 4.6×10-4 plus log of conjugate base amount which is going to be 1.48 molar divided by weak acid amount which is 2.0 molar.

When we plug this in, we get 3.21 as the pH for this buffer solution.

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

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

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

Calculating Buffer Range

Now, when it comes to calculating buffer range, first it's important to remember that buffers are effective at a specific pH range. To figure that out, we see that pH equals PKA ±1. That's the range, the pH range in which a buffer will act most effectively at resisting a sharp change in pH. Now recall that a buffer is ideal though when the concentration of weak acid is equal to the concentration of conjugate base. Or you could say when the concentration of weak base is equal to the concentration of conjugate acid, same thing.

Now this is because the pH of the buffer will be equal to the PKA of the weak acid and this will resist a pH change the best. Now here if we add an example we have pH=PKA+log(conjugate base)weak acid. Remember, when it comes to the Henderson-Hasselbalch equation, we can observe it two different ways. If they're giving you ka, you can use the top version where pH=PKA+log(conjugate base)weak acid. The bottom one you use if they give you KB here, pH=PKB+log(conjugate acid)weak base.

Well, going back to this, we can see that both the conjugate base and weak acid amounts are equal to one another. So .40 / .40 is just equal to 1. And remember, we're dealing with log of one, which this is. If you punch that into your calculator, log of one gives you 0. So the equation simplifies to pH=PKA+0, and if you drop off the 0 then pH=PKA. So when the amount of conjugate base and weak acid are equal to each other, or when the amount of conjugate acid is equal to weak base, we have an ideal buffer and then the equation, the Henderson-Hasselbalch equation, simplifies down to pH=PKA or pH=PKB depending on which one you're using.

Right? So keep that in mind. The effectiveness of a buffer happens best within a pH range of pH=PKA±1 and it's most ideal when the concentrations of these species are equal to one another.

Henderson-Hasselbalch Equation Example

Here in this example it says determine the buffering range of a solution containing lactic acid which has Ka=1.4×10-4 and sodium lactate, it's conjugate base. Now here we're looking for a buffering range. Remember that when it comes to your buffer range, the pH range, so the range in which the buffer works most effectively is pH=pKa±1.

Remember that pKa=−log(Ka). So just take the negative log of this Ka value. When we do that, we have 3.85, so that means our buffering range or pH range for the effectiveness of a buffer is equal to 3.85±1.

That would mean our range is 3.85−1 to 3.85+1, which will translate to a range of 2.85 pH to 4.85 pH. OK, this would be our buffering range in which this particular buffer will be most effective.

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)

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.