Calculate pH of (CH3)2NH2+ Given 1.86 × 10-2 M
This premium calculator solves the pH of dimethylammonium, the conjugate acid of dimethylamine, using equilibrium chemistry. Enter the concentration, choose units, confirm the pKb of dimethylamine, and get an exact weak-acid pH result with a chart and step-by-step values.
Weak Acid pH Calculator
Click Calculate pH to compute the acidity of dimethylammonium at the specified concentration.
Visual Equilibrium Snapshot
The chart compares the initial dimethylammonium concentration, the equilibrium hydronium concentration, and the remaining acid after dissociation. Because weak acids dissociate only slightly, the graph uses a logarithmic concentration scale.
Species
(CH3)2NH2+ is the conjugate acid of dimethylamine.
Type
It behaves as a weak acid in water and partially donates H+.
Core relation
Ka = Kw / Kb, then solve for [H3O+].
How to Calculate pH of (CH3)2NH2+ Given 1.86 × 10-2 M
When a chemistry problem asks you to calculate the pH of (CH3)2NH2+ at 1.86 × 10-2 M, the key idea is that you are not dealing with a strong acid. Instead, dimethylammonium is the conjugate acid of dimethylamine, which is a weak base. That means its proton donation in water is partial, not complete. To find pH correctly, you must use an equilibrium approach rather than simply assuming the hydronium concentration equals the initial solute concentration.
This distinction matters. If a student mistakenly treated dimethylammonium as a strong acid, the predicted pH would be far too low. In reality, the species dissociates only slightly because its acid dissociation constant is small. The proper calculation starts from the base constant of dimethylamine, converts that value into the conjugate acid constant, and then solves for the hydronium ion concentration generated at equilibrium. Once [H3O+] is known, pH follows from the familiar negative logarithm relation.
Step 1: Identify the acid-base pair
Dimethylamine is written as (CH3)2NH. Its conjugate acid is dimethylammonium, written as (CH3)2NH2+. In water, the conjugate acid establishes the equilibrium:
(CH3)2NH2+ + H2O ⇌ (CH3)2NH + H3O+Because the acid is weak, the equilibrium lies strongly to the left. Only a small fraction of the initial dimethylammonium concentration produces hydronium ions. That is why the pH is mildly acidic rather than strongly acidic.
Step 2: Convert pKb to Ka
Most references list dimethylamine as a weak base and report its pKb, not the Ka of the conjugate acid. At 25°C, a typical value is pKb = 3.27. From this:
Kb = 10^-pKb = 10^-3.27 ≈ 5.37 × 10^-4Then use the water ion-product relationship:
Ka = Kw / Kb = (1.0 × 10^-14) / (5.37 × 10^-4) ≈ 1.86 × 10^-11This very small Ka tells you immediately that dimethylammonium is a weak acid. The acid dissociation is limited, so the hydronium concentration will be much smaller than 1.86 × 10-2 M.
Step 3: Set up the ICE table
Let the initial concentration be C = 1.86 × 10-2 M. Suppose x mol/L dissociates:
- Initial: [(CH3)2NH2+] = 1.86 × 10-2, [(CH3)2NH] = 0, [H3O+] = 0
- Change: acid decreases by x, base increases by x, hydronium increases by x
- Equilibrium: [acid] = 1.86 × 10-2 – x, [base] = x, [H3O+] = x
The equilibrium expression becomes:
Ka = x^2 / (C – x)Substituting the values gives:
1.86 × 10^-11 = x^2 / (1.86 × 10^-2 – x)Step 4: Solve for x
Since Ka is tiny, many instructors allow the approximation C – x ≈ C. Then:
x ≈ √(Ka × C) = √((1.86 × 10^-11)(1.86 × 10^-2)) ≈ 5.88 × 10^-7 MThat means:
[H3O+] ≈ 5.88 × 10^-7 MFinally:
pH = -log(5.88 × 10^-7) ≈ 6.23If you use the exact quadratic equation, the value is essentially the same for this concentration because x is extremely small relative to the starting concentration. In other words, the approximation is very reliable here.
Why the pH is only slightly acidic
Students often expect any positively charged ammonium-type ion to produce a strongly acidic solution, but that is not generally true. The strength of the acid depends on how stable the protonated amine is and how strongly the neutral amine acts as a base. Dimethylamine is a fairly strong weak base compared with ammonia, so its conjugate acid is correspondingly weak. As a result, dimethylammonium only modestly lowers the pH below neutral.
