Calculate the pH of a 5.7 m Solution of Aniline
Use this premium weak-base calculator to estimate the pH, pOH, hydroxide concentration, degree of ionization, and molarity of an aniline solution. It is preloaded for a 5.7 m aniline solution and uses the weak-base equilibrium relation for accurate chemistry workflow support.
Aniline pH Calculator
Click the button to compute the pH of the 5.7 m aniline solution using weak-base equilibrium.
Visual Concentration Breakdown
The chart compares the estimated formal concentration of aniline, the hydroxide concentration produced by base ionization, and the remaining un-ionized base concentration.
How to Calculate the pH of a 5.7 m Solution of Aniline
Calculating the pH of a 5.7 m solution of aniline is a classic weak-base equilibrium problem, but it becomes more interesting because the concentration is given in molality rather than molarity. In chemistry, this matters. Molality, written as m, means moles of solute per kilogram of solvent. Molarity, written as M, means moles of solute per liter of solution. Since acid-base equilibrium expressions are usually applied using concentration in terms of volume, molarity is generally the form we want for direct pH work.
Aniline, with formula C6H5NH2, is an aromatic amine and a weak base. It does not dissociate completely in water like sodium hydroxide. Instead, it accepts a proton from water only to a limited extent:
This equilibrium is governed by the base dissociation constant Kb. For aniline at room temperature, a commonly used value is approximately 4.3 × 10^-10. That very small Kb immediately tells you aniline is a weak base and that only a tiny amount of hydroxide forms relative to the total amount of dissolved aniline.
Step 1: Understand What 5.7 m Means
A 5.7 m solution contains 5.7 moles of aniline per 1 kilogram of solvent. If the solvent is water, the setup is:
- Molality of aniline = 5.7 mol/kg solvent
- Mass of solvent = 1000 g water
- Moles of aniline = 5.7 mol
By itself, this does not directly give liters of solution, so it does not directly give molarity. To estimate pH rigorously from the usual weak-base equation, we often convert to molarity using the solution density and the molar mass of aniline.
Step 2: Convert 5.7 m to an Approximate Molarity
The molar mass of aniline is about 93.13 g/mol. If you have 5.7 mol aniline, the mass of aniline is:
Then the total mass of solution is approximately:
If we assume a solution density of about 1.02 g/mL, the approximate total volume is:
So the estimated molarity becomes:
This is an important point. A 5.7 m solution is not automatically a 5.7 M solution. Depending on density, the actual molarity can be notably different.
Step 3: Set Up the Weak Base Equilibrium
Let the initial molarity of aniline be C. The equilibrium is:
Using an ICE framework:
- Initial: [C6H5NH2] = C, [C6H5NH3+] = 0, [OH-] = 0
- Change: -x, +x, +x
- Equilibrium: [C6H5NH2] = C – x, [C6H5NH3+] = x, [OH-] = x
The base dissociation expression is:
For aniline, using Kb = 4.3 × 10^-10 and C ≈ 3.80 M:
Since Kb is very small, x is much smaller than 3.80, so many textbooks make the approximation:
That means:
- [OH-] ≈ 4.04 × 10^-5 M
- pOH ≈ 4.39
- pH ≈ 14.00 – 4.39 = 9.61
The calculator above uses the quadratic form, which is more rigorous than relying only on the small-x shortcut. At this Kb and concentration, both methods give essentially the same answer.
Estimated Final Answer
Under the stated assumptions and using a typical room-temperature Kb for aniline, the pH of a 5.7 m solution of aniline is approximately 9.6. A more precise answer depends on the exact density, temperature, and activity effects at such a high concentration, but for standard educational chemistry calculations, a value around 9.60 to 9.61 is a strong estimate.
Why the pH Is Not Extremely High
Many learners expect a highly concentrated base to have a pH near 13 or 14. That would be true for a strong base, but not for aniline. Aniline is weak because the lone pair on the nitrogen is partially delocalized into the aromatic ring. That delocalization makes the nitrogen less eager to bind a proton, which lowers basicity. As a result, even several molar aniline solution generates only a relatively small hydroxide concentration compared with strong bases.
