Calculate pH of 0.136 M HC5H9O2 and 0.136 M NaC5H9O2
Use this interactive calculator to find the pH of a buffer made from HC5H9O2 and its conjugate base NaC5H9O2. For equal concentrations of weak acid and conjugate base, the Henderson-Hasselbalch equation predicts a pH equal to the acid pKa. The default chemistry values here are set for a 0.136 M acid and 0.136 M sodium salt mixture.
Calculator
Enter the molarity of the weak acid HC5H9O2.
Enter the molarity of the conjugate base salt NaC5H9O2.
Default pKa is a typical literature value used for pentanoic acid calculations.
This is the corresponding weak acid dissociation constant.
Used for display context. This calculator uses Kw = 1.0 × 10^-14 at 25 C.
If both acid and base are present, the buffer equation is used. Equal concentrations give pH = pKa.
With 0.136 M HC5H9O2 and 0.136 M NaC5H9O2 at equal concentrations, the buffer ratio is 1.00, so pH = pKa = 4.82.
Quick Chemistry Insight
- HC5H9O2 is a weak acid and NaC5H9O2 provides its conjugate base C5H9O2-.
- For a buffer, use pH = pKa + log([A-]/[HA]).
- If [A-] = [HA], then log(1) = 0 and pH = pKa.
- For the default values, [A-]/[HA] = 0.136 / 0.136 = 1.000.
- That makes the predicted pH approximately 4.82, assuming pKa = 4.82.
How to calculate the pH of 0.136 M HC5H9O2 and 0.136 M NaC5H9O2
To calculate the pH of a solution containing 0.136 M HC5H9O2 and 0.136 M NaC5H9O2, the most direct approach is to recognize that this is a buffer system. HC5H9O2 is the weak acid, and NaC5H9O2 dissociates in water to supply the conjugate base C5H9O2-. When a weak acid and its conjugate base are both present in appreciable amounts, the Henderson-Hasselbalch equation is the standard tool:
pH = pKa + log([A-]/[HA])
In this expression, [A-] is the concentration of the conjugate base and [HA] is the concentration of the weak acid. For this problem, both concentrations are 0.136 M. That means the ratio is:
[A-]/[HA] = 0.136 / 0.136 = 1.000
Since log(1) = 0, the equation simplifies to:
pH = pKa
If you use a typical pKa value of about 4.82 for HC5H9O2 in this type of calculation, then the pH of the solution is approximately 4.82. That is the central result students and lab workers are usually expected to report for an equimolar weak acid and conjugate base buffer.
Why this mixture is a buffer
A buffer forms when you combine a weak acid with its conjugate base, or a weak base with its conjugate acid. In this case, HC5H9O2 can donate H+, while C5H9O2- can accept H+. Because both species are present, the solution can resist abrupt pH changes when small amounts of acid or base are added. This resistance is strongest when the acid and base concentrations are similar, and it reaches its best balance when the ratio is exactly 1:1.
That is why the mixture of 0.136 M HC5H9O2 and 0.136 M NaC5H9O2 is so straightforward to evaluate. The chemistry is not dominated by just the weak acid, and it is not dominated by just the conjugate base. The system is balanced, which makes the pH land near the pKa. In many textbook and exam settings, this is one of the most important buffer patterns to recognize quickly.
Step by step setup
- Identify HC5H9O2 as the weak acid, HA.
- Identify NaC5H9O2 as the source of the conjugate base, A-.
- Write the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]).
- Insert the concentrations: pH = pKa + log(0.136/0.136).
- Simplify the ratio: 0.136/0.136 = 1.
- Use log(1) = 0.
- Conclude that pH = pKa.
- If pKa = 4.82, report pH = 4.82.
Key data table for the default calculation
| Parameter | Value | Meaning | Effect on pH |
|---|---|---|---|
| HC5H9O2 concentration | 0.136 M | Weak acid concentration, [HA] | Forms the acidic half of the buffer pair |
| NaC5H9O2 concentration | 0.136 M | Conjugate base concentration, [A-] | Forms the basic half of the buffer pair |
| Buffer ratio [A-]/[HA] | 1.000 | Equal acid and base levels | Drives log term to zero |
| Typical pKa used | 4.82 | Acid strength reference value | Becomes the final pH when ratio = 1 |
| Calculated pH | 4.82 | Predicted buffer pH | Final answer for the default setup |
What happens if the concentrations are not equal?
