Chegg: Calculate the pH in a 1.00×10-2 Morpholine Solution
Use this interactive weak-base calculator to solve the pH of a 0.0100 M morpholine solution, inspect the equilibrium chemistry, and visualize how much hydroxide forms at equilibrium. The default settings are built for the classic textbook and homework-style prompt, but you can also change concentration, method, units, and displayed precision.
Morpholine pH Calculator
Enter the morpholine concentration and the conjugate-acid pKa value. By default, the calculator uses pKa = 8.36 for morpholinium at 25 degrees C and solves the weak-base equilibrium exactly.
Reaction used
Morpholine acts as a weak Brønsted base: B + H2O ⇌ BH+ + OH–.
Key assumption
The default model assumes dilute aqueous solution at 25 degrees C, where pH + pOH = 14.00.
Why exact mode matters
The quadratic solution is more rigorous, especially when concentration becomes small enough that the common square-root shortcut is less reliable.
How to calculate the pH of a 1.00×10-2 morpholine solution
If you are working through a chemistry problem that asks you to calculate the pH in a 1.00×10-2 morpholine solution, you are dealing with a classic weak-base equilibrium. Morpholine is an amine-containing heterocycle, so it accepts a proton from water instead of dissociating completely the way a strong base would. That means the pH is not found by simply setting the hydroxide concentration equal to 0.0100 M. Instead, you must use an equilibrium constant, usually the base dissociation constant Kb or the conjugate-acid pKa.
In many textbook, Chegg, homework, and exam settings, morpholine is treated with a conjugate-acid pKa near 8.36 at 25 degrees C. Because the conjugate acid and the base are related through KaKb = Kw, you can convert pKa into Kb and then solve for the equilibrium hydroxide concentration. For the default values used in this calculator, the answer is a pH of about 10.18. Depending on the exact pKa source and rounding rules, you may also see answers in the narrow range of about 10.16 to 10.18.
Step-by-step solution
1. Write the base equilibrium
Let morpholine be represented as B. In water, the equilibrium is:
B + H2O ⇌ BH+ + OH–
The corresponding equilibrium expression is:
Kb = ([BH+][OH–]) / [B]
2. Convert pKa to Kb
Many references list the acidity of the conjugate acid rather than Kb directly. If pKa of morpholinium is 8.36, then:
- Ka = 10-8.36
- Kw = 1.0×10-14 at 25 degrees C
- Kb = Kw / Ka = 10-14 / 10-8.36 = 10-5.64
- So Kb ≈ 2.29×10-6
3. Set up the ICE table
Start with 0.0100 M morpholine and no products:
- Initial: [B] = 0.0100, [BH+] = 0, [OH–] = 0
- Change: [B] decreases by x, [BH+] increases by x, [OH–] increases by x
- Equilibrium: [B] = 0.0100 – x, [BH+] = x, [OH–] = x
Substitute into the Kb expression:
2.29×10-6 = x2 / (0.0100 – x)
4. Solve for x
Because this is a weak base and x will be much smaller than 0.0100, the approximation method gives:
x ≈ √(KbC) = √((2.29×10-6)(1.00×10-2)) = √(2.29×10-8) ≈ 1.51×10-4
So [OH–] ≈ 1.51×10-4 M. If you solve the quadratic exactly, you obtain essentially the same result: about 1.50×10-4 M.
5. Convert hydroxide concentration into pH
Now calculate pOH:
pOH = -log(1.50×10-4) ≈ 3.82
Then use pH + pOH = 14.00:
pH = 14.00 – 3.82 = 10.18
Why the answer is not 12 or higher
Students often overestimate the pH because morpholine is a base. The key point is that morpholine is not a strong base like sodium hydroxide. It only partially reacts with water, so the hydroxide concentration generated at equilibrium is much smaller than the analytical concentration of the morpholine itself. A 0.0100 M strong base would have [OH–] = 0.0100 M and pH = 12.00, but 0.0100 M morpholine gives [OH–] of only around 1.5×10-4 M and a pH near 10.18.