| Property | Aniline | Ammonia | Sodium hydroxide |
|---|---|---|---|
| Base type | Weak aromatic amine | Weak base | Strong base |
| Typical Kb at 25 degrees C | 4.3 × 10^-10 | 1.8 × 10^-5 | Effectively complete dissociation |
| Relative basic strength vs aniline | 1× | About 42,000× larger Kb | Much stronger than weak-base equilibrium scale |
| Expected pH behavior at high concentration | Moderately basic | More strongly basic | Very strongly basic |
Exact vs Approximate Treatment
In introductory chemistry, weak-base problems are often solved using the square-root approximation. That is usually acceptable when the percent ionization is small. For a base like aniline, it is extremely small indeed. The degree of ionization for a several-molar solution is far below 1 percent. However, at a concentration as large as 5.7 m, real solutions can deviate from ideal behavior. That means:
- The density assumption matters because it changes molarity.
- Activities may differ from concentrations at high ionic strength.
- The Kb value itself can vary slightly with temperature and source.
- Published data may be based on dilute-solution approximations.
For classroom and general chemistry use, the standard equilibrium method is still the expected approach. For advanced physical chemistry or industrial process design, you would move beyond concentration-only expressions and incorporate activity coefficients.
Worked Example Summary
- Start with 5.7 mol aniline in 1.000 kg water.
- Compute aniline mass: 5.7 × 93.13 = 530.84 g.
- Total mass of solution: 1530.84 g.
- Assume density 1.02 g/mL, so volume ≈ 1.50082 L.
- Convert to molarity: 5.7 ÷ 1.50082 ≈ 3.80 M.
- Use Kb = 4.3 × 10^-10 and solve for [OH-].
- Find pOH, then pH.
- Result: pH ≈ 9.61.
Reference Data Table for This Calculator
| Input or output | Value used | Units | Why it matters |
|---|---|---|---|
| Molality | 5.7 | mol/kg solvent | Defines how much aniline is dissolved per kilogram of water |
| Molar mass of aniline | 93.13 | g/mol | Needed to convert moles of aniline into mass |
| Assumed density | 1.02 | g/mL | Allows conversion from total mass to total solution volume |
| Estimated molarity | ≈ 3.80 | mol/L | Used in the weak-base equilibrium expression |
| Kb of aniline | 4.3 × 10^-10 | dimensionless | Determines the extent of proton acceptance from water |
| [OH-] | ≈ 4.04 × 10^-5 | mol/L | Directly determines pOH |
| pOH | ≈ 4.39 | pOH units | Intermediate acid-base quantity |
| pH | ≈ 9.61 | pH units | Final result |
Common Mistakes When Solving This Problem
- Treating 5.7 m as 5.7 M without checking. Molality and molarity are not the same.
- Using Ka instead of Kb. Aniline is a base, so Kb is the natural starting constant.
- Assuming complete dissociation. Aniline is weak, so the hydroxide concentration is much smaller than the formal concentration.
- Forgetting the density step. Without density, converting molality to molarity is only an approximation.
- Ignoring temperature. pKw changes with temperature, so pH from pOH should be adjusted if not at 25 degrees C.
How Reliable Is the Calculator?
This calculator is reliable for educational and planning use. It applies accepted weak-base equilibrium chemistry, performs the molality-to-molarity conversion using the entered density, and then solves the hydroxide concentration with the quadratic expression. The main source of uncertainty is not the algebra. It is the real physical behavior of a concentrated aniline solution. If you need publication-grade values, you should use experimentally measured density data, confirm the literature Kb for your exact conditions, and consider nonideal solution effects.
Authoritative Chemistry References
For users who want to verify equilibrium concepts, acid-base definitions, and molecular data, these authoritative sources are helpful:
- NIST Chemistry WebBook
- Chemistry LibreTexts
- PubChem entry for aniline from the U.S. National Library of Medicine
Practical Takeaway
If your assignment or lab asks you to calculate the pH of a 5.7 m solution of aniline, the core idea is simple: convert molality into an approximate molarity, apply the weak-base equilibrium formula, and then convert hydroxide concentration into pOH and pH. Using standard values, you should expect a final pH near 9.6, not a strongly caustic value. That moderate basicity reflects the weak-base nature of aniline and the resonance stabilization of its unprotonated form.
The calculator above makes the process faster and clearer. It is especially useful if you want to test different density assumptions, compare temperatures, or see how small changes in Kb influence the final pH. For students, this helps connect symbolic equilibrium equations with the physical meaning of concentration units. For practitioners, it provides a convenient first-pass estimate before more advanced thermodynamic treatment.