The nice shortcut of pH = pKa only works when the acid and conjugate base concentrations are equal. If they are different, you still use the same equation, but the logarithm term no longer disappears. For example, if the base concentration becomes larger than the acid concentration, the log term is positive, and the pH rises above the pKa. If the acid concentration becomes larger than the base concentration, the log term is negative, and the pH drops below the pKa.
This is one reason the Henderson-Hasselbalch equation is so powerful. It allows you to estimate pH shifts simply from the concentration ratio. In practical buffer design, the ratio determines the final pH, while the total concentration controls buffer capacity, meaning how strongly the solution resists added acid or base.
Comparison table for common ratio changes
| [A-]/[HA] ratio | log([A-]/[HA]) | If pKa = 4.82 | Interpretation |
|---|---|---|---|
| 0.10 | -1.000 | pH = 3.82 | Acid dominates strongly |
| 0.50 | -0.301 | pH = 4.52 | Moderately more acid than base |
| 1.00 | 0.000 | pH = 4.82 | Equal acid and base, ideal midpoint |
| 2.00 | 0.301 | pH = 5.12 | Moderately more base than acid |
| 10.00 | 1.000 | pH = 5.82 | Base dominates strongly |
Why pKa matters so much
The pKa is a compact way to describe the strength of a weak acid. Lower pKa values correspond to stronger acids, while higher pKa values correspond to weaker acids. In an equimolar buffer, the pH equals the pKa because the acid and base contributions are balanced perfectly. This relationship is so fundamental that one of the easiest ways to build a buffer near a target pH is to select a weak acid whose pKa is near that target.
For HC5H9O2 and NaC5H9O2, once you accept the working pKa value being used for the calculation, the pH follows directly. If your instructor, text, or lab manual supplies a slightly different pKa, use that value. The answer may shift a little, but the method remains exactly the same. In other words, the logic does not change, only the constant does.
Can you solve this without Henderson-Hasselbalch?
Yes, but it is usually unnecessary for this kind of problem. You could write the acid dissociation equilibrium, include the initial concentrations of both acid and conjugate base, set up an ICE table, and derive the hydrogen ion concentration from the equilibrium expression. However, for a standard buffer where both components are present in significant amounts, the Henderson-Hasselbalch equation is the accepted simplification. It gives a reliable result quickly and clearly.
The direct equilibrium method becomes more useful when concentrations are extremely low, when ionic strength effects are important, or when the problem specifically asks for an exact equilibrium treatment. For general chemistry and most analytical chemistry examples, the buffer equation is the standard answer.
When the simple buffer equation works best
- Both acid and conjugate base are present in measurable amounts.
- The ratio [A-]/[HA] is within a practical buffer range, often about 0.1 to 10.
- The solution is not so dilute that water autoionization becomes important.
- The problem is designed as a buffer pH question rather than a full activity correction problem.
Common mistakes students make
- Forgetting that NaC5H9O2 contributes the conjugate base, not the acid.
- Using concentration values in the wrong order inside the logarithm.
- Mixing up pKa and Ka without converting properly.
- Trying to treat the solution as only a weak acid solution even though the conjugate base is clearly present.
- Ignoring that equal concentrations force the ratio to 1 and therefore make pH equal to pKa.
A fast way to check your answer is to ask whether the final pH should be acidic, neutral, or basic. Since this system is built around a weak acid buffer, the result should be acidic but not extremely acidic. A value around 4.8 fits that expectation very well.
Buffer capacity and why 0.136 M is useful
The number 0.136 M does not change the pH in an equimolar buffer ratio because the ratio remains 1. Still, it does matter for buffer capacity. Buffer capacity refers to how much added acid or base the solution can absorb before the pH changes substantially. A 0.136 M and 0.136 M buffer has more capacity than a very dilute version of the same ratio. So while the pH may still equal the pKa, the higher total concentration makes the solution more resistant to disturbance.
In laboratory practice, this distinction is important. Two buffers can share the same pH but perform differently when reagents are added. That is why both the ratio and the absolute concentrations matter in real applications.
Authoritative references for pH and buffer chemistry
If you want to verify the chemistry concepts behind this calculation, these sources are useful starting points:
- USGS: pH and Water
- NCBI Bookshelf: Acid Base Physiology and Buffer Concepts
- Chemistry LibreTexts at an .edu-hosted network: Henderson-Hasselbalch Approximation
Final answer summary
For a solution containing 0.136 M HC5H9O2 and 0.136 M NaC5H9O2, the acid and conjugate base concentrations are equal. Therefore, the Henderson-Hasselbalch ratio term is zero, and the pH equals the pKa. With the default pKa value used in this calculator, the result is:
pH = 4.82