At 1.86 × 10-2 M, the initial concentration is thousands of times larger than the hydronium concentration actually produced. This is classic weak-acid behavior. The equilibrium concentration of the undissociated acid remains close to the initial value, while the amount converted to neutral dimethylamine is very small.
Worked answer for the given example
- Given: C = 1.86 × 10-2 M
- Use pKb((CH3)2NH) = 3.27
- Calculate Kb = 10-3.27 ≈ 5.37 × 10-4
- Calculate Ka = 1.0 × 10-14 / 5.37 × 10-4 ≈ 1.86 × 10-11
- Solve x ≈ √(KaC) ≈ 5.88 × 10-7 M
- Calculate pH = -log(x) ≈ 6.23
| Parameter | Value | Meaning |
|---|---|---|
| Initial concentration, C | 1.86 × 10^-2 M | Starting concentration of dimethylammonium |
| pKb of dimethylamine | 3.27 | Base strength of the conjugate base |
| Kb | 5.37 × 10^-4 | Base dissociation constant |
| Ka of dimethylammonium | 1.86 × 10^-11 | Acid dissociation constant |
| [H3O+] | 5.88 × 10^-7 M | Hydronium formed at equilibrium |
| pH | 6.23 | Final answer |
Comparison with related weak-acid systems
One helpful way to understand the answer is to compare dimethylammonium with other familiar conjugate acids. A lower pKa means a stronger acid and a lower pH at the same concentration. Dimethylammonium has a high pKa, so it is weak. This keeps the pH close to neutral even when the formal concentration is on the order of 10-2 M.
| Conjugate acid | Typical pKa | Approximate acid strength trend | Expected pH behavior at similar concentration |
|---|---|---|---|
| NH4+ | 9.25 | Stronger than dimethylammonium | Somewhat more acidic |
| (CH3)2NH2+ | 10.73 | Weak acid | Mildly acidic, near neutral |
| CH3COOH | 4.76 | Much stronger weak acid | Significantly lower pH |
Common mistakes to avoid
- Treating the acid as strong. Do not set [H3O+] equal to 1.86 × 10-2 M.
- Using Kb directly in the acid expression. You must convert Kb to Ka for the conjugate acid.
- Ignoring units. The given concentration 1.86 × 10-2 M equals 0.0186 M, not 1.86 M.
- Mixing pK values. pKa + pKb = 14 only at the standard temperature assumption of 25°C.
- Rounding too early. Keep extra digits until the final pH step.
When the exact quadratic method is better
For this problem, the square-root approximation works extremely well because x is tiny compared with 0.0186 M. However, not every weak-acid problem behaves so conveniently. If the acid is stronger, or if the concentration is much lower, the approximation can become less accurate. In those cases, solving the quadratic equation is the safest route:
x = (-Ka + √(Ka^2 + 4KaC)) / 2The calculator above can use either method. The exact mode is especially useful for checking homework steps, preparing lab reports, or validating class examples.
What the chemistry means in practical terms
A pH of about 6.23 indicates a mildly acidic solution. This is far from the acidity of strong mineral acids and much closer to neutral water. In real laboratory practice, amine salts and ammonium-like species often produce solutions in this region because their protonated forms are weak acids. Understanding this helps in buffer preparation, salt hydrolysis questions, and organic chemistry problems involving amines and amine salts.
It also explains why acid-base equilibria are best understood as connected systems. Dimethylamine as a base and dimethylammonium as an acid are two sides of the same equilibrium pair. Knowing one equilibrium constant allows you to determine the other, which is a recurring theme in general chemistry, analytical chemistry, and biochemistry.
Authoritative references for pH and acid-base principles
For broader background on pH, water chemistry, and equilibrium concepts, these authoritative sources are useful:
Final answer summary
To calculate the pH of (CH3)2NH2+ given 1.86 × 10-2 M, first recognize it as a weak acid. Use the pKb of dimethylamine to find Ka for dimethylammonium. Then apply the weak-acid equilibrium expression and solve for the hydronium concentration. For the standard 25°C constants, the solution gives a hydronium concentration of about 5.88 × 10-7 M, which corresponds to a pH of approximately 6.23. That is the correct chemistry-based result for this problem.