Henderson-Hasselbalch Equation: Videos & Practice Problems

The Henderson-Hasselbalch equation is essential for calculating the pH of buffer solutions, particularly those composed of a conjugate acid-base pair. When given the acid dissociation constant (K_a), the equation is pH = pK_a + log([conjugate base]/[weak acid]). Conversely, if provided with the base dissociation constant (K_b), use pH = pK_b + log([conjugate acid]/[weak base]). Buffers are most effective within a pH range of pK_a ± 1, ideally when the concentrations of the weak acid and conjugate base are equal.

Henderson-Hasselbalch Equation

Now, the Henderson-Hasselbalch 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-Hasselbalch equation as two different formulas. The formula that you use is based on if they give you the Ka or Kb of your buffer solution. If they give you the Ka of your buffer solution, you can say that pH equals pKa plus logconjugate baseweak acid. Now, if they give you Kb, then you could use this formula: pH equals pKb plus logconjugate acidweak base. Now, when it comes to the sign, the brackets, we know that it tends to mean molarity or concentration. For the Henderson-Hasselbalch equations, it could also be used for moles. Okay. So, just remember that the units 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-Hasselbalch 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_a value is less than 1, we know that nitrous acid is a weak acid. Lithium nitrate 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, and 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. 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, negative log 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. 4.38
• D. 9.29

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)

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 say that pH=pKa±1. That's the range, the pH range, in which a buffer will act most effectively in resisting a sharp change in pH.

Now recall that a buffer is ideal 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. 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 had an example, we have pH=pKa+log⁡(conjugatebaseweakacid). Remember, when it comes to the Henderson-Hasselbalch equation we can observe it in 2 different ways. If they're giving you Ka, you can use the top version where pH=pKa+log⁡(conjugatebase/weakacid). The bottom one you use if they give you Kb. Here pH=pKb+log⁡(conjugateacid/weakbase). Well, going back to this, we can see that both the conjugate base and weak acid amounts are equal to one another. So 0.40 divided by 0.40 is just equal to 1. And remember, if we're dealing with log of 1, which this is, you punch that into your calculator, log of 1 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 Henderson-Hasselbalch equations simplify down to pH=pKa or pH=pKb, depending on which one you're using.

Alright? 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 the 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 a Ka1.4 times 10-4, and sodium lactate, its 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=-logKa. 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 3.85-1to3.85+1, which will translate to a range of 2.85pHto4.85pH. Okay. 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)

Answer: (C)

Here’s what students ask on this topic:

What is the Henderson-Hasselbalch equation and how is it used?

The Henderson-Hasselbalch equation is used to calculate the pH of buffer solutions, particularly those composed of a conjugate acid-base pair. The equation is:

$\mathrm{pH}=\mathrm{pK}{a}_{}+log\left(\frac{\mathrm{[conjugate\; base]}}{\mathrm{[weak\; acid]}}\right)$

Alternatively, if given the base dissociation constant (K_b), the equation is:

$\mathrm{pH}=\mathrm{pK}{b}_{}+log\left(\frac{\mathrm{[conjugate\; acid]}}{\mathrm{[weak\; base]}}\right)$

This equation helps in determining the pH without using an ICE chart, making it a valuable tool in chemistry.

How do you calculate the pH of a buffer solution using the Henderson-Hasselbalch equation?

To calculate the pH of a buffer solution using the Henderson-Hasselbalch equation, follow these steps:

1. Identify whether you have the acid dissociation constant (K_a) or the base dissociation constant (K_b).

2. Use the appropriate equation:

If given K_a: $\mathrm{pH}=\mathrm{pK}{a}_{}+log\left(\frac{\mathrm{[conjugate\; base]}}{\mathrm{[weak\; acid]}}\right)$

If given K_b: $\mathrm{pH}=\mathrm{pK}{b}_{}+log\left(\frac{\mathrm{[conjugate\; acid]}}{\mathrm{[weak\; base]}}\right)$

3. Plug in the concentrations (or moles) of the conjugate base and weak acid (or conjugate acid and weak base).

4. Calculate the pH.

What is the effective pH range of a buffer according to the Henderson-Hasselbalch equation?

The effective pH range of a buffer according to the Henderson-Hasselbalch equation is pK_a ± 1. This means that the buffer will effectively resist changes in pH within one pH unit above and below the pK_a of the weak acid. For example, if the pK_a is 4.75, the buffer will be effective in the pH range of 3.75 to 5.75. This range ensures that the buffer can neutralize added acids or bases, maintaining a relatively stable pH.

Why is a buffer most effective when the concentrations of the weak acid and conjugate base are equal?

A buffer is most effective when the concentrations of the weak acid and conjugate base are equal because, at this point, the pH of the buffer is equal to the pK_a of the weak acid. This is derived from the Henderson-Hasselbalch equation:

$\mathrm{pH}=\mathrm{pK}{a}_{}+log\left(\frac{\mathrm{[conjugate\; base]}}{\mathrm{[weak\; acid]}}\right)$

When [conjugate base] = [weak acid], the log term becomes log(1), which is 0. Thus, pH = pK_a. At this point, the buffer has the maximum capacity to neutralize added acids or bases, making it most effective in resisting pH changes.

Can the Henderson-Hasselbalch equation be used with moles instead of molarity?

Yes, the Henderson-Hasselbalch equation can be used with moles instead of molarity. The equation:

$\mathrm{pH}=\mathrm{pK}{a}_{}+log\left(\frac{\mathrm{[conjugate\; base]}}{\mathrm{[weak\; acid]}}\right)$

or

$\mathrm{pH}=\mathrm{pK}{b}_{}+log\left(\frac{\mathrm{[conjugate\; acid]}}{\mathrm{[weak\; base]}}\right)$

can be applied using the concentrations in moles, as long as the volumes of the solutions are the same. This is because the ratio of moles will be the same as the ratio of molarities, simplifying the calculation process.