Exact versus approximate calculation
For this problem, the shortcut x ≈ √(KbC) works very well because x is only about 1.5 percent of the starting concentration. That means replacing 0.0100 – x with 0.0100 introduces very little error. Still, when you want a rigorous answer, especially at lower concentrations, the quadratic formula is better. This calculator lets you switch between both methods so you can see how close they are under different conditions.
| Morpholine concentration | Method | Estimated [OH-] (M) | Estimated pH | Approximation error in pH |
|---|---|---|---|---|
| 1.00×10^-1 M | Exact quadratic | 4.77×10^-4 | 10.679 | Baseline |
| 1.00×10^-1 M | Square-root approximation | 4.79×10^-4 | 10.680 | About 0.001 |
| 1.00×10^-2 M | Exact quadratic | 1.50×10^-4 | 10.177 | Baseline |
| 1.00×10^-2 M | Square-root approximation | 1.51×10^-4 | 10.180 | About 0.003 |
| 1.00×10^-4 M | Exact quadratic | 1.40×10^-5 | 9.146 | Baseline |
| 1.00×10^-4 M | Square-root approximation | 1.51×10^-5 | 9.180 | About 0.034 |
Comparison with other weak bases
One useful way to understand morpholine is to compare it with other familiar weak bases. The stronger the base, the more hydroxide it produces at the same formal concentration. Morpholine is noticeably more basic than pyridine and aniline, but less basic than ammonia. That makes a 0.0100 M morpholine solution moderately basic, with a pH above 10 but not as high as a similarly concentrated ammonia solution.
| Base | Conjugate-acid pKa at about 25 degrees C | Approximate Kb | pH at 0.0100 M | Relative basicity versus morpholine |
|---|---|---|---|---|
| Morpholine | 8.36 | 2.29×10^-6 | 10.18 | Reference |
| Ammonia | 9.25 | 1.78×10^-5 | 10.63 | Stronger base |
| Pyridine | 5.23 | 1.70×10^-9 | 8.61 | Much weaker base |
| Aniline | 4.60 | 3.98×10^-10 | 8.30 | Much weaker base |
Common mistakes in this type of problem
- Treating morpholine as a strong base. This produces a pH that is much too high.
- Using pKa directly as Kb. You must first convert through Kw.
- Forgetting to calculate pOH first. Since morpholine produces OH–, the direct logarithm gives pOH, not pH.
- Dropping the negative sign in logarithms. pOH = -log[OH–].
- Ignoring temperature assumptions. The relation pH + pOH = 14.00 is specific to 25 degrees C in the standard classroom treatment.
- Rounding too early. Carry extra digits through Kb, x, and pOH before reporting the final pH.
How this calculator helps with homework and exam prep
The calculator above is built around the exact chemistry of the problem statement “calculate the pH in a 1.00×10-2 morpholine solution.” It does more than return a final number. It also shows the converted Kb, equilibrium hydroxide concentration, pOH, and the percent of morpholine that becomes protonated. That last number is especially useful because it helps you judge whether the approximation method is valid. If only a small fraction reacts, the square-root shortcut is justified.
The chart is another practical feature. In weak-base problems, the quantities can look abstract because the equilibrium shift is small relative to the starting concentration. A bar chart of initial concentration, equilibrium hydroxide, equilibrium conjugate acid, and remaining base makes the chemistry tangible. You can instantly see that only a small amount of morpholine reacts, even though the resulting pH shift is still significant.
Interpretation of the final answer
A pH of about 10.18 tells you the solution is basic but not aggressively alkaline. In practical terms, morpholine is widely used in industrial and chemical applications because it is a basic amine, yet its behavior in water still follows the equilibrium logic of a weak base. From an educational standpoint, it is a near-ideal example for learning how to move between pKa, Kb, equilibrium expressions, approximations, and logarithmic pH calculations.
If your instructor or textbook uses a slightly different pKa value, your final pH may shift by a few hundredths. That is normal. Chemistry references can differ a little depending on ionic strength, source tables, and data conventions. The important part is showing the correct setup and solving the equilibrium consistently.
Authoritative references for morpholine and acid-base chemistry
For deeper validation and background reading, consult: PubChem morpholine record, NIST Chemistry WebBook entry for morpholine, and MIT OpenCourseWare acid-base equilibria materials.
Final takeaway
To solve the pH of a 1.00×10-2 morpholine solution, treat morpholine as a weak base, convert the conjugate-acid pKa to Kb, solve the equilibrium for [OH–], then convert through pOH to pH. With pKa = 8.36, the best standard answer is pH ≈ 10.18. If you need to verify a homework result, compare your setup to the steps above and use the calculator to check both the exact and approximate